# categories

id: Set
name: category of sets
notation: $\mathbf{Set}$
objects: sets
morphisms: maps
nlab_link: https://ncatlab.org/nlab/show/Set
dual_category_id: Set_op
description:
The category of sets plays a fundamental role in category theory. Due to the Yoneda embedding, many results about general categories can be reduced to the category of sets. It is also usually the first example of a category that one encounters.


id: SetxSet
name: category of pairs of sets
notation: $\mathbf{Set} \times \mathbf{Set}$
objects: pairs $(A,B)$ of sets $A$ and $B$
morphisms: A morphism $(A,B) \to (C,D)$ consists of a map $A \to C$ and a map $B \to D$.
description:
This is an example of the <a href="https://ncatlab.org/nlab/show/product+category" target="_blank">product of categories</a>. It inherits most (but not all) properties from $\mathbf{Set}$. It can also be seen as the category $\mathbf{Sh}(1+1)$ of sheaves on a discrete space with two points, and also as the slice category $\mathbf{Set}/(1+1)$.


id: Set_op
name: dual of the category of sets
notation: $\mathbf{Set}^{\mathrm{op}}$
objects: sets
morphisms: A morphism $f : X \to Y$ is a map of sets $Y \to X$.
nlab_link: https://ncatlab.org/nlab/show/opposite+category
dual_category_id: Set
description:
By definition, this category is the dual (or opposite) of <a href="/category/Set">the category of sets</a>.


id: Set*
name: category of pointed sets
notation: $\mathbf{Set}_*$
objects: pointed sets
morphisms: pointed maps
nlab_link: https://ncatlab.org/nlab/show/pointed+set
description:
This is the category of sets with a distinguished element, often called the base point. A map is called pointed when it preserves the base point.


id: Setne
name: category of non-empty sets
notation: $\mathbf{Set}_{\neq \varnothing}$
objects: non-empty sets
morphisms: maps
nlab_link: https://ncatlab.org/nlab/show/inhabited+set
description:
This entry demonstrates that removing an object (the empty set) can drastically change the properties of a category. In particular, this category is neither complete nor cocomplete.


id: Rel
name: category of sets and relations
notation: $\mathbf{Rel}$
objects: sets
morphisms: A morphism from $A$ to $B$ is a relation, i.e. a subset of $A \times B$.
nlab_link: https://ncatlab.org/nlab/show/Rel
description:
This category is self-dual as it can be: There is an isomorphism $\mathbf{Rel} \cong \mathbf{Rel}^{\mathrm{op}}$ that is the identity on objects and maps a relation to its opposite relation. It is the prototype of a dagger-category.


id: Set_c
name: category of countable sets
notation: $\mathbf{Set}_\mathrm{c}$
objects: countable sets
morphisms: maps
description:
A set is countable if it admits a surjection from $\mathbb{N}$. In particular, every finite set is countable.


id: Set_f
name: category of sets with finite-to-one maps
notation: $\mathbf{Set}_\mathrm{f}$
objects: sets
morphisms:
maps $f : X \to Y$ with the property that for every $y \in Y$ the fiber $f^*(\{y\})$ is a finite set

description:
In this variant of $\mathbf{Set}$ we only consider maps with finite fibers, which are commonly called <i>finite-to-one</i>. Equivalently, every preimage of a finite set is again finite, and this description makes it obvious that composition is well-defined.


id: Vect
name: category of vector spaces
notation: $\mathbf{Vect}_K$
objects: vector spaces over a field $K$
morphisms: linear maps
nlab_link: https://ncatlab.org/nlab/show/Vect
description:
This is a special case of the category of modules over a ring, where the ring is a field. It is the prototype of a split abelian category.


id: Alg(R)
name: category of algebras
notation: $\mathbf{Alg}(R)$
objects: algebras over a commutative ring $R \neq 0$
morphisms: maps preserving the ring and module structure
nlab_link: https://ncatlab.org/nlab/show/Alg
description:
This is a generalization of the category of rings, which we get for $R = \mathbb{Z}$. We assume our rings (and algebras) to be unital. For $R = 0$ we would get the trivial category, which is why we exclude this here.


id: CAlg(R)
name: category of commutative algebras
notation: $\mathbf{CAlg}(R)$
objects: commutative algebras over a commutative ring $R \neq 0$
morphisms: maps preserving the ring and module structure
nlab_link: https://ncatlab.org/nlab/show/CommAlg
description:
This is a generalization of the category of commutative rings, which we get for $R = \mathbb{Z}$. In general, $\mathbf{CAlg}(R) \cong R \,/\, \mathbf{CRing}$. We assume our rings (and algebras) to be unital. For $R = 0$ we would get the trivial category, which is why we exclude this here.


id: M-Set
name: category of M-sets
notation: $M{-}\mathbf{Set}$
objects: sets with a left action of a monoid $M$
nlab_link: https://ncatlab.org/nlab/show/MSet
morphisms:
maps that are compatible with the $M$-action, meaning $f(m \cdot x)=m \cdot f(x)$, also called $M$-maps

description:
Here, $M$ can be any monoid. But the most important special case is that of a group. To settle (future) non-properties, we assume that $M$ is non-trivial, since otherwise we just get the <a href="/category/Set">category of sets</a>.


id: R-Mod
name: category of left modules over a ring
notation: $R{-}\mathbf{Mod}$
objects: left $R$-modules
morphisms: $R$-linear maps
nlab_link: https://ncatlab.org/nlab/show/module
description:
This is the prototype of an abelian category. The category of right modules is the same with the opposite ring $R^{\mathrm{op}}$, hence not listed here.<br>
	To settle the unsatisfied properties, we make the assumption that $R$ is <i>not</i> semisimple: If $R$ is semisimple, then by the Artin-Wedderburn theorem, the category is equivalent to a finite direct product of categories $D{-}\mathbf{Mod}$ for division rings $D$, and the case of division rings is in a separate entry. In particular, $R \neq 0$ and $R$ is not a field.


id: R-Mod_div
name: category of left modules over a division ring
notation: $R{-}\mathbf{Mod}$
objects: left $R$-modules
morphisms: $R$-linear maps
nlab_link: https://ncatlab.org/nlab/show/module
description:
Here, we assume that $R$ is a non-commutative division ring, i.e. a skew-field which is not a field. The category of modules behaves mostly the same as in the commutative case.


id: Cat
name: category of small categories
notation: $\mathbf{Cat}$
objects: small categories
morphisms: functors
nlab_link: https://ncatlab.org/nlab/show/Cat
description:
This is the category of small categories and functors between them. It is the prototype of a 2-category, but here we only treat it as a 1-category.


id: Ban
name: category of Banach spaces with linear contractions
notation: $\mathbf{Ban}$
objects: Banach spaces over $\mathbb{C}$
morphisms: linear contractions, i.e. linear maps of norm $\leq 1$
nlab_link: https://ncatlab.org/nlab/show/Banach+space
description:
The choice of morphisms is similar to that of $\mathbf{Met}$ which yields better categorical properties.


id: Meas
name: category of measurable spaces
notation: $\mathbf{Meas}$
objects: measurable spaces
morphisms: measurable maps
nlab_link: https://ncatlab.org/nlab/show/Meas
description:
This is very similar to the category of topological spaces. Accordingly, limits and colimits can be constructed in the same way.


id: Met
name: category of metric spaces with non-expansive maps
notation: $\mathbf{Met}$
objects: metric spaces
morphisms: non-expansive maps $f$, meaning $d(f(x),f(y)) \leq d(x,y)$ for all $x,y$
nlab_link: https://ncatlab.org/nlab/show/Met
description:
In contrast to continuous maps, which only refer to the induced topology, non-expansive maps are closer related to the metrics themselves. This category is badly-behaved, though, especially when compared with $\mathbf{Met}_{\infty}$.


id: Met_oo
name: category of metric spaces with ∞ allowed
notation: $\mathbf{Met}_{\infty}$
objects: metric spaces, where the metric is allowed to assume the value $\infty$
morphisms: non-expansive maps $f$, meaning $d(f(x),f(y)) \leq d(x,y)$ for all $x,y$
nlab_link: https://ncatlab.org/nlab/show/Met
description:
The fact that we allow $\infty$ means that universal constructions work much better when compared to $\mathbf{Met}$.


id: Met_c
name: category of metric spaces with continuous maps
notation: $\mathbf{Met}_c$
objects: metric spaces
morphisms: continuous maps
nlab_link: https://ncatlab.org/nlab/show/metrisable+topological+space
description:
This category is equivalent to the subcategory of $\mathbf{Top}$ (or $\mathbf{Haus}$) that consists of metrizable topological spaces. Hence, the metrics only play a secondary role here.


id: Top*
name: category of pointed topological spaces
notation: $\mathbf{Top}_*$
objects: pointed topological spaces
morphisms: pointed continuous functions
nlab_link: https://ncatlab.org/nlab/show/pointed+topological+space
description:
This category plays an important role in algebraic topology and homotopy theory. Although it may appear similar to $\mathbf{Top}$, adding a base point drastically changes its categorical properties. In particular, it introduces a zero object.


id: Haus
name: category of Hausdorff spaces
notation: $\mathbf{Haus}$
objects: Hausdorff spaces
morphisms: continuous functions
nlab_link: https://ncatlab.org/nlab/show/Hausdorff+space
description:
This is the full subcategory of $\mathbf{Top}$ consisting of those spaces that are <a href="https://en.wikipedia.org/wiki/Hausdorff_space" target="_blank">Hausdorff</a>.


id: sSet
name: category of simplicial sets
notation: $\mathbf{sSet}$
morphisms: natural transformations
nlab_link: https://ncatlab.org/nlab/show/SimpSet
objects:
simplicial sets, i.e. functors $\Delta^{\mathrm{op}} \to \mathbf{Set}$ where $\Delta$ is the <a href="/category/Delta">simplex category</a>


id: Man
name: category of smooth manifolds
notation: $\mathbf{Man}$
objects: smooth manifolds
morphisms: smooth maps
nlab_link: https://ncatlab.org/nlab/show/Diff
description:
Here, a smooth manifold is defined as a second-countable Hausdorff space with a smooth atlas. The dimension is locally constant, not necessarily constant.


id: Delta
name: simplex category
notation: $\Delta$
objects: the non-empty ordered sets $[n] := \{0 < \cdots < n\}$ for $n \in \mathbb{N}$
morphisms: order-preserving maps
nlab_link: https://ncatlab.org/nlab/show/simplex+category
description:
The simplex category is a skeleton of $\mathbf{FinOrd} \setminus \{\varnothing\}$. It plays an important role in topology and is used to define the <a href="/category/sSet">category of simplicial sets</a>.


id: Sh(X)
name: category of sheaves
notation: $\mathrm{Sh}(X)$
objects: sheaves of sets on a topological space $X$
morphisms: morphisms of sheaves
nlab_link: https://ncatlab.org/nlab/show/category+of+sheaves
description:
Here, we assume that the topological space $X$ is neither discrete nor indiscrete, since otherwise this category is just a product of copies of $\mathbf{Set}$. Another valid notation is $\mathrm{Sh}(X,\mathbf{Set})$.


id: Sh(X,Ab)
name: category of abelian sheaves
notation: $\mathrm{Sh}(X,\mathbf{Ab})$
objects: sheaves of abelian groups on a topological space $X$
morphisms: morphisms of sheaves
nlab_link: https://ncatlab.org/nlab/show/sheaf+of+abelian+groups
description:
Here, we assume that the topological space $X$ is neither discrete nor indiscrete, since otherwise this category is just a product of copies of $\mathbf{Ab}$.


id: LRS
name: category of locally ringed spaces
notation: $\mathbf{LRS}$
objects: locally ringed spaces
nlab_link: https://ncatlab.org/nlab/show/locally+ringed+topological+space
morphisms:
morphisms of locally ringed spaces, thus consisting of a continuous map and a homomorphism of sheaves that induces local ring homomorphisms in the stalks


id: Z
name: category of Z-functors
notation: $[\mathbf{CRing}, \mathbf{Set}]$
objects: Z-functors, i.e. functors from commutative rings to sets
morphisms: natural transformations
description:
This category is used in functorial algebraic geometry. It also provides a typical example of a functor category that is not locally small, but nevertheless relevant. Most of its properties are directly derived from the category of sets, so other functor categories $[\mathbf{C}, \mathbf{Set}]$ for large categories $\mathbf{C}$ will be similar.


id: Pos
name: category of posets
notation: $\mathbf{Pos}$
morphisms: order-preserving functions
nlab_link: https://ncatlab.org/nlab/show/Pos
objects:
partially ordered sets (aka posets), i.e. sets equipped with a reflexive, transitive, antisymmetric relation

description:
Even though there are many similarities with $\mathbf{Prost}$, the main difference is that the forgetful functor $\mathbf{Pos} \to \mathbf{Set}$ has no right adjoint.


id: Prost
name: category of prosets
notation: $\mathbf{Prost}$
objects: preordered sets (aka prosets), i.e. sets equipped with a reflexive, transitive relation
morphisms: order-preserving functions
nlab_link: https://ncatlab.org/nlab/show/Prost
description:
Even though there are many similarities with $\mathbf{Pos}$, the main difference is that the forgetful functor $\mathbf{Prost} \to \mathbf{Set}$ has a right adjoint, mapping $X$ to $(X , X \times X)$ (chaotic preorder).


id: FinOrd
name: category of finite ordered sets
notation: $\mathbf{FinOrd}$
objects: finite (totally) ordered sets
morphisms: order-preserving maps
nlab_link: https://ncatlab.org/nlab/show/total+order
description:
The finite ordered sets of the form $\{1 < \dotsc < n\}$ for $n \in \mathbb{N}$ provide a skeleton (including the empty set for $n = 0$), the augmented simplex category.


id: B
name: category of finite sets and bijections
notation: $\mathbb{B}$
objects: finite sets
morphisms: bijective maps
nlab_link: https://ncatlab.org/nlab/show/permutation+groupoid
description:
This category is also known as the permutation groupoid. It appears in the definition of a combinatorial species.


id: FS
name: category of finite sets and surjections
notation: $\mathbf{FS}$
objects: finite sets
morphisms: surjective maps
description:
This category is badly-behaved in itself, but it appears in representation theory. It has two connected components, consisting of the empty set and the non-empty finite sets.


id: Sp
name: category of combinatorial species
notation: $\mathbf{Sp}$
morphisms: natural transformations
nlab_link: https://ncatlab.org/nlab/show/species
objects:
combinatorial species, defined as functors $\mathbb{B} \to \mathbf{FinSet}$, where $\mathbb{B}$ is the category of finite sets and bijections

description:
Most categorical properties are immediately inferred from $\mathbf{FinSet}$. Notice that this category is not locally small; it is just equivalent to a locally small category.


id: 0
name: empty category
notation: $\mathbf{0}$
objects: no objects
morphisms: no morphisms
nlab_link: https://ncatlab.org/nlab/show/empty+category
description:
This is the category with no objects and no morphisms. It is the initial object in the category of small categories.


id: 1
name: trivial category
notation: $\mathbf{1}$
objects: a single object $0$
morphisms: only the identity morphism
nlab_link: https://ncatlab.org/nlab/show/terminal+category
description:
This is the simplest category, consisting of a single object $0$ and its identity morphism $0 \to 0$. A concrete representation is the full subcategory of $\mathbf{Set}$ consisting of the empty set. It is the terminal object in the category of small categories.


id: 2
name: discrete category on two objects
notation: $\mathbf{2}$
objects: two objects $0$ and $1$
morphisms: only the two identity morphisms
description:
A concrete representation is the full subcategory of $\mathbf{CRing}$ consisting of the two fields $\mathbb{F}_2$ and $\mathbb{F}_3$.


id: walking_morphism
name: walking morphism
notation: $\{0 \to 1\}$
objects: $0$, $1$
morphisms: the two identities and a single morphism from $0$ to $1$
nlab_link: https://ncatlab.org/nlab/show/interval+category
description:
This is also known as the interval category. It has the property that functors $\{0 \to 1\} \to \mathcal{C}$ are the same as morphisms in $\mathcal{C}$.


id: walking_pair
name: walking parallel pair
notation: $\{0 \rightrightarrows 1 \}$
objects: two objects $0$ and $1$
morphisms: the two identities and two parallel morphisms from $0$ to $1$
nlab_link: https://ncatlab.org/nlab/show/walking+structure
description:
The name of this category comes from the fact that it consists of two parallel morphisms, and a functor out of this category is the same as a parallel pair of morphisms in the target category.


id: walking_isomorphism
name: walking isomorphism
notation: $\{0 \rightleftarrows 1\}$
objects: two objects $0$ and $1$
morphisms: identities, and two morphisms $0 \to 1$ and $1 \to 0$ that are mutually inverse
nlab_link: https://ncatlab.org/nlab/show/walking+isomorphism
description:
The name of this category comes from the fact that it consists of two objects and an isomorphism between them, and a functor out of this category is the same as an isomorphism in the target category. The walking isomorphism is actually equivalent to the trivial category.


id: walking_composable_pair
name: walking composable pair
notation: $\{ 0 \to 1 \to 2 \}$
objects: three objects $0,1,2$
morphisms: a single morphism from each natural number to one greater than or equal to it
nlab_link: https://ncatlab.org/nlab/show/composable+pair
description:
The name of this category comes from the fact that a functor out of it is the same as a composable pair of morphisms.


id: walking_fork
name: walking fork
notation: $\{0 \to 1 \rightrightarrows 2\}$
objects: three objects $0,1,2$
morphisms:
one morphism $i : 0 \to 1$, two morphisms $f,g : 1 \rightrightarrows 2$, one morphism $0 \to 2$ (namely, $f \circ i = g \circ i$), and the identities

description:
The name of this category comes from the fact that a functor out of it is the same as a <a href="https://ncatlab.org/nlab/show/fork" target="_blank">fork</a>.


id: walking_commutative_square
name: walking commutative square
notation: $\{0 \to 1\}^2$
objects: four objects $a,b,c,d$
morphisms: morphisms $a \to b$, $b \to d$, $a \to c$, $c \to d$, identities, and one morphism $a \to d$
nlab_link: https://ncatlab.org/nlab/show/commutative+square
description:
This category consists of a commutative square:
	$$\begin{CD} a @>>> b \\ @VVV @VVV \\ c @>>> d \end{CD}$$
	Its name comes from the fact that a functor out of it is the same as a <a href="https://ncatlab.org/nlab/show/commutative+square" target="_blank">commutative square</a> in the target category. Notice that the category is isomorphic to the product category $\{0 \to 1\} \times \{0 \to 1\}$ of the <a href="/category/walking_morphism">walking morphism</a> with itself. Hence, most (but not all) properties are inherited from it. It is also isomorphic to the partial order of positive divisors of $6$.


id: walking_span
name: walking span
notation: $\{1 \leftarrow 0 \rightarrow 2\}$
objects: $0,1,2$
morphisms: $0 \to 1$, $0 \to 2$, and the identities
nlab_link: https://ncatlab.org/nlab/show/span#the_walking_span
description:
The name of this category comes from the fact that a functor out of it is the same as a <a href="https://ncatlab.org/nlab/show/span" target="_blank">span</a> in the target category. It is isomorphic to the partial order of proper positive divisors of $6$.


id: walking_idempotent
name: walking idempotent
notation: $\mathrm{Idem}$
objects: a single object $0$
morphisms: two morphisms $\mathrm{id}_0,e : 0 \to 0$ with $e^2=e$
description:
The name of this category comes from the fact that a functor out of it is the same as an <a href="https://ncatlab.org/nlab/show/idempotent" target="_blank">idempotent morphism</a> in the target category. It can also be seen as the delooping of the monoid $\{1,e\}$ in which $e^2=e$.


id: walking_splitting
name: walking splitting
notation: $\mathrm{Split}$
objects: two objects $0$ and $1$
morphisms:
the identities, morphisms $i : 0 \to 1$, $p : 1 \to 0$ satisfying $pi = \mathrm{id}_0$, and the idempotent $ip : 1 \to 1$.

description:
This category could also be called the "walking split idempotent" (or "walking section", "walking retraction"), but we chose a name that emphasizes that the splitting belongs to the data. Notice that the $5$ given morphisms are indeed closed under composition. For example, $p \circ ip = p$ and $ip \circ i = i$.<br>The walking splitting can be interpreted as a skeleton of the category of $\mathbb{F}_2$-vector spaces of dimension $\leq 1$.


id: BG_c
name: delooping of an infinite countable group
notation: $BG$
objects: a single object
morphisms: the elements of an infinite countable group $G$
nlab_link: https://ncatlab.org/nlab/show/delooping
description:
Every group $G$ yields a groupoid $BG$ with a single object $*$, morphisms given by the elements of $G$, and composition given by the group operation. In this example, we consider the case of an infinite countable group $G$ (such as $G = \mathbb{Z}$).


id: BG_f
name: delooping of a non-trivial finite group
notation: $BG$
objects: a single object
morphisms: the elements of a non-trivial finite group $G$
nlab_link: https://ncatlab.org/nlab/show/delooping
description:
Every group $G$ yields a groupoid $BG$ with a single object $*$, morphisms given by the elements of $G$, and composition given by the group operation. In this example, we consider the case of a non-trivial finite group $G$ (such as $G = C_2$).


id: BN
name: delooping of the additive monoid of natural numbers
notation: $B\mathbb{N}$
objects: a single object
morphisms: the natural numbers, with addition serving as composition
description:
Every monoid $M$ induces a category $BM$ with a single object $*$, morphisms given by the elements of $M$, and composition given by the monoid operation. Some of the properties of this category depend on the specific monoid. In this example, we take the commutative monoid $M = (\mathbb{N},+,0)$, so composition is $n \circ m = n + m$.


id: BOn
name: delooping of the additive monoid of ordinal numbers
notation: $B\mathbf{On}$
objects: a single object
morphisms: ordinal numbers, with addition as composition
description:
Every monoid $M$ induces a category $BM$ with a single object $*$. This also works when $M$ is large, in which case $BM$ is not locally small. In this example, we apply this construction to the large monoid of ordinal numbers with respect to addition, so composition is $\alpha \circ \beta = \alpha + \beta$.


id: N
name: poset of natural numbers
notation: $(\mathbb{N},\leq)$
objects: natural numbers $0, 1, 2, \dotsc$
morphisms: a unique morphism $(n,m) : n \to m$ if $n \leq m$
description:
This can also be seen as the path category of the infinite linear graph $\bullet \to \bullet \to \bullet \to \cdots$.


id: real_interval
name: poset [0,1]
notation: $([0,1],\leq)$
objects: real numbers between $0$ and $1$
morphisms: a unique morphism $(s,t) : s \to t$ when $s \leq t$
nlab_link: https://ncatlab.org/nlab/show/interval
description:
Every poset can be regarded as a small thin category. This is a specific example. This category is locally $\aleph_1$-presentable (in fact, <i>every</i> object is $\aleph_1$-presentable), but not locally finitely presentable (in fact, only $0$ is finitely presentable).


id: Z_div
name: proset of integers w.r.t. divisibility
notation: $(\mathbb{Z},\mid)$
objects: integers
morphisms: a unique morphism $(a,b) : a \to b$ if $a$ divides $b$
description:
This is a proset, not a poset, because $a$ and $-a$ divide each other, but are not equal for $a \neq 0$. Notice that this category is equivalent (but not isomorphic) to $(\mathbb{N},\mid)$.


# category_comments

category_id: FinSet
comment:
For the non-existence of sequential (co-)limits it is <i>not</i> sufficient to take a diagram of finite sets whose (co-)limit in $\mathbf{Set}$ is not contained in $\mathbf{FinSet}$.


category_id: Ring
comment:
It is likely that the epimorphisms can be described as in the <a href="/category/CRing">commutative case</a>.


category_id: Rng
comment:
It is likely that the epimorphisms can be described as in the <a href="/category/CRing">commutative, unital case</a>.


category_id: CRing
comment:
Regular monomorphisms are discussed in <a href="https://math.stackexchange.com/questions/695685" target="_blank">MSE/695685</a>, but probably they cannot be classified.


category_id: Fld
comment:
Limits and colimits are discussed in <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.


category_id: Sch
comment:
Monomorphisms are discussed at <a href="https://mathoverflow.net/questions/56591" target="_blank">MO/56591</a>. At least the case of morphisms of locally finite type is understood.


category_id: Sch
comment:
Regular monomorphisms are discussed at <a href="https://mathoverflow.net/questions/66279" target="_blank">MO/66279</a>.


category_id: Sch
comment:
Epimorphisms are discussed at <a href="https://mathoverflow.net/questions/56564" target="_blank">MO/56564</a>. Probably they cannot be classified.


category_id: Sh(X)
comment:
It is likely that neither of the currently remaining unknown properties (finitary algebraic, locally ℵ₁-presentable, exact filtered colimits, etc.) are satisfied for a <i>generic</i> space $X$, but we need to make this precise by adding additional requirements to $X$. Maybe we need to create separate entries for specific spaces $X$.


category_id: Sh(X,Ab)
comment:
It is likely that neither of the currently remaining unknown properties (finitary algebraic, locally ℵ₁-presentable, CSP, etc.) are satisfied for a <i>generic</i> space $X$, but we need to make this precise by adding additional requirements to $X$. Maybe we need to create separate entries for specific spaces $X$.


category_id: M-Set
comment:
If this category is semi-strongly connected depends on the choice of $M$. For $M = 1$ it is, for $M = \mathbb{Z}$ it is not. In general, if $G$ is a group, then $G{-}\mathbf{Set}$ is semi-strongly connected if and only if for all subgroups $H,K \subseteq G$, $H$ is subconjugated to $K$ or $K$ is subconjugated to $H$. If $G$ is abelian, this means that the poset of subgroups is linear, in which case $G$ is either isomorphic to $\mathbb{Z}/p^n$ or to $\mathbb{Z}/p^{\infty}$ for a prime $p$. See also <a href="https://math.stackexchange.com/questions/5129804" target="_blank">MSE/5129804</a>.


category_id: Meas
comment:
The thread <a href="https://math.stackexchange.com/questions/5024471/">MSE/5024471</a> asks for the finitely presentable objects of this category.


# category_properties

id: pullbacks
relation: has
nlab_link: https://ncatlab.org/nlab/show/pullback
dual_property_id: pushouts
description:
A category $\mathcal{C}$ has <i>pullbacks</i> if every cospan of morphisms $X \rightarrow S \leftarrow Y$ has a pullback $X \times_S Y$. This is also known as a fiber product. Equivalently, the slice category $\mathcal{C}/S$ has binary products.


id: pushouts
relation: has
nlab_link: https://ncatlab.org/nlab/show/pushout
dual_property_id: pullbacks
description:
A category $\mathcal{C}$ has <i>pushouts</i> if every span of morphisms $X \leftarrow S \rightarrow Y$ has a pushout $X \sqcup_S Y$. This is also known as a fiber coproduct. Equivalently, the coslice category $S/\mathcal{C}$ has binary coproducts.


id: initial object
relation: has an
nlab_link: https://ncatlab.org/nlab/show/initial+object
dual_property_id: terminal object
description:
An <i>initial object</i> is an object that has a unique morphism to every object in the category. This property refers to the existence of an initial object.


id: terminal object
relation: has a
nlab_link: https://ncatlab.org/nlab/show/terminal+object
dual_property_id: initial object
description:
A <i>terminal object</i> (or <i>final object</i>) is an object that has a unique morphism from every object in the category. This property refers to the existence of a terminal object.


id: products
relation: has
nlab_link: https://ncatlab.org/nlab/show/cartesian+product
dual_property_id: coproducts
description:
Given a family of objects $(A_i)_{i \in I}$, a <i>product</i> $\prod_{i \in I} A_i$ is defined as an object with morphisms $p_i : \prod_{i \in I} A_i \to A_i$ satisfying the following universal property: For every object $T$ and every family of morphisms $(f_i : T \to A_i)_{i \in I}$ there is a unique morphism $f : T \to \prod_{i \in I} A_i$ such that $p_i \circ f = f_i$ for all $i \in I$. This property refers to the existence of <i>small products</i>, i.e., products of small families of objects.


id: coproducts
relation: has
nlab_link: https://ncatlab.org/nlab/show/coproduct
dual_property_id: products
description:
Given a family of objects $(A_i)_{i \in I}$, a <i>coproduct</i> $\coprod_{i \in I} A_i$ is defined as an object with morphisms $i_i : A_i \to \coprod_{i \in I} A_i$ satisfying the following universal property: For every object $T$ and every family of morphisms $(f_i : A_i \to T)_{i \in I}$ there is a unique morphism $f : \coprod_{i \in I} A_i \to T$ such that $f \circ i_i = f_i$ for all $i \in I$. This property refers to the existence of <i>small coproducts</i>, i.e., coproducts of small families of objects.


id: finite products
relation: has
nlab_link: https://ncatlab.org/nlab/show/finite+product
dual_property_id: finite coproducts
description:
A category has <i>finite products</i> if it has products for finite families of objects. Equivalently, it has a terminal object and binary products.


id: finite coproducts
relation: has
nlab_link: https://ncatlab.org/nlab/show/finite+coproduct
dual_property_id: finite products
description:
A category has <i>finite coproducts</i> if it has coproducts for finite families of objects. Equivalently, it has an initial object and binary coproducts.


id: binary coproducts
relation: has
nlab_link: https://ncatlab.org/nlab/show/coproduct
dual_property_id: binary products
description:
A category has <i>binary coproducts</i> if every pair $A,B$ of objects has a coproduct $A \sqcup B$.


id: powers
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: copowers
description:
If $X$ is an object and $I$ is a set, the <i>power</i> is defined as the product $X^I := \prod_{i \in I} X$. This property refers to the existence of powers.


id: copowers
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: powers
description:
If $X$ is an object and $I$ is a set, the <i>copower</i> is defined as the coproduct $I \otimes X := \coprod_{i \in I} X$. This property refers to the existence of copowers.


id: countable powers
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: countable copowers
description:
A category has <i>countable powers</i> when for every object $X$ and every countable set $I$ the power $X^I$ exists.


id: countable copowers
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: countable powers
description:
A category has <i>countable copowers</i> when for every object $X$ and every countable set $I$ the copower $I \otimes X$ exists.


id: finite powers
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: finite copowers
description:
A category has <i>finite powers</i> when for every object $X$ and every finite set $I$ the power $X^I$ exists. Equivalently, for every $n \in \mathbb{N}$ the power $X^n$ exists.


id: finite copowers
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: finite powers
description:
A category has <i>finite copowers</i> when for every object $X$ and every finite set $I$ the copower $I \otimes X$ exists. Equivalently, for every $n \in \mathbb{N}$ the copower $n \otimes X$ exists.


id: binary powers
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: binary copowers
description:
A category has <i>binary powers</i> when for every object $X$ the product $X \times X$  exists. These objects might also be called <i>squares</i>.


id: binary copowers
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: binary powers
description:
A category has <i>binary copowers</i> when for every object $X$ and every binary set $I$ the coproduct $X \sqcup X$ exists. These objects might also be called <i>doubles</i>.


id: equalizers
relation: has
nlab_link: https://ncatlab.org/nlab/show/equalizer
dual_property_id: coequalizers
description:
An <i>equalizer</i> of a pair of morphisms $f,g : A \rightrightarrows B$ is an object $E$ with a morphism $e : E \to A$ such that $f \circ e = g \circ e$ and which is universal with respect to this property. This property refers to the existence of equalizers.


id: coequalizers
relation: has
nlab_link: https://ncatlab.org/nlab/show/coequalizer
dual_property_id: equalizers
description:
A <i>coequalizer</i> of a pair of morphisms $f,g : A \rightrightarrows B$ is an object $C$ with a morphism $c : B \to C$ such that $c \circ f = c \circ g$ and which is universal with respect to this property. This property refers to the existence of coequalizers.


id: kernels
relation: has
nlab_link: https://ncatlab.org/nlab/show/kernel
dual_property_id: cokernels
description:
A category has <i>kernels</i> if it has zero morphisms and every morphism $f : A \to B$ has a kernel, i.e. an equalizer of $f$ with the zero morphism $0_{A,B} : A \to B$.


id: cokernels
relation: has
nlab_link: https://ncatlab.org/nlab/show/cokernel
dual_property_id: kernels
description:
A category has <i>cokernels</i> if it has zero morphisms and every morphism $f : A \to B$ has a cokernel, i.e. a coequalizer of $f$ with the zero morphism $0_{A,B} : A \to B$.


id: cofiltered limits
relation: has
nlab_link: https://ncatlab.org/nlab/show/filtered+limit
dual_property_id: filtered colimits
description:
A category has <i>cofiltered limits</i> if it has limits of diagrams indexed by small <a href="/category-property/cofiltered">cofiltered</a> categories. This is actually equivalent to having directed limits.


id: filtered colimits
relation: has
nlab_link: https://ncatlab.org/nlab/show/filtered+colimit
dual_property_id: cofiltered limits
description:
A category has <i>filtered colimits</i> if it has colimits of diagrams indexed by small <a href="/category-property/filtered">filtered</a> categories. This is actually equivalent to having directed colimits.


id: directed limits
relation: has
nlab_link: https://ncatlab.org/nlab/show/directed+limit
dual_property_id: directed colimits
description:
A category has <i>directed limits</i> if it has limits of diagrams indexed by codirected (small) posets. This is actually equivalent to having cofiltered limits.


id: directed colimits
relation: has
nlab_link: https://ncatlab.org/nlab/show/directed+colimit
dual_property_id: directed limits
description:
A category has <i>directed colimits</i> if it has colimits of diagrams indexed by directed (small) posets. This is actually equivalent to having filtered colimits. Directed colimits are (somewhat confusingly) also known as <i>inverse limits</i>.


id: sequential limits
relation: has
nlab_link: https://ncatlab.org/nlab/show/sequential+limit
dual_property_id: sequential colimits
description:
A category has <i>sequential limits</i> if it has limits of diagrams of the form $\cdots \bullet \to \bullet \to \bullet$.


id: sequential colimits
relation: has
nlab_link: https://ncatlab.org/nlab/show/sequential+colimit
dual_property_id: sequential limits
description:
A category has <i>sequential colimits</i> if it has colimits of diagrams of the form: $\bullet \to \bullet \to \bullet \to \cdots$.


id: connected limits
relation: has
nlab_link: https://ncatlab.org/nlab/show/connected+limit
dual_property_id: connected colimits
description:
A category has <i>connected limits</i> if it has limits of diagrams indexed by connected small categories.


id: connected colimits
relation: has
nlab_link: https://ncatlab.org/nlab/show/connected+colimit
dual_property_id: connected limits
description:
A category has <i>connected colimits</i> if it has colimits of diagrams indexed by connected small categories.


id: wide pullbacks
relation: has
nlab_link: https://ncatlab.org/nlab/show/wide+pullback
dual_property_id: wide pushouts
description:
A category $\mathcal{C}$ has <i>wide pullbacks</i> if for every object $S$ the slice category $\mathcal{C}/S$ has arbitrary small products.


id: wide pushouts
relation: has
nlab_link: https://ncatlab.org/nlab/show/wide+pushout
dual_property_id: wide pullbacks
description:
A category $\mathcal{C}$ has <i>wide pushouts</i> if for every object $S$ the coslice category $S/\mathcal{C}$ has arbitrary small coproducts.


id: coreflexive equalizers
relation: has
nlab_link: https://ncatlab.org/nlab/show/reflexive+coequalizer
dual_property_id: reflexive coequalizers
description:
A <i>coreflexive equalizer</i> is a limit of a diagram consisting of a parallel pair of morphisms with a common retraction (left inverse), which is the same concept as an equalizer of such a parallel pair. This property refers to the existence of coreflexive equalizers.


id: reflexive coequalizers
relation: has
nlab_link: https://ncatlab.org/nlab/show/reflexive+coequalizer
dual_property_id: coreflexive equalizers
description:
A <i>reflexive coequalizer</i> is a colimit of a diagram consisting of a parallel pair of morphisms with a common section (right inverse), which is the same concept as a coequalizer of such a parallel pair. This property refers to the existence of reflexive coequalizers.


id: quotients of congruences
relation: has
nlab_link: https://ncatlab.org/nlab/show/congruence
dual_property_id: coquotients of cocongruences
description:
A <i>congruence</i> (or internal equivalence relation) on an object $X$ of a category is a parallel pair $p_1, p_2 : E \rightrightarrows X$ which is jointly monomorphic, and such that for every object $T$, the image of $(p_1 \circ {-}, p_2 \circ {-}) : \operatorname{Hom}(T, E) \to \operatorname{Hom}(T, X)^2$ is an equivalence relation. The category has <i>quotients of congruences</i> if for each such congruence, there exists a coequalizer of $p_1$ and $p_2$. Note that in the case of a category with binary powers, the corresponding subobjects of $X \times X$ are also commonly referred to as congruences, or as internal equivalence relations.


id: coquotients of cocongruences
relation: has
dual_property_id: quotients of congruences
description:
A <i>cocongruence</i> (or internal equivalence corelation) on an object $X$ of a category is a parallel pair $i_1, i_2 : X \rightrightarrows E$ which is jointly epimorphic, and such that for every object $T$, the image of $({-} \circ i_1, {-} \circ i_2) : \operatorname{Hom}(E, T) \to \operatorname{Hom}(X, T)^2$ is an equivalence relation. The category has <i>coquotients of cocongruences</i> if for each such cocongruence, there exists an equalizer of $i_1$ and $i_2$. Note that in the case of a category with binary copowers, the corresponding quotients of $X + X$ are also commonly referred to as cocongruences, or as internal equivalence corelations.


id: cosifted limits
relation: has
nlab_link: https://ncatlab.org/nlab/show/sifted+colimit
dual_property_id: sifted colimits
description:
A category has <i>cosifted limits</i> if it has limits of diagrams indexed by small <a href="/category-property/cosifted">cosifted</a> categories.


id: sifted colimits
relation: has
nlab_link: https://ncatlab.org/nlab/show/sifted+colimit
dual_property_id: cosifted limits
description:
A category has <i>sifted colimits</i> if it has colimits of diagrams indexed by small <a href="/category-property/sifted">sifted</a> categories.


id: multi-complete
relation: is
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-cocomplete
description:
A <i>multi-limit</i> of a diagram $D\colon \mathcal{S} \to \mathcal{C}$ is a set $I$ of cones over $D$ such that every cone over $D$ uniquely factors through a unique cone belonging to $I$. This property refers to the existence of multi-limits of small diagrams. Note that any diagram with no cone admits a multi-limit, which is the empty set of cones.


id: multi-cocomplete
relation: is
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-complete
description:
A <i>multi-colimit</i> of a diagram $D\colon \mathcal{S} \to \mathcal{C}$ is a set $I$ of cocones under $D$ such that every cocone under $D$ uniquely factors through a unique cocone belonging to $I$. This property refers to the existence of multi-colimits of small diagrams. Note that any diagram with no cocone admits a multi-colimit, which is the empty set of cocones.


id: multi-terminal object
relation: has a
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-initial object
description:
This property refers to the existence of a multi-limit of the empty diagram. A category has a multi-terminal object if and only if the collection of all connected components is isomorphic to a set, and each connected component has a terminal object.


id: multi-initial object
relation: has a
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-terminal object
description:
This property refers to the existence of a multi-colimit of the empty diagram. A category has a multi-initial object if and only if the collection of all connected components is isomorphic to a set, and each connected component has a initial object.


id: biproducts
relation: has
nlab_link: https://ncatlab.org/nlab/show/biproduct
dual_property_id: biproducts
description:
A category has <i>biproducts</i> when it has zero morphisms, finite products (denoted $\times$), finite coproducts (denoted $\oplus$), and for every finite family of objects $A_1,\dotsc,A_n$ the canonical morphism
	$$\mu : A_1 \oplus \cdots \oplus A_n \to A_1 \times \cdots \times A_n$$
	is an isomorphism. Such a category is also called <i>semi-additive</i>, and it is automatically enriched over commutative monoids: the sum of $f,g : A \rightrightarrows B$ is defined as:
	$$A \xrightarrow{(f,g)} B \times B \xrightarrow{\mu^{-1}} B \oplus B \xrightarrow{\nabla} B$$


id: pointed
relation: is
nlab_link: https://ncatlab.org/nlab/show/pointed+category
dual_property_id: pointed
description:
A category is <i>pointed</i> when it has a zero object, i.e. an object which is both initial and terminal.


id: strict initial object
relation: has a
nlab_link: https://ncatlab.org/nlab/show/strict+initial+object
dual_property_id: strict terminal object
description:
A <i>strict initial object</i> is an initial object $0$ such that every morphism $A \to 0$ is an isomorphism. This property refers to the existence of a strict initial object.


id: strict terminal object
relation: has a
nlab_link: https://ncatlab.org/nlab/show/strict+terminal+object
dual_property_id: strict initial object
description:
A <i>strict terminal object</i> is a terminal object $1$ such that every morphism $1 \to A$ is an isomorphism. This property refers to the existence of a strict terminal object.


id: distributive
relation: is
nlab_link: https://ncatlab.org/nlab/show/distributive+category
dual_property_id: codistributive
description:
A category is <i>distributive</i> if it has finite products, finite coproducts, and for every object $A$ the functor $A \times -$ preserves finite coproducts. Concretely, for every finite family of objects $(B_i)$ the canonical morphism $\coprod_i (A \times B_i) \to A \times \coprod_i B_i$ must be an isomorphism.


id: infinitary distributive
relation: is
nlab_link: https://ncatlab.org/nlab/show/distributive+category
dual_property_id: infinitary codistributive
description:
A category is <i>infinitary distributive</i> if it has finite products, all coproducts, and for every object $A$ the functor $A \times -$ preserves all coproducts. Concretely, for every family of objects $(B_i)$ the canonical morphism $\coprod_i (A \times B_i) \to A \times \coprod_i B_i$ must be an isomorphism.


id: countably distributive
relation: is
dual_property_id: countably codistributive
description:
A category is <i>countably distributive</i> if it has finite products, countable coproducts, and for every object $A$ the functor $A \times -$ preserves countable coproducts. Concretely, for every countable family of objects $(B_i)$ the canonical morphism $\coprod_i (A \times B_i) \to A \times \coprod_i B_i$ must be an isomorphism.


id: countably codistributive
relation: is
dual_property_id: countably distributive
description:
A category is <i>countably codistributive</i> if it has finite coproducts, countable products, and for every object $A$ the functor $A \sqcup -$ preserves countable products. Concretely, for every countable family of objects $(B_i)$ the canonical morphism $A \sqcup \prod_i B_i \to \prod_i (A \sqcup B_i)$ must be an isomorphism.


id: codistributive
relation: is
dual_property_id: distributive
description:
A category is <i>codistributive</i> if it has finite coproducts, finite products, and for every object $A$ the functor $- \sqcup A$ preserves finite products. Concretely, for every finite family of objects $(B_i)$ the canonical morphism $A \sqcup \prod_i B_i \to \prod_i (A \sqcup B_i)$ must be an isomorphism.


id: infinitary codistributive
relation: is
dual_property_id: infinitary distributive
description:
A category is <i>infinitary codistributive</i> if it has finite coproducts, all products, and for every object $A$ the functor $A \sqcup -$ preserves all products. Concretely, for every family of objects $(B_i)$ the canonical morphism $A \sqcup \prod_i B_i \to \prod_i (A \sqcup B_i)$ must be an isomorphism.


id: exact filtered colimits
relation: has
nlab_link: https://ncatlab.org/nlab/show/commutativity+of+limits+and+colimits
dual_property_id: exact cofiltered limits
description:
In a category $\mathcal{C}$, which we assume to have filtered colimits and finite limits, we say that <i>filtered colimits are exact</i> if the following equivalent conditions are satisfied:
	<ol>
		<li>For every finite category $\mathcal{I}$ the functor $\lim : [\mathcal{I}, \mathcal{C}] \to \mathcal{C}$ preserves filtered colimits.</li>
		<li>For every small filtered category $\mathcal{J}$ the functor $\mathrm{colim} : [\mathcal{J},\mathcal{C}] \to \mathcal{C}$ preserves finite limits.</li>
		<li>For every diagram $X : \mathcal{I} \times \mathcal{J} \to \mathcal{C}$, where $\mathcal{I}$ is finite and $\mathcal{J}$ is small filtered, the canonical morphism $\mathrm{colim}_j \lim_i X(i,j) \to \lim_i \mathrm{colim}_j X(i,j)$ is an isomorphism.</li>
	</ol>


id: exact cofiltered limits
relation: has
nlab_link: https://ncatlab.org/nlab/show/commutativity+of+limits+and+colimits
dual_property_id: exact filtered colimits
description:
In a category $\mathcal{C}$, which we assume to have cofiltered limits and finite colimits, we say that <i>cofiltered limits are exact</i> if the following equivalent conditions are satisfied:
	<ol>
		<li>For every finite category $\mathcal{I}$ the functor $\mathrm{colim} : [\mathcal{I}, \mathcal{C}] \to \mathcal{C}$ preserves cofiltered limits.</li>
		<li>For every small cofiltered category $\mathcal{J}$ the functor $\lim : [\mathcal{J},\mathcal{C}] \to \mathcal{C}$ preserves finite colimits.</li>
		<li>For every diagram $X : \mathcal{I} \times \mathcal{J} \to \mathcal{C}$, where $\mathcal{I}$ is finite and $\mathcal{J}$ is small cofiltered, the canonical morphism $\mathrm{colim}_i \lim_j X(i,j) \to \lim_j \mathrm{colim}_i X(i,j)$ is an isomorphism.</li>
	</ol>


id: cartesian filtered colimits
relation: has
dual_property_id: cocartesian cofiltered limits
description:
In a category $\mathcal{C}$, which we assume to have filtered colimits and finite products, we say that <i>filtered colimits are cartesian</i> if for every finite set $I$ the product functor $\prod : \mathcal{C}^I \to \mathcal{C}$ preserves filtered colimits. Equivalently, for every $X \in \mathcal{C}$ the functor $X \times - : \mathcal{C} \to \mathcal{C}$ preserves filtered colimits.<br>
	This is no standard terminology, it has been suggested in <a href="https://mathoverflow.net/questions/510240" target="_blank">MO/510240</a>. We have added it to the database since it clarifies the relationship between many related properties.


id: cocartesian cofiltered limits
relation: has
dual_property_id: cartesian filtered colimits
description:
In a category $\mathcal{C}$, which we assume to have cofiltered limits and finite coproducts, we say that <i>cofiltered limits are cocartesian</i> if for every finite set $I$ the coproduct functor $\coprod : \mathcal{C}^I \to \mathcal{C}$ preserves cofiltered limits. Equivalently, for every $X \in \mathcal{C}$ the functor $X \sqcup - : \mathcal{C} \to \mathcal{C}$ preserves cofiltered limits.<br>
	This is no standard terminology, its dual has been suggested in <a href="https://mathoverflow.net/questions/510240" target="_blank">MO/510240</a>. We have added it to the database since it clarifies the relationship between many related properties.


id: disjoint finite coproducts
relation: has
nlab_link: https://ncatlab.org/nlab/show/disjoint+coproduct
dual_property_id: disjoint finite products
description:
A category has <i>disjoint finite coproducts</i> if it has finite coproducts, for every pair of objects $A,B$ the coproduct inclusions $A \rightarrow A+B \leftarrow B$ are monomorphisms, and the pullback $A \times_{A + B} B$ exists and is given by the initial object $0$.


id: disjoint coproducts
relation: has
nlab_link: https://ncatlab.org/nlab/show/disjoint+coproduct
dual_property_id: disjoint products
description:
A category has <i>disjoint coproducts</i> if it has coproducts, the coproduct inclusions $A_i \to \coprod_{i \in I} A_i$ are monomorphisms, and the pullback of the inclusions $A_i \to \coprod_{i \in I} A_i$ and $A_j \to \coprod_{i \in I} A_i$ for $i \neq j$ exists and is given by the initial object $0$.


id: disjoint finite products
relation: has
dual_property_id: disjoint finite coproducts
description:
A category has <i>disjoint finite products</i> if it has finite products, for every pair of objects $A,B$ the product projections $A \leftarrow A \times B \rightarrow B$ are epimorphisms, and the pushout $A \sqcup_{A \times B} B$ exists and is given by the terminal object $1$.<br>
	This terminology does not seem to be common, but we have added it as a dual for the more commonly known property of having disjoint finite coproducts.


id: disjoint products
relation: has
dual_property_id: disjoint coproducts
description:
A category has <i>disjoint products</i> if it has products, the product projections $\prod_{i \in I} A_i \to A_i$ are epimorphisms, and the pushout of the projections $\prod_{i \in I} A_i \to A_i$ and $\prod_{i \in I} A_i \to A_j$ for $i \neq j$ exists and is given by the terminal object $1$.<br>
	This terminology does not seem to be common, but we have added it as a dual for the more commonly known property of having disjoint coproducts.


id: regular
relation: is
nlab_link: https://ncatlab.org/nlab/show/regular+category
dual_property_id: coregular
description:
A category is <i>regular</i> when it is finitely complete, for every morphism $X \to Y$ its kernel pair $X \times_Y X \rightrightarrows X$ has a coequalizer, and regular epimorphisms are stable under pullbacks.


id: coregular
relation: is
dual_property_id: regular
description:
A category is <i>coregular</i> when its dual is regular, i.e. it is finitely cocomplete, for every morphism $Y \to X$ its cokernel pair $X \rightrightarrows X \sqcup_Y X$ has an equalizer, and regular monomorphisms are stable under pushouts.


id: extensive
relation: is
nlab_link: https://ncatlab.org/nlab/show/extensive+category
dual_property_id: coextensive
description:
A category $\mathcal{C}$ is <i>extensive</i> when it has finite coproducts and for all objects $A,B \in \mathcal{C}$ the coproduct functor $\mathcal{C}/A \times \mathcal{C}/B \to \mathcal{C}/(A+B)$ is an equivalence of categories. Equivalently, pullbacks of finite coproduct inclusions along arbitrary morphisms exist and finite coproducts are disjoint and stable under pullback.


id: coextensive
relation: is
dual_property_id: extensive
description:
A category $\mathcal{C}$ is <i>coextensive</i> when it has finite products and for all objects $A,B \in \mathcal{C}$ the product functor $A/\mathcal{C} \times B/\mathcal{C} \to (A \times B)/\mathcal{C}$ is an equivalence of categories. The prototypical example is the category of commutative rings.


id: infinitary extensive
relation: is
nlab_link: https://ncatlab.org/nlab/show/extensive+category
dual_property_id: infinitary coextensive
description:
A category $\mathcal{C}$ is <i>infinitary extensive</i> when it has coproducts and for all families of objects $(A_i)_{i \in I}$ the coproduct functor $\prod_{i \in I} \mathcal{C}/A_i \to \mathcal{C}/(\coprod_{i \in I} A_i)$ is an equivalence of categories. Equivalently, pullbacks of coproduct inclusions along arbitrary morphisms exist and coproducts are disjoint and stable under pullback.


id: infinitary coextensive
relation: is
dual_property_id: infinitary extensive
description:
A category $\mathcal{C}$ is <i>infinitary coextensive</i> when it has products and for all families of objects $(A_i)_{i \in I}$ the product functor $\prod_{i \in I} A_i / \mathcal{C}/A_i \to \prod_{i \in I} A_i / \mathcal{C}$ is an equivalence of categories. <br>
	This terminology does not seem to be common, but we have added it as a dual for the more commonly known property of being infinitary extensive.


id: unital
relation: is
nlab_link: https://ncatlab.org/nlab/show/unital+category
dual_property_id: counital
description:
A category is <i>unital</i> if it has a zero object, finite limits, and for all objects $X,Y$ the two morphisms $(\mathrm{id}_X,0) : X \hookrightarrow X \times Y$ and $(0,\mathrm{id}_Y) : Y \hookrightarrow X \times Y$ are jointly strongly epimorphic. This means: there is no proper subobject of $X \times Y$ that contains $X$ and $Y$. When coproducts exist, the canonical morphism $X \sqcup Y \to X \times Y$ therefore must be a strong epimorphism.


id: counital
relation: is
dual_property_id: unital
description:
A category is <i>counital</i> if its dual is unital, i.e., it has a zero object, finite colimits, and for all objects $X,Y$ the two morphisms $(\mathrm{id}_X;0) : X \sqcup Y \twoheadrightarrow X$ and $(0;\mathrm{id}_Y) : X \sqcup Y \twoheadrightarrow Y$ are jointly strongly monomorphic. When products exist, the canonical morphism $X \sqcup Y \to X \times Y$ therefore must be a strong monomorphism.


id: CIP
relation: satisfies
dual_property_id: CSP
description:
A category satisfies <i>CIP</i> ("coproducts inject into products") if it has zero morphisms, products, coproducts, and for every family of objects $(X_i)_{i \in I}$ the canonical morphism
	$$\textstyle \alpha : \coprod_i X_i \to \prod_{i \in I} X_i$$
	defined by $p_j \circ \alpha \circ \iota_i = \delta_{i,j}$ is a monomorphism. This is no standard terminology. This property has been added to clarify relationships between other properties, in particular those concerning the commutation between limits and colimits.


id: CSP
relation: satisfies
dual_property_id: CIP
description:
A category satisfies <i>CSP</i> ("coproducts surject onto products") if it has zero morphisms, products, coproducts, and for every family of objects $(X_i)_{i \in I}$ the canonical morphism
	$$\textstyle \alpha : \coprod_i X_i \to \prod_{i \in I} X_i$$
	defined by $p_j \circ \alpha \circ \iota_i = \delta_{i,j}$ is an epimorphism. This is no standard terminology. This property has been added to clarify relationships between other properties, in particular those concerning the commutation between limits and colimits.


id: inhabited
relation: is
nlab_link: https://ncatlab.org/nlab/show/inhabited+set
dual_property_id: inhabited
description:
A category is <i>inhabited</i> if it has at least one object. In classical logic, this is equivalent to being <i>non-empty</i> (which is a double negation).


id: small
relation: is
nlab_link: https://ncatlab.org/nlab/show/small+category
dual_property_id: small
description:
A category is <i>small</i> when the collection of objects and the collection of morphisms are sets, i.e. small.


id: essentially small
relation: is
nlab_link: https://ncatlab.org/nlab/show/small+category
dual_property_id: essentially small
description:
A category is <i>essentially small</i> when it is equivalent to a small category. In particular, there is a set of objects such that every object is isomorphic to an object in this set. In contrast to the property of being small, being essentially small is invariant under equivalences of categories.


id: locally small
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+small+category
dual_property_id: locally small
description:
A category is <i>locally small</i> when for every pair of objects $A,B$ the collection of morphisms $A \to B$ is a set. Here, we work with a set-theoretic foundation in which there are sets and collections. Categories are based on collections of objects and morphisms.


id: locally essentially small
relation: is
dual_property_id: locally essentially small
description:
A category is <i>locally essentially small</i> when for every pair of objects $A,B$ the collection of morphisms $A \to B$ is isomorphic to a set. (Here, we work with a set-theoretic foundation in which there are sets and collections. Categories are based on collections of objects and morphisms.) Equivalently, the category is equivalent to a locally small category. In contrast to being locally small, this condition is invariant under equivalences of categories. This is why we have added it to the database. For instance, every algebraic category is locally essentially small, but not necessarily locally small. This indicates that this is the "right" notion to work with.


id: essentially finite
relation: is
dual_property_id: essentially finite
description:
A category is <i>essentially finite</i> if it is equivalent to a finite category. Equivalently, there are only finitely many objects up to isomorphism, and the collection of morphisms between any two objects is isomorphic to a finite set. In contrast to being finite, this property is invariant under equivalences of categories.


id: well-powered
relation: is
nlab_link: https://ncatlab.org/nlab/show/well-powered+category
dual_property_id: well-copowered
description:
A category is <i>well-powered</i> if the collection of subobjects of any object is isomorphic to a set.


id: well-copowered
relation: is
nlab_link: https://ncatlab.org/nlab/show/well-powered+category
dual_property_id: well-powered
description:
A category is <i>well-copowered</i> if the collection of quotients of any object is isomorphic to a set.


id: generator
relation: has a
nlab_link: https://ncatlab.org/nlab/show/separator
dual_property_id: cogenerator
description:
An object $G$ of a category is called a <i>generator</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$ the equation $f = g$ holds if for every morphism $h : G \to A$ we have $f \circ h = g \circ h$. Equivalently, the functor $\mathrm{Hom}(G,-) : \mathcal{C} \to \mathbf{Set}^+$ is faithful. This property refers to the existence of a generator. By definition, $G$ is a generator if and only if $\{G\}$ is a <a href="/category-property/generating_set">generating set</a>.


id: cogenerator
relation: has a
nlab_link: https://ncatlab.org/nlab/show/cogenerator
dual_property_id: generator
description:
An object $Q$ of a category is called a <i>cogenerator</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$ the equation $f = g$ holds if for every morphism $h : B \to Q$ we have $h \circ f = h \circ g$. Equivalently, the functor $\mathrm{Hom}(-,Q) : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}^+$ is faithful. This property refers to the existence of a cogenerator. By definition, $Q$ is a cogenerator if and only if $\{Q\}$ is a <a href="/category-property/cogenerating_set">cogenerating set</a>.


id: generating set
relation: has a
nlab_link: https://ncatlab.org/nlab/show/separator
dual_property_id: cogenerating set
description:
A set of objects $S$ is called a <i>generating set</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$, $f = g$ holds if and only if for every morphism $h : G \to A$ with $G \in S$ we have $f \circ h = g \circ h$. Equivalently, the functor $(\hom(G,-))_{G \in S} : \mathcal{C} \to (\mathbf{Set}^+)^S$ is faithful. This property refers to the existence of a generating set.


id: cogenerating set
relation: has a
nlab_link: https://ncatlab.org/nlab/show/cogenerator
dual_property_id: generating set
description:
A set of objects $S$ is called a <i>cogenerating set</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$, $f = g$ holds if and only if for every morphism $h : B \to G$ with $G \in S$ we have $h \circ f = h \circ g$. Equivalently, the functor $(\hom(-,G))_{G \in S} : \mathcal{C}^{\mathrm{op}} \to (\mathbf{Set}^+)^S$ is faithful. This property refers to the existence of a cogenerating set.


id: mono-regular
relation: is
nlab_link: https://ncatlab.org/nlab/show/regular+monomorphism
dual_property_id: epi-regular
description:
A category is <i>mono-regular</i> when every monomorphism is regular, i.e. the equalizer of a pair of morphisms. Notice that this is not standard terminology, <a href="https://math.stackexchange.com/questions/5031588" target="_blank">apparently</a> the literature has no name for this yet. A <i>preadditive</i> category is mono-regular iff it is normal. The notion of a normal category is reserved for categories with zero morphisms, while mono-regular applies to all categories.


id: epi-regular
relation: is
nlab_link: https://ncatlab.org/nlab/show/regular+epimorphism
dual_property_id: mono-regular
description:
A category is <i>epi-regular</i> when every epimorphism is regular, i.e. the coequalizer of a pair of morphisms. Notice that this is not standard terminology, <a href="https://math.stackexchange.com/questions/5031588" target="_blank">apparently</a> the literature has no name for this yet. A <i>preadditive</i> category is epi-regular iff it is conormal. The notion of a conormal category is reserved for categories with zero morphisms, while epi-regular applies to all categories.


id: normal
relation: is
nlab_link: https://ncatlab.org/nlab/show/normal+monomorphism
dual_property_id: conormal
description:
A category is <i>normal</i> if it has zero morphisms and every monomorphism is a kernel of some morphism (in which case case it is also called a <i>normal monomorphism</i>). The assumption of having zero morphisms makes it possible to talk about kernels.


id: conormal
relation: is
nlab_link: https://ncatlab.org/nlab/show/normal+epimorphism
dual_property_id: normal
description:
A category is <i>conormal</i> if it has zero morphisms and every epimorphism is a cokernel of some morphism (in which case case it is also called a <i>normal epimorphism</i>). The assumption of having zero morphisms makes it possible to talk about cokernels.


id: left cancellative
relation: is
nlab_link: https://ncatlab.org/nlab/show/cancellative+category
dual_property_id: right cancellative
description:
A category is <i>left cancellative</i> if for every morphism $f : A \to B$ and every parallel pair of morphisms $g,h : T \rightrightarrows A$ with $f \circ g = f \circ h$ we have $g = h$. Equivalently, every morphism is a monomorphism.


id: right cancellative
relation: is
nlab_link: https://ncatlab.org/nlab/show/cancellative+category
dual_property_id: left cancellative
description:
A category is <i>right cancellative</i> if for every morphism $f : A \to B$ and every parallel pair of morphisms $g,h : B \rightrightarrows T$ with $g \circ f = h \circ f$ we have $g = h$. Equivalently, every morphism is an epimorphism.


id: Cauchy complete
relation: is
nlab_link: https://ncatlab.org/nlab/show/Cauchy+complete+category
dual_property_id: Cauchy complete
description:
A category is <i>Cauchy complete</i> if every idempotent splits. That is, every endomorphism $e : X \to X$ with $e^2 = e$ may be written as $e = i \circ p$ for some morphisms $p : X \to Y$ and $i : Y \to X$ with $p \circ i = \mathrm{id}_Y$. Equivalently, the pair $e,\mathrm{id}_X : X \rightrightarrows X$ has an equalizer (or an coequalizer).


id: balanced
relation: is
nlab_link: https://ncatlab.org/nlab/show/balanced+category
dual_property_id: balanced
description:
A category is <i>balanced</i> if every morphism which is a monomorphism and an epimorphism must be an isomorphism.


id: direct
relation: is
nlab_link: https://ncatlab.org/nlab/show/direct+category
dual_property_id: inverse
description:
A category is <i>direct</i> if it contains no infinite sequence of non-identity morphisms of the form
	$$\cdots \to A_2 \to A_1 \to A_0.$$
	For example, a poset is direct iff it is well-founded.


id: inverse
relation: is
nlab_link: https://ncatlab.org/nlab/show/inverse+category
dual_property_id: direct
description:
A category is <i>inverse</i> if its dual is direct, i.e., if it contains no infinite sequence of non-identity morphisms of the form
	$$A_0 \to A_1 \to A_2 \to \cdots.$$


id: filtered-colimit-stable monomorphisms
relation: has
dual_property_id: cofiltered-limit-stable epimorphisms
description:
A category has <i>filtered-colimit-stable monomorphisms</i> if it has filtered colimits and for every filtered diagram of monomorphisms $(X_i \to Y_i)$ also their colimit $\mathrm{colim}_i X_i \to \mathrm{colim}_i Y_i$ is a monomorphism.


id: cofiltered-limit-stable epimorphisms
relation: has
dual_property_id: filtered-colimit-stable monomorphisms
description:
A category has <i>cofiltered-limit-stable epimorphisms</i> if it has cofiltered limits and for every cofiltered diagram of epimorphisms $(X_i \to Y_i)$ also their limit $\lim_i X_i \to \lim_i Y_i$ is an epimorphism.


id: core-thin
relation: is
nlab_link: https://ncatlab.org/nlab/show/gaunt+category
dual_property_id: core-thin
description:
A category is <i>core-thin</i> if it satisfies the following equivalent conditions:
	<ol>
		<li>Its <a href="https://ncatlab.org/nlab/show/core+groupoid" target="_blank">core</a> is thin.</li>
		<li>Parallel isomorphisms are equal.</li>
		<li>Every automorphism is the identity.</li>
	</ol>


id: gaunt
relation: is
nlab_link: https://ncatlab.org/nlab/show/gaunt+category
dual_property_id: gaunt
description:
A category is <i>gaunt</i> when every isomorphism $f : A \to B$ must be the identity (in particular, $A = B$). This is the "skeletal variant" of being core-thin.


id: finitary algebraic
relation: is
nlab_link: https://ncatlab.org/nlab/show/algebraic+category
description:
A category is <i>finitary algebraic</i> if it is equivalent to the category of models of a one-sorted finitary algebraic theory. This is also known as a variety of one-sorted finitary algebras.


id: zero morphisms
relation: has
nlab_link: https://ncatlab.org/nlab/show/zero+morphism
dual_property_id: zero morphisms
description:
A category has <i>zero morphisms</i> if for every pair of objects $A,B$ there is a distinguished morphism $0_{A,B} : A \to B$, called the zero morphism, such that we have $f \circ 0_{A,B} = 0_{A,C}$ and $0_{B,C} \circ g = 0_{A,C}$ for all morphisms $f : B \to C$ and $g : A \to B$. The zero morphisms are unique if they exist, hence this is actually a property of the category.


id: preadditive
relation: is
nlab_link: https://ncatlab.org/nlab/show/Ab-enriched+category
dual_property_id: preadditive
description:
A category is <i>preadditive</i> when it is locally essentially small* and each hom-set carries the structure of an abelian group such that the composition is bilinear. Notice that "preadditive" is an extra structure. The property here just says that some preadditive structure exists.<br>
	*We demand this instead of the more common "locally small" to ensure that preadditive categories are invariant under equivalences of categories.


id: additive
relation: is
nlab_link: https://ncatlab.org/nlab/show/additive+category
dual_property_id: additive
description:
A category is <i>additive</i> if it is preadditive and has finite products (equivalently, finite coproducts). Note that in the context of finite products, the preadditive structure <a href="/lemma/preadditive_structure_unique">is unique</a>.


id: abelian
relation: is
nlab_link: https://ncatlab.org/nlab/show/abelian+category
dual_property_id: abelian
description:
A category is <i>abelian</i> if it is additive, every morphism has a kernel and a cokernel, and every monomorphism and epimorphism is normal. Equivalently, it is additive, has equalizers and coequalizers, and it is mono-regular and epi-regular. As opposed to other types of categories (such as monoidal categories), being abelian turns out to be a mere property.


id: split abelian
relation: is
nlab_link: https://ncatlab.org/nlab/show/split+exact+sequence
dual_property_id: split abelian
description:
A category is <i>split abelian</i> if it is abelian and every short exact sequence splits. Equivalently, every object is projective. Equivalently, every object is injective.


id: Grothendieck abelian
relation: is
nlab_link: https://ncatlab.org/nlab/show/Grothendieck+category
description:
A category is <i>Grothendieck abelian</i> if it is abelian, has coproducts (and is therefore cocomplete), a generator, and filtered colimits commute with finite limits. These categories play an important role in homological algebra.


id: trivial
relation: is
nlab_link: https://ncatlab.org/nlab/show/terminal+category
dual_property_id: trivial
description:
A category is <i>trivial</i> if it is equivalent to the <a href="/category/1">trivial category</a>. Equivalently, there is an initial object $0$ such that for every object $A$ the unique morphism $0 \to A$ is an isomorphism. Notice that we do not demand that the category is <i>isomorphic</i> to the trivial category. As a consequence, every inhabited indiscrete category is trivial in our sense.


id: thin
relation: is
nlab_link: https://ncatlab.org/nlab/show/thin+category
dual_property_id: thin
description:
A category is <i>thin</i> when between any pair of objects there is at most one morphism. Such categories correspond to preordered collections.


id: discrete
relation: is
nlab_link: https://ncatlab.org/nlab/show/discrete+category
dual_property_id: discrete
description:
A category is <i>discrete</i> when every morphism is an identity morphism. Thus, a discrete category is merely a collection of objects.


id: essentially discrete
relation: is
nlab_link: https://ncatlab.org/nlab/show/discrete+category
dual_property_id: essentially discrete
description:
A category is <i>essentially discrete</i> if it is equivalent to a discrete category. Equivalently, it is a thin groupoid. Notice that the nLab calls this property simply "discrete". In contrast to being discrete, this property is invariant under equivalences of categories. An essentially discrete category is the same as a <i>setoid</i> (a set equipped with an equivalence relation).


id: skeletal
relation: is
nlab_link: https://ncatlab.org/nlab/show/skeleton
dual_property_id: skeletal
description:
A category is <i>skeletal</i> when isomorphic objects are already equal. Every category is equivalent to a skeletal category (using the axiom of choice).


id: locally finitely presentable
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+finitely+presentable+category
description:
A category is <i>locally finitely presentable</i> if it satisfies one of the following equivalent conditions:
	<ol>
		<li>It is finitely accessible and cocomplete.</li>
		<li>It is finitely accessible and complete.</li>
		<li>It is equivalent to the category of finite-limit-preserving functors to $\mathbf{Set}$ from a small category with finite limits.</li>
		<li>It is equivalent to the category of models of a small finite-limit sketch.</li>
	</ol>
	For equivalence of conditions above, see Cor. 2.47, Thm. 1.46, and Cor. 1.52 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. This is the same as being locally $\aleph_0$-presentable.


id: locally presentable
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+presentable+category
dual_property_id: locally copresentable
description:
Let $\kappa$ be a regular cardinal. A category is <i>locally $\kappa$-presentable</i> if it satisfies one of the following equivalent conditions:
	<ol>
		<li>It is $\kappa$-accessible and cocomplete.</li>
		<li>It is $\kappa$-accessible and complete.</li>
		<li>It is equivalent to the category of $\kappa$-limit-preserving functors to $\mathbf{Set}$ from a small category with $\kappa$-limits.</li>
		<li>It is equivalent to the category of models of a small $\kappa$-limit sketch.</li>
	</ol>
	For equivalence of conditions above, see Cor. 2.47, Thm. 1.46, and Cor. 1.52 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.
	A category is <i>locally presentable</i> if it is locally $\kappa$-presentable for some regular cardinal $\kappa$.


id: locally ℵ₁-presentable
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+presentable+category
description:
This is the special case of the notion of locally $\kappa$-presentable categories, where $\kappa = \aleph_1$ is the first uncountable cardinal.


id: locally strongly finitely presentable
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+strongly+finitely+presentable+category
description:
A category is <i>locally strongly finitely presentable</i> if it is cocomplete and there is a set $G$ of strongly finitely presentable objects such that every object is a sifted colimit of objects from $G$.
	There are several equivalent conditions:
	<ol>
		<li>It is equivalent to the category of models of a many-sorted finitary algebraic theory.</li>
		<li>It is equivalent to the category of finite-product-preserving functors to $\mathbf{Set}$ from a small category with finite products (=Lawvere theory).</li>
		<li>It is equivalent to the category of models of a small finite-product sketch.</li>
		<li>It is equivalent to the Eilenberg&ndash;Moore category of a finitary (=filtered-colimit-preserving) monad on $\mathbf{Set}^S$ for some set $S$.</li>
		<li>It is equivalent to the Eilenberg&ndash;Moore category of a sifted-colimit-preserving monad on $\mathbf{Set}^S$ for some set $S$. (cf. [<a href="https://doi.org/10.2168/LMCS-8(3:14)2012" target="_blank">KR12</a>, Proposition 3.3])</li>
	</ol>
	A category satisfying this property is simply called a <i>variety</i> (of algebras) by some authors, although one should be aware that this term is sometimes used only for the one-sorted case.


id: accessible
relation: is
nlab_link: https://ncatlab.org/nlab/show/accessible+category
dual_property_id: coaccessible
description:
Let $\kappa$ be a regular cardinal. A category is <i>$\kappa$-accessible</i> if it has $\kappa$-filtered colimits and there is a (small) set $G$ of $\kappa$-presentable objects such that every object is a $\kappa$-filtered colimit of objects in $G$. A category is <i>accessible</i> if it is $\kappa$-accessible for some regular cardinal $\kappa$.


id: finitely accessible
relation: is
nlab_link: https://ncatlab.org/nlab/show/accessible+category
description:
A category is <i>finitely accessible</i> if it has filtered colimits and there is a (small) set $G$ of finitely presentable objects such that every object is a filtered colimit of objects in $S$.


id: ℵ₁-accessible
relation: is
nlab_link: https://ncatlab.org/nlab/show/accessible+category
description:
This is the special case of the notion of $\kappa$-accessible categories, where $\kappa = \aleph_1$ is the first uncountable cardinal.


id: generalized variety
relation: is a
description:
A category is a <i>generalized variety</i> if it has sifted colimits and there is a (small) set $G$ of strongly finitely presentable objects such that every object is a sifted colimit of objects from $G$. Generalized varieties are like locally strongly finitely presentable categories but without colimits. The relation is similar as between finitely accessible and locally finitely presentable categories. This notion is defined in <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01, Def. 3.6]</a>.


id: multi-algebraic
relation: is
description:
A category is <i>multi-algebraic</i> if it satisfies one of the following equivalent conditions:
	<ol>
		<li>It is a multi-cocomplete generalized variety, that is, it has multi-colimits and sifted colimits of all small diagrams, and there is a (small) set $G$ of strongly finitely presentable objects such that every object is a sifted colimit of objects from $G$.</li>
		<li>It is equivalent to the category of models of a small (finite product, coproduct)-sketch, shortly small <i>FPC-sketch</i>.</li>
		<li>It is equivalent to the category of multi-finite-product-preserving functors to $\mathbf{Set}$ from a small category with multi-finite-products (<i>multi-algebraic theory</i>). Here, multi-finite-products means multi-limits of finite discrete diagrams.</li>
		<li>It is equivalent to the category of models of a small multi-finite-product sketch.</li>
	</ol>
	Multi-algebraic categories are like locally strongly finitely presentable categories but only with multi-colimits. The relation is similar as between locally finitely multi-presentable and locally finitely presentable categories.
	For equivalence of conditions above, see [<a href="https://doi.org/10.1016/S0022-4049(01)00015-9" target="_blank">AR01a</a>, Lem. 1] and [<a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">AR01b</a>, Thm. 4.4].
	This notion was originally introduced by <a href="https://doi.org/10.1007/BF01224953" target="_blank">Diers</a>.


id: locally multi-presentable
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+multipresentable+category
description:
Let $\kappa$ be a regular cardinal. A category is <i>locally $\kappa$-multi-presentable</i> if it satisfies one of the following equivalent conditions:
	<ol>
		<li>It is $\kappa$-accessible and multi-cocomplete.</li>
		<li>It is $\kappa$-accessible and has connected limits.</li>
		<li>It is equivalent to the category of models of a small ($\kappa$-limit, coproduct)-sketch.</li>
	</ol>
	For equivalence of conditions above, see Thm. 4.30, Thm. 4.32, and the remark below in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. A category is <i>locally multi-presentable</i> if it is locally $\kappa$-multi-presentable for some $\kappa$.


id: locally finitely multi-presentable
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+multipresentable+category
description:
A category is <i>locally finitely multi-presentable</i> if it satisfies one of the following equivalent conditions:
	<ol>
		<li>It is finitely accessible and multi-cocomplete.</li>
		<li>It is finitely accessible and has connected limits.</li>
		<li>It is equivalent to the category of models of a small (finite limit, coproduct)-sketch.</li>
	</ol>
	For equivalence of conditions above, see Thm. 4.30, Thm. 4.32, and the remark below in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.


id: cartesian closed
relation: is
nlab_link: https://ncatlab.org/nlab/show/cartesian+closed+category
dual_property_id: cocartesian coclosed
description:
A category is <i>cartesian closed</i> if all finite products and exponentials $[X,Y]$ exist, defined by the adjunction $\mathrm{Hom}(T,[X,Y]) \cong \mathrm{Hom}(T \times X,Y)$.


id: cocartesian coclosed
relation: is
nlab_link: https://ncatlab.org/nlab/show/cocartesian+coclosed+category
dual_property_id: cartesian closed
description:
A category is <i>cocartesian coclosed</i> if its dual category is cartesian closed, i.e. if all finite coproducts and coexponentials $\mathrm{Coexp}(X,Y)$ exist, defined by the adjunction $\mathrm{Hom}(\mathrm{Coexp}[X,Y],T) \cong \mathrm{Hom}(Y,T \sqcup X)$.


id: locally cocartesian coclosed
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+cocartesian+coclosed+category
dual_property_id: locally cartesian closed
description:
A category is <i>locally cocartesian coclosed</i> if its dual is locally cartesian closed, i.e. if each of its coslice categories is cocartesian coclosed.


id: subobject classifier
relation: has a
nlab_link: https://ncatlab.org/nlab/show/subobject+classifier
dual_property_id: quotient object classifier
description:
A category $\mathcal{C}$ has a <i>subobject classifier</i> if it has finite limits and a monomorphism* $\top : 1 \hookrightarrow \Omega$ such that for every monomorphism $m : A \hookrightarrow B$ there is a unique morphism $\chi_m : B \to \Omega$ such that
	$$\begin{CD} A @>{m}>> B \\ @V{!}VV @VV{\chi_m}V \\ 1 @>>{\top}> \Omega \end{CD}$$
	is a pullback diagram. Equivalently, the functor $\mathrm{Sub} : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}^+$ is representable.<br>
	*Every morphism $1 \to \Omega$ is a split monomorphism anyway.


id: quotient object classifier
relation: has a
dual_property_id: subobject classifier
description:
A category $\mathcal{C}$ has a <i>quotient object classifier</i> if its dual has a subobject classifier. This means that it has finite colimits and an epimorphism* $\top : \Psi \twoheadrightarrow 0$ such that for every epimorphism $e : A \twoheadrightarrow B$ there is a unique morphism $\psi_e : \Psi \to A$ such that
	$$\begin{CD} \Psi @>{\top}>> 0 \\ @V{\psi_e}VV @VV{!}V \\ A @>>{e}> B \end{CD}$$
	is a pushout diagram. Equivalently, the functor $\mathrm{Quot} : \mathcal{C} \to \mathbf{Set}^+$ is representable.<br>
	*Every morphism $\Psi \to 0$ is a split epimorphism anyway.


id: regular subobject classifier
relation: has a
nlab_link: https://ncatlab.org/nlab/show/subobject+classifier
dual_property_id: regular quotient object classifier
description:
A category $\mathcal{C}$ has a <i>regular subobject classifier</i> if it has finite limits and a regular monomorphism* $\top : 1 \hookrightarrow \Omega$ such that for every regular monomorphism $m : A \hookrightarrow B$ there is a unique morphism $\chi_m : B \to \Omega$ such that
	$$\begin{CD} A @>{m}>> B \\ @V{!}VV @VV{\chi_m}V \\ 1 @>>{\top}> \Omega \end{CD}$$
	is a pullback diagram. Equivalently, the functor $\mathrm{Sub}_{\mathrm{reg}} : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}^+$ is representable.<br>
	*Every morphism $1 \to \Omega$ is a split monomorphism and hence regular anyway.


id: regular quotient object classifier
relation: has a
dual_property_id: regular subobject classifier
description:
A category $\mathcal{C}$ has a <i>regular quotient object classifier</i> if its dual has a regular subobject classifier. This means that it has finite colimits and a regular epimorphism* $\top : \Psi \twoheadrightarrow 0$ such that for every regular epimorphism $e : A \twoheadrightarrow B$ there is a unique morphism $\psi_e : \Psi \to A$ such that
	$$\begin{CD} \Psi @>{\top}>> 0 \\ @V{\psi_e}VV @VV{!}V \\ A @>>{e}> B \end{CD}$$
	is a pushout diagram. Equivalently, the functor $\mathrm{Quot}_{\mathrm{reg}} : \mathcal{C} \to \mathbf{Set}^+$ is representable.<br>
	*Every morphism $\Psi \to 0$ is a split epimorphism anyway.


id: elementary topos
relation: is an
nlab_link: https://ncatlab.org/nlab/show/topos
description:
An <i>elementary topos</i> is a cartesian closed category with finite limits and a subobject classifier.


id: natural numbers object
relation: has a
nlab_link: https://ncatlab.org/nlab/show/natural+numbers+object
description:
A <i>natural numbers object</i> (NNO) in a category with finite products is a triple
	$$(N,\, z : 1 \to N,\, s : N \to N)$$
	satisfying the following universal property: for all $f : A \to X$,  $g : X \to X$ there is a unique $\Phi : A \times N \to X$ such that $\Phi(a,z)=f(a)$ and $\Phi(a,s(n)) = g(\Phi(a,n))$ in element notation.
	<br>This concept is an abstraction of the set of natural numbers, which indeed provide a NNO for the category of sets. We have used the parametrized definition here which is more natural (sic!) for categories that are not cartesian closed (cf. <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Remark 2.5.3).


id: connected
relation: is
nlab_link: https://ncatlab.org/nlab/show/connected+category
dual_property_id: connected
description:
A category is <i>connected</i> if it is inhabited and every two objects can be joined via a zig-zag path of morphisms. Equivalently, $\mathcal{C}$ is connected if $\mathcal{C} \simeq \coprod_{i \in I} \mathcal{C}_i$ implies $\mathcal{C}_i \simeq 0$ for some $i$.


id: semi-strongly connected
relation: is
nlab_link: https://ncatlab.org/nlab/show/strongly+connected+category
dual_property_id: semi-strongly connected
description:
A category is <i>semi-strongly connected</i> if it is inhabited and for every two objects $A,B$ there is a morphism $A \to B$ or there is a morphism $B \to A$. Notice that this is stronger than being connected, and that posets with this property are precisely the inhabited totally ordered sets.


id: strongly connected
relation: is
nlab_link: https://ncatlab.org/nlab/show/strongly+connected+category
dual_property_id: strongly connected
description:
A category is <i>strongly connected</i> if it is inhabited and for every two objects $A,B$ there is a morphism $A \to B$. In other words, each hom-set is inhabited. Notice that when a terminal object $1$ exists, this property means that every object $A$ admits a morphism $1 \to A$.


id: filtered
relation: is
nlab_link: https://ncatlab.org/nlab/show/filtered+category
dual_property_id: cofiltered
description:
A category is <i>filtered</i> if every finite diagram admits a cocone. Equivalently, it is inhabited, for every two objects $x,y$ there is a cospan $x \rightarrow s \leftarrow y$ (not necessarily universal), and every parallel pair $x \rightrightarrows y$ is coequalized by some morphism $y \to c$ (not necessarily universal).


id: cofiltered
relation: is
nlab_link: https://ncatlab.org/nlab/show/cofiltered+category
dual_property_id: filtered
description:
A category is <i>cofiltered</i> if every finite diagram admits a cone. Equivalently, it is inhabited, for every two objects $x,y$ there is a span $x \leftarrow p \rightarrow y$ (not necessarily universal), and every parallel pair $x \rightrightarrows y$ is equalized by some morphism $e \to x$ (not necessarily universal).


id: sifted
relation: is
nlab_link: https://ncatlab.org/nlab/show/sifted+category
dual_property_id: cosifted
description:
A category $\mathcal{C}$ is <i>sifted</i> if it is inhabited and the diagonal functor $\Delta : \mathcal{C} \to \mathcal{C} \times \mathcal{C}$ is final, i.e. if it is non-empty and for any two objects $X,Y \in \mathcal{C}$ the category of cospans
	$$X \rightarrow Z \leftarrow Y$$
	is connected. Equivalently, a small category $\mathcal{C}$ is sifted if $\mathrm{colim} : \mathbf{Set}^{\mathcal{C}} \to \mathbf{Set}$ preserves finite products. This property is a weaker notion than being filtered.


id: cosifted
relation: is
nlab_link: https://ncatlab.org/nlab/show/sifted+category
dual_property_id: sifted
description:
A category $\mathcal{C}$ is <i>cosifted</i> if it is inhabited and the diagonal functor $\Delta : \mathcal{C} \to \mathcal{C} \times \mathcal{C}$ is initial, i.e. if it is non-empty and for any two objects $X,Y \in \mathcal{C}$ the category of spans
	$$X \leftarrow Z \rightarrow Y$$
	is connected. Equivalently, a small category $\mathcal{C}$ is cosifted if $\mathrm{colim} : \mathbf{Set}^{{\mathcal{C}}^\mathrm{op}} \to \mathbf{Set}$ preserves finite products. This property is a weaker notion than being cofiltered.


id: Malcev
relation: is
nlab_link: https://ncatlab.org/nlab/show/Malcev+category
dual_property_id: co-Malcev
description:
A category is <i>Malcev</i> when it has finite limits and every internal reflexive relation is an internal equivalence relation. That is, if $R \subseteq X^2$ is a subobject with $\Delta_X \subseteq R$, then $R$ is symmetric and transitive.


id: co-Malcev
relation: is
dual_property_id: Malcev
description:
A category is <i>co-Malcev</i> when its dual is Malcev, i.e., it has finite colimits and if $X \sqcup X \twoheadrightarrow R$ is a coreflexive corelation, then it is cosymmetric and cotransitive.<br>
	This terminology is not standard, but we have added it to properly formulate the interesting theorem that the dual of an elementary topos is Malcev, i.e., that every elementary topos is co-Malcev.<br>
	To settle this property, we often use that $\mathcal{C}$ is co-Malcev if and only if the category of representable functors $\mathcal{C} \to \mathbf{Set}^+$ is Malcev.


# category_implications

id: sifted_colimits_consequence
reason:
This is because filtered categories are sifted and because the index category for a reflexive coequalizer is sifted.


id: sifted_colimits_criterion
reason:
See Cor. 5.2 in <a href="http://www.tac.mta.ca/tac/volumes/37/35/37-35abs.html" target="_blank">Chen's paper</a>.


id: equalizers_via_coreflexive_equalizers
reason:
If $f,g : X \rightrightarrows Y$ are two morphisms, we have a coreflexive pair $(\mathrm{id}_X,f), (\mathrm{id}_X,g) : X \rightrightarrows X \times Y$. A morphism with codomain $X$ equalizes $f$ and $g$ if and only if it equalizes $(\mathrm{id}_X,f)$ and $(\mathrm{id}_X,g)$. Thus, their equalizers agree.


id: congruence_quotients_are_reflexive_coequalizers
reason:
A congruence $E \rightrightarrows X$ has a common section $X \to E$ given by the reflexivity morphism.


id: cokernels_via_congruence_quotients
reason:
By the regularity assumption, it suffices to consider cokernels of subobjects. Given a subobject $Y$ of $X$, we have the congruence on $X$ given by the pullback of ${-} : X \times X \to X$ and $Y$. The quotient of this congruence is a cokernel of $Y \hookrightarrow X$.


id: congruence_quotients_via_cokernels
reason:
For any congruence $E$ on an object $X$ of a preadditive category, let $E_0$ be the kernel of $p_2 : E \to X$.  The restriction of $p_1$ to $E_0$ is a monomorphism.  We can then see that $E$ must be the pullback of $p_1 - p_2 : E \to X$ and $E_0 \hookrightarrow X$.  Then the cokernel of $E_0 \hookrightarrow X$ is a quotient of $E$.


id: finite_products_characterization
reason:
The non-trivial direction follows since finite products can be constructed recursively via $X_1 \times \cdots \times X_{n+1} = (X_1 \times \cdots \times X_n) \times X_{n+1}$.


id: products_criterion
reason:
The product $\prod_{i \in I} X_i$ is the cofiltered limit of the finite partial products $\prod_{i \in E} X_i$ where $E$ ranges over the finite subsets of $I$.


id: complete_characterization
reason:
See <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, V.2, Thm. 2.


id: finitely_complete_characterization
reason:
See <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, V.2, Cor. 1.


id: pullbacks_criterion
reason:
The pullback of $f : X \to S$ and $g : Y \to S$ is the equalizer of $p_1 \circ f, \, p_2 \circ g : X \times Y \rightrightarrows S$.


id: equalizers_criterion
reason:
The equalizer of $f,g : X \rightrightarrows Y$ is the pullback of $(f,g) : X \to Y \times Y$ with the diagonal $Y \to Y \times Y$.


id: equalizers_consequence
reason:
If $e : X \to X$ is an idempotent, then the equalizer of $e, \mathrm{id}_X : X \rightrightarrows X$ provides a splitting of $e$.


id: sequential_colimits_consequence
reason:
Assume that $e : X \to X$ is an idempotent morphism. Consider the sequence
	$$X \xrightarrow{e} X \xrightarrow{e} X \to \cdots.$$
	A cocone under this sequence is a family of morphisms $f_n : X \to Y$ satisfying $$f_n = f_{n+1} e.$$
	Then
	$$f_n = f_{n+1} e = f_{n+2} e^2 = f_{n+2} e = f_{n+1}$$
	shows that all the morphisms are equal. Thus, a cocone is the same as a morphism $f_0 : X \to Y$ with $f_0 = f_0 e$, meaning it coequalizes $\mathrm{id}_X,e : X \rightrightarrows X$. Hence, if a colimit exists, $e$ splits.


id: finite_filtered_colimits
reason:
See <a href="https://mathoverflow.net/questions/509853" target="_blank">MO/509853</a>, where it is in fact shown that the ind-completion of any finite Cauchy-complete category becomes itself.


id: directed_colimits_suffice
reason:
This is Theorem 1.5. in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.


id: connected_limits_characterization
reason:
The direction $\Rightarrow$ is trivial. The direction $\Leftarrow$ can be found at the <a href="https://ncatlab.org/nlab/show/connected+limit" target="_blank">nLab</a>.


id: wide_pullbacks_criterion
reason:
To prove $\Leftarrow$, a wide pullback can be constructed as a cofiltered limit of finite pullbacks, and finite pullbacks can be reduced to binary pullbacks (the empty-indexed pullback always exists). Conversely, assume that wide pullbacks exist in $\mathcal{C}$. For every object $A$ then the slice category $\mathcal{C} / A$ has wide pullbacks and a terminal object, hence is complete. Since a cofiltered limit can be finally reduced to such a slice, we are done.


id: wide_pullbacks_finite
reason:
Each slice category has finite products and is essentially finite, hence has all products by <a href="/category-implication/freyd_finite">this result</a> followed by <a href="/category-implication/thin_finite_product_reduction">this result</a>.


id: sequential_limits_criterion
reason:
See <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, V.2, Prop. 3. The proof can easily be adapted to this case. Namely, the limit of $\cdots \to X_2 \to X_1 \to X_0$ is the equalizer of two suitable endomorphisms of $\prod_{n \geq 0} X_n$.


id: countable_products_criterion
reason:
If $X_1,X_2,\dotsc$ is an infinite sequence of objects, then their product is the limit of the sequence $\cdots \to X_2 \times X_1 \to X_1$.


id: countable_powers_criterion
reason:
We can write $X^{\mathbb{N}}$ as the limit of the sequence $\cdots \to X^3 \to X^2 \to X \to 1$ with transition morphisms $f_n : X^{n+1} \to X^n$, $(x_1,\dotsc,x_{n+1}) \mapsto (x_1,\dotsc,x_n)$, i.e., $p_i f_n = p_i$ for $1 \leq i \leq n$.


id: powers_criterion
reason:
The product $X^I$ is the cofiltered limit of the finite powers $X^E$, where $E$ ranges over the finite subsets of $I$.


id: multi-terminal_with_connected
reason:
Let $(T_i)_{i\in I}$ be a multi-terminal object in a connected category $\mathcal{C}$. By definition of multi-terminal objects, for each object $C$, there are a unique index $i_C\in I$ and a unique morphism $C \to T_{i_C}$. Since the index $i_C$ is invariant under connected components, $I$ must be a singleton. The converse is trivial.


id: multi-complete_with_finite_coproducts
reason:
Let $D\colon \mathcal{S} \to \mathcal{C}$ be a small diagram in a category $\mathcal{C}$. Since $\mathcal{C}$ has finite coproducts, the category $\mathbf{Cone}(D)$ of cones over $D$ has finite coproducts. In particular, $\mathbf{Cone}(D)$ is connected, hence a multi-terminal object in it automatically becomes a terminal object.


id: biproducts_unital
reason:
For all objects $X,Y$ the morphism $X \sqcup Y \to X \times Y$ is an isomorphism, hence a strong epimorphism.


id: biproducts_cartesian_filtered_colimits
reason:
If $I$ is a finite set, the product functor $\mathcal{C}^I \to \mathcal{C}$ is isomorphic to the coproduct functor $\mathcal{C}^I \to \mathcal{C}$, hence preserves <i>all</i> colimits that exist in $\mathcal{C}$.


id: disjoint_coproducts_thin_trivial
reason:
For every object $A$ the two inclusions $A \rightrightarrows A + A$ must be equal, so their equalizer is $A$, but also $0$ since the coproduct is disjoint. Hence $A = 0$.


id: disjoint_coproducts_strict
reason:
Let $1$ be the strict terminal object, and let $A$ be any object. Then $1 \to A + 1$ is an isomorphism, since $1$ is strict. Also, $A \to A + 1$ is a monomorphism by assumption. It follows that the unique morphism $u : A \to 1$ is a monomorphism. For all $f,g : B \to A$ we have $uf = ug$ (since $1$ is terminal), hence $f = g$.


id: exact_filtered_colimits_monos
reason:
This is because $f : X \longrightarrow Y$ is a monomorphism iff the diagram
	$$\begin{CD} X @>{\mathrm{id}}>> X \\ @V{\mathrm{id}}VV @VV{f}V \\ X @>>{f}> Y \end{CD}$$
	is a pullback, and if a functor preserves finite limits, it preserves pullbacks in particular.


id: distributive_duality
reason:
This is equivalent to the classical result that the <a href="https://en.wikipedia.org/wiki/Distributive_lattice" target="_blank">distributivity law in a lattice</a> can be defined in two ways, see e.g. Birkhoff, <i>Lattice Theory</i>, Ch. IX, Thm. 1.


id: strict_initial_trivial
reason:
If $0$ is the zero object, then for every object $A$ the unique morphism $A \to 0$ is an isomorphism by assumption.


id: strict_initial_left_criterion
reason:
It suffices to prove that in general any monomorphism $f : A \to 0$ into an initial object is an isomorphism. If $g : 0 \to A$ is the unique morphism, then $f \circ g = \mathrm{id}_0$ since $0$ is initial. But then $f$ is a split epimorphism and a monomorphism, hence an isomorphism.


id: strict_initial_right_criterion
reason:
Let $f : A \to 0$ be a morphism. Let $g : 0 \to A$ be the unique morphism. It is an epimorphism by assumption. Also, $f \circ g = \mathrm{id}_0$ since $0$ is initial. But then $g$ is a split monomorphism and an epimorphism, hence an isomorphism.


id: extensive_cocartesian_cofiltered_limits
reason:
Let $\mathcal{C}$ be an extensive category with cofiltered limits and a terminal object. Then the coproduct functor $\mathcal{C} \times \mathcal{C} \cong \mathcal{C}/1 \times \mathcal{C}/1 \to \mathcal{C}/(1+1)$ is an equivalence. The forgetful functor $\mathcal{C}/A \to \mathcal{C}$ creates connected limits, and hence preserves cofiltered limits. For every $X \in \mathcal{C}$ the functor $(X,-) : \mathcal{C} \to \mathcal{C} \times \mathcal{C}$ also preserves cofiltered limits. The composition of these functors is $X \sqcup - : \mathcal{C} \to \mathcal{C}$ and therefore also preserves cofiltered limits.


id: distributive_consequence
reason:
See the <a href="https://ncatlab.org/nlab/show/distributive+category" target="_blank">nLab</a> or Prop. 3.4 in <a href="https://doi.org/10.1016/0022-4049(93)90035-R" target="_blank">Introduction to extensive and distributive categories</a>.


id: distributive_criterion
reason:
Each functor $A \times -$ is left adjoint and hence preserves finite coproducts (in fact, all colimits).


id: countably_distributive_criterion
reason:
Each functor $A \times -$ is left adjoint and hence preserves countable coproducts (in fact, all colimits).


id: countably_distributive_filtered_criterion
reason:
Each functor $A \times -$ preserves finite coproducts and filtered colimits, hence all countable coproducts.


id: extensive_consequences
reason:
These are Prop. 2.6 and 2.8 in <a href="https://doi.org/10.1016/0022-4049(93)90035-R" target="_blank">Introduction to extensive and distributive categories</a>.


id: extensive_distributivity
reason:
This is Prop. 4.5 in <a href="https://doi.org/10.1016/0022-4049(93)90035-R" target="_blank">Introduction to extensive and distributive categories</a>.


id: infinitary_extensive_distributivity
reason:
One can adjust the proof of Prop. 4.5 in <a href="https://doi.org/10.1016/0022-4049(93)90035-R" target="_blank">Introduction to extensive and distributive categories</a> (which deals with the finite case).


id: power_construction
reason:
The power $X^I$ can be constructed as $[I \otimes 1, X]$ because
	$$\begin{align*}
	\mathrm{Hom}(T,[I \otimes 1, X]) & \cong \mathrm{Hom}(T \times (I \otimes 1),X) \\
	& \cong \mathrm{Hom}(I \otimes (T \times 1),X) \\
	& \cong \mathrm{Hom}(I \otimes T,X) \\
	& \cong \mathrm{Hom}(T,X)^I.
	\end{align*}$$
	In the second isomorphism we have used that $T \times -$ preserves copowers, which is true because it is a left adjoint.


id: CIP_criterion
reason:
Let $(X_i)_{i \in I}$ be a family of objects. For every finite subset $E \subseteq I$ the canonical morphism $\coprod_{i \in E} X_i = \prod_{i \in E} X_i \to \prod_{i \in I} X_i$ is a (split) monomorphism. Hence, their colimit is also a monomorphism, which is the canonical morphism $\coprod_{i \in I} X_i \to \prod_{i \in I} X_i$.


id: generator_via_coproduct
reason:
If $S$ is a generating set, we claim that $U := \coprod_{G \in S} G$ is a generator. Let $f,g : A \rightrightarrows B$ be two morphisms with $f h = g h$ for all $h : U \to A$. If $G \in S$, any morphism $G \to A$ extends to $U$ by using zero morphisms outside of $G$. Thus, $fh = gh$ holds for all $h : G \to A$ and $G \in S$. Since $S$ is a generating set, this implies $f = g$.


id: free-algebra-generates
reason:
Pick an algebraic theory that represents the category. The free algebra $F(1)$ on one generator is a generator since morphisms $F(1) \to X$ correspond to the elements of (the underlying set of) the algebra $X$.


id: cancellative_consequence
reason:
If $f,g$ are two parallel morphisms, then their equalizer is a regular monomorphism, but also an epimorphism by assumption, so it must be an isomorphism. But this means that $f = g$.


id: zero-no-mono
reason:
If $f,g : A \rightrightarrows B$ are two morphisms, then $0_{B,B} \circ f = 0_{A,B} = 0_{B,B} \circ g$, so that $f = g$.


id: codiagonal-no-mono
reason:
For every object $A$ the codiagonal $A + A \to A$ is a split epimorphism, and by assumption a monomorphism, hence an isomorphism. Hence, the two inclusions $i_1,i_2 : A \rightrightarrows A + A$ coincide. Now, if $f, g : A \rightrightarrows B$ are two morphisms, consider the induced morphism $h : A + A \to B$ and compute $f = h \circ i_1 = h \circ i_2 = g$.


id: reflexive_pair_trivial
reason:
Any parallel pair of morphisms with a common section (or retraction) must be a pair of equal isomorphisms.


id: direct_implies_one-way
reason:
The category is one-way since any non-identity endomorphism yields an infinite sequence of equal non-identity morphisms. The category is skeletal since any non-identity isomorphism $f : A \to B$ yields the infinite sequence $\dotsc,f^{-1},f,f^{-1},f$.


id: one-way_zero
reason:
If $f,g : A \rightrightarrows B$ are two morphisms, then since $0_{B,B} = \mathrm{id}_B$ we have
	$$f = 0_{B,B} \circ f = 0_{A,B} = 0_{B,B} \circ g = g.$$


id: direct_criterion
reason:
See the <a href="https://ncatlab.org/nlab/show/direct+category#direct_versus_oneway_categories" target="_blank">nLab</a> for a proof.


id: core-thin_products_thin
reason:
Let $X$ be any object. The swap $\tau : X \times X \to X \times X$ is an automorphism, hence equal to the identity. It follows that the projections $p_1,p_2 : X \times X \rightrightarrows X$ are the same. And this means that every two morphisms $Y \rightrightarrows X$ are the same.


id: direct_implies_sequential_limits
reason:
Assume that $\cdots \to A_2 \to A_1 \to A_0$ is a sequence of morphisms. We will prove that almost all of them are identities, so that the sequence is eventually constant and the limit exists. Assume the opposite, i.e. that there are infinitely many $A_k \to A_{k-1}$ which are not the identity. Pick some $n_1$ such that $A_{n_1} \to A_{n_1 - 1}$ is not the identity, and let $n_0 := n_1 - 1$. If $A_{n_i} \to A_{n_{i-1}}$ has been constructed, there is some $n_{i+1} > n_i$ such that the composite $A_{n_{i+1}} \to A_{n_i}$ is not the identity, because otherwise it would follow inductively that all $A_{k+1} \to A_k$, $k \geq n_i$ would be identities, which would contradict our infiniteness assumption. This way we construct an infinite sequence of non-identity morphisms $A_{n_{i+1}} \to A_{n_i}$, a contradiction.


id: one-way_reflexive
reason:
Every reflexive pair is equal: If $f s = g s = \mathrm{id}$, then since $s f  = \mathrm{id}$ (one-way), we must have $f = s^{-1}$, and likewise $g = s^{-1}$.


id: core-hin_quotients
reason:
If $p_1, p_2 : E \rightrightarrows X$ is a congruence, the symmetry morphism $s : E \to E$ is an automorphism of $E$, hence equal to $\mathrm{id}_E$ by assumption. But then $p_1 = p_2 \circ s = p_2$, and simply $\mathrm{id}_X$ is a coequalizer.


id: finitary_algebraic_consequence
reason:
This is trivial because a locally strongly finitely presentable category is a variety of many-sorted finitary algebras.


id: additive_biproducts
reason:
This is standard, see e.g. Prop. 2.1 on the <a href="https://ncatlab.org/nlab/show/additive+category" target="_blank">nLab</a>.


id: abelian_implies_regular
reason:
In an abelian category, every epimorphism is regular, and epimorphisms are pullback-stable, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. VIII.


id: preadditive_products_criterion
reason:
See <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, VIII.2., Theorem 2.


id: algebraic_well-copowered
reason:
See <a href="https://mathoverflow.net/questions/486607" target="_blank">MSE/486607</a>. Alternatively, one may combine the facts that finitary algebraic categories are locally (finitely) presentable and that locally presentable categories are well-copowered, both of which are saved in the database. But we include the direct proof here as well, since it makes it easier to deduce the property for finitary algebraic categories appearing in practice.


id: many_sorted_algebraic_implies_regular
reason:
The regular epimorphisms are precisely the sort-wise surjective homomorphisms, which are clearly stable under pullbacks.


id: algebras_with_0_disjoint_products
reason:
We have a constant in every algebra, let us denoted it by $0$. Then the projection $A \times B \to A$ is clearly surjective, hence an epimorphism. To show that $A \sqcup_{A \times B} B$ is trivial, let $R$ be an algebra which admits homomorphisms $f : A \to R$, $g : B \to R$ such that $f(p_1(a,b)) = g(p_2(a,b))$ for all $(a,b) \times A \times B$. This means $f(a) = g(b)$. In particular, $f(a) = g(0) = 0$. Likewise, $g(b) = 0$, and we are done.


id: grothendieck_abelian_cogenerator
reason:
See <a href="https://ncatlab.org/nlab/show/Categories+and+Sheaves" target="_blank">Kashiwara-Schapira</a>, Thm. 9.6.3.


id: grothendieck_abelian_self-dual
reason:
This follows since the dual of a non-trivial Grothendieck abelian category cannot be Grothendieck abelian. See Peter Freyd, <i>Abelian categories</i>, p. 116.


id: groupoid_with_multi-terminal
reason:
Let $f,g\colon A \rightrightarrows B$ be a parallel pair of morphisms. Since the category has a multi-terminal object, the connected component containing $A$ and $B$ has a terminal object. But since the category is a groupoid, both $A$ and $B$ are terminal objects in the connected component, hence $f=g$.


id: thin_consequence
reason:
Any two parallel morphisms are equal, so their equalizer is the identity, and every morphism is a monomorphism as well.


id: freyd_small
reason:
See <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, V.2, Prop. 3. The proof works for any category with powers.


id: freyd_countable
reason:
Adjust the proof of <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, V.2, Prop. 3.


id: thin_finite_product_reduction
reason:
The category is equivalent to a finite preordered set. But then products are just infima, so that repetitions of objects do not matter, and every product can be reduced to a finite one.


id: thin_countable_product_reduction
reason:
The category is equivalent to a countable preordered set. But then products are just infima, so that repetitions of objects do not matter, and every product can be reduced to a countable one.


id: cartesian_closed_thin_criterion
reason:
This is an application of the adjoint functor theorem. Specifically, if $P$ is a complete lattice in which $\sup_i \inf(t,x_i) = \inf(t, \sup_i y_i)$ always holds, then the functor $\inf(t,-)$ is a left adjoint because it preserves all suprema.


id: sequential_implies_lcc
reason:
Each slice is thin, semi-strongly connected, and has a terminal object. Thus, it corresponds to a linear order with a largest element $1$. Every such category is cartesian closed, where the exponential $a \Rightarrow b$ (Heyting implication) is $1$ when $a \leq b$ and otherwise $b$.


id: accessible_trivial_consequence
reason:
For a $\kappa$-accessible category, the set $G$ appearing in the definition gives a small dense full subcategory, which is in particular a generating set.


id: accessible_well-powered
reason:
See <a href="https://ncatlab.org/nlab/show/accessible+category#wellpoweredness_and_wellcopoweredness" target="_blank">nLab</a>.


id: accessible_locally_small
reason:
See the proof of Prop. 2.1.5 in <a href="https://bookstore.ams.org/conm-104" target="_blank">Makkai-Pare</a>.


id: accessible_well-copowered
reason:
See Thm. 2.49 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a> or Prop. 6.1.3 in <a href="https://bookstore.ams.org/conm-104" target="_blank">Makkai-Pare</a>.


id: locally_finitely_presentable_consequence
reason:
Special case of <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Prop. 1.59 with $\lambda = \aleph_0$.


id: locally_presentable_essentially_small
reason:
This follows from <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Thm. 1.64.


id: grothendieck_abelian_presentable
reason:
See <a href="https://arxiv.org/abs/1409.7051" target="_blank">Deriving Auslander's formula</a>, Cor. 5.2, or <a href="https://arxiv.org/abs/math/0102087" target="_blank">Sheafifiable homotopy model categories</a>, Prop. 3.10.


id: locally_strongly_finitely_presentable_raise
reason:
See <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Cor. 3.7.


id: finitely_accessible_raise
reason:
This is because any regular cardinal is strictly smaller than its successor cardinal. See <a href="https://ncatlab.org/nlab/show/sharply+smaller+cardinal" target="_blank">nLab</a>.


id: countably_accessible_thin
reason:
In general, every $\kappa$-filtered diagram in a poset whose elements are less than $\kappa$ admits the greatest element. Therefore, all the elements are $\kappa$-presentable, and the poset is $\kappa$-accessible.


id: generalized_variety_implies_accessible
reason:
See <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01, Remark 4.8(2)]</a>.


id: multi-algebraic_implies_locally_finitely_multi-presentable
reason:
This is because that every (finite product, coproduct)-sketch is clearly a (finite limit, coproduct)-sketch.


id: locally-finitely-multi-presentable_stable-monos
reason:
Every locally finitely multi-presentable category is a multi-reflective full subcategory of a presheaf category closed under filtered colimits (<a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, 4.30). Since multi-reflective full subcategories are in general closed under connected limits (<a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Thm. 4.26), in particular, we can calculate not only filtered colimits but also kernel pairs as well as in a presheaf category.


id: subobject_classifier_well-powered
reason:
See <a href="https://ncatlab.org/nlab/show/Sheaves+in+Geometry+and+Logic" target="_blank">Mac Lane & Moerdijk</a>, Prop. I.3.1.


id: ccc_no_strict_terminal
reason:
If a morphism $X \to Y$ exists, we get a morphism $1 \to [X,Y]$, which forces $[X,Y]$ to be a terminal object by assumption. But then any two morphisms $1 \rightrightarrows [X,Y]$ are equal, so that any two morphisms $X \rightrightarrows Y$ are equal.


id: subobject_classifier_consequence
reason:
The first part holds by convention, and the second part: any monomorphism $U \to X$ is the equalizer of $\chi_U,\chi_X : X \to \Omega$.


id: subobject_classifier_pointed_case
reason:
The universal property of $\top : 0 \to \Omega$ precisely says that every monomorphism $A \to B$ is the kernel of a unique morphism $B \to \Omega$, so it is normal.


id: regular_subobjects_trivial
reason:
In a right cancellative category, every regular monomorphism is an isomorphism, so that a terminal object is a regular subobject classifier.


id: regular_subobject_classifier_strict
reason:
Let $\Omega$ be a regular subobject classifier. Since $1$ is a strict terminal object, $\top : 1 \to \Omega$ is an isomorphism. This implies that every regular monomorphism is an isomorphism. Hence, by taking the equalizer of two parallel morphisms, we see that the category is thin.


id: topos_consequence
reason:
See <a href="https://ncatlab.org/nlab/show/Sheaves+in+Geometry+and+Logic" target="_blank">Mac Lane & Moerdijk</a>, Cor. IV.5.4, Cor. IV.10.5, Thm. 4.7.8.


id: topos_well-copowered_criterion
reason:
This follows from <a href="https://ncatlab.org/nlab/show/Sheaves+in+Geometry+and+Logic" target="_blank">Mac Lane & Moerdijk</a>, Theorem IV.7.8 (and Prop. I.3.1).


id: grothendieck_topos_definition
reason:
See <a href="https://ncatlab.org/nlab/show/Sheaves+in+Geometry+and+Logic" target="_blank">Mac Lane & Moerdijk</a>, Appendix, Prop. 4.4.


id: grothendieck_topos_consequence
reason:
A Grothendieck topos is locally presentable by Prop. 3.4.16 in <a href="https://www.cambridge.org/core/books/handbook-of-categorical-algebra/5033A02442342401E7BCC26A042DAB94" target="_blank">Handbook of Categorical Algebra Vol. 3</a>, has a cogenerator (see <a href="https://ncatlab.org/nlab/show/cogenerator" target="_blank">nLab</a>) and is infinitary extensive by <a href="https://ncatlab.org/nlab/show/Giraud%27s+theorem" target="_blank">Giraud's Theorem</a>. To show that it has exact filtered colimits, first observe that this is clearly true in every presheaf topos (since $\mathbf{Set}$ has the property). Every Grothendieck topos is a full reflective subcategory of a presheaf topos such that the reflector preserves finite limits (<a href="https://ncatlab.org/nlab/show/sheaf+toposes+are+equivalently+the+left+exact+reflective+subcategories+of+presheaf+toposes" target="_blank">nLab</a>), so we conclude with <a href="/lemma/exact_filtered_colimits_descend">this lemma</a>.


id: topos_is_locally_cartesian_closed
reason:
See <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Cor. A2.3.4.


id: lcc_implies_regular
reason:
See <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Lemma A1.5.13. From this it follows also that every elementary topos is regular.


id: topos_implies_coregular
reason:
This is proven in <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, A2.6.3 (for every quasitopos).


id: lcc_implies_extensive
reason:
The pullback functor preserves finite coproducts because it has a right adjoint. Remark: In combination with other implication, this result implies that every elementary topos is extensive.


id: lcc_extensive_yields_infinitary
reason:
The pullback functor preserves coproducts because it has a right adjoint. See also Remark 2.6 at the <a href="https://ncatlab.org/nlab/show/extensive+category" target="_blank">nLab</a>.


id: topos_is_malcev
reason:
This is Example 2.2.18 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>. An alternative proof is given later in A.5.17.


id: nno_criterion
reason:
Consider the copower $N := \coprod_{n \in \mathbb{N}} 1$ with inclusions $i_n : 1 \to N$ for $n \in \mathbb{N}$. We define $z := i_1 : 1 \to N$ and $s : N \to N$ by $s \circ i_n = i_{n+1}$. Since the category is countably distributive, we have $A \times N \cong \coprod_{n \in \mathbb{N}} A$ for every object $A$. Given morphisms $f : A \to X$, $g : X \to X$, a morphism $\Phi : A \times N \to X$ therefore corresponds to a family of morphisms $\phi_n : A \to X$ for $n \in \mathbb{N}$. The condition $\Phi(a,z)=f(a)$ becomes $\phi_0 = f$. The condition $\Phi(a,s(n)) = g(\Phi(a,n))$ becomes $\phi_{n+1} = g \circ \phi_n$. This recursively defines the morphisms $\phi_n$. (We are basically using that $\mathbb{N}$ is a natural numbers object in $\mathbf{Set}$.) Concretely, $\phi_n = g^n \circ f$.


id: nno_pointed_case
reason:
Let $(N,z,s)$ be a natural numbers object in a category with a zero object, denoted $0$. The morphism $z : 0 \to N$ must be zero. The universal property applied to $A=1$ implies that $s : N \to N$ is an initial object in the category of endomorphisms. This exists, it is given by the identity $0 \to 0$. Therefore, $N = 0$. The general universal property now becomes: For all $f : A \to X$, $g : X \to X$ there is a unique $\Phi : A \to X$ such that $\Phi(a) = f(a)$ and $\Phi(a)=g(\Phi(a))$. Apply this to $g = 0$ to conclude $f = 0$.


id: nno_terminal
reason:
By assumption, $z : 1 \to N$ is an isomorphism. Therefore, the terminal object $1$ is a NNO with $z = \mathrm{id}_1$ and $s = \mathrm{id}_1$. This precisely means that for all $f : A \to X$ and $g : X \to X$ there is a unique $\Phi : A \to X$ with $\Phi = f$ and $\Phi = g \circ \Phi$. In other words, we have $f = g \circ f$, and therefore $g = \mathrm{id}_X$ (take $f = \mathrm{id}_X$), which proves the claim. (From here one can further deduce that the category is thin.)


id: strongly_connected_pointed_criterion
reason:
By assumption there is a morphism $1 \to 0$. There is also a unique morphism $0 \to 1$. They are necessarily inverse to each other.


id: strongly_connected_disjoint_products
reason:
See <a href="https://math.stackexchange.com/a/5132300/1650" target="_blank">MSE/5130190</a> for a proof.


id: malcev_additive_criterion
reason:
See Prop. 2.2.13. in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.


id: malcev_implies_unital
reason:
This follows from Corollary 2.2.10 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>. The proof is also written down in <a href="https://math.stackexchange.com/a/5034834/1650" target="_blank">MSE/5033161</a>.


id: filtered_is_sifted
reason:
Every filtered category $\mathcal C$ is inhabited and has final diagonal functors $\Delta:\mathcal C\to\mathcal C^J$ for all finite index categories $J$; in particular, it is inhabited and its diagonal $\Delta:\mathcal C\to\mathcal C\times\mathcal C$ is final.


id: coproducts_implies_sifted
reason:
The category is inhabited by assumption, and the coproduct of two objects is initial in the corresponding category of cospans.


id: sifted_left_cancellative_implies_thin
reason:
For any object $X$ in a left-cancellative category, the connected component containing
	$$X \xrightarrow{\mathrm{id}} X \xleftarrow{\mathrm{id}} X$$
	in the category of cospans from $X$ to $X$ consists only of cospans
	$$X \xrightarrow{f} Y \xleftarrow{g} X$$
	where $f=g$; hence when the category is also sifted, all cospans must be of this form, and so any two parallel morphisms are equal.


# special_objects

category_id: Sh(X)
type: initial object
description:
constant sheaf with value $\varnothing$, sending all non-empty open sets to $\varnothing$ and the empty set to a singleton


category_id: Ban
type: coproducts
description:
The coproduct of a family of Banach spaces $(B_i)$ is the subspace $\bigl\{x \in \prod_i B_i : \sum_i |x_i| < \infty\bigr\}$ equipped with the $1$-norm $|x| := \sum_i |x_i|$.


category_id: Met_c
type: coproducts
description:
Given metric spaces $(X_i,d_i)$ with $d_i \leq 1$ w.l.o.g, we endow the disjoint union $\coprod_i X_i$ with the metric $d$ that extends the metrics $d_i$ and satisfies $d(x,y) = 1$ when $x,y$ are in different $X_i$.


category_id: Met_oo
type: coproducts
description:
disjoint union with the metric that extends the given ones and gives points in different spaces the distance $\infty$


category_id: Haus
type: products
description:
direct product with the <a href="https://en.wikipedia.org/wiki/Product_topology" target="_blank">product topology</a>


category_id: LRS
type: products
description:
See <a href="https://arxiv.org/abs/1103.2139" target="_blank">Localization of ringed spaces</a> by W. Gillam. See also <a href="https://math.stackexchange.com/questions/1033675" target="_blank">MSE/1033675</a>.


category_id: Meas
type: products
description:
direct products with the <a href="https://de.wikipedia.org/wiki/Produkt-%CF%83-Algebra" target="_blank">product $\sigma$-algebra</a>


category_id: Top*
type: products
description:
direct product with the <a href="https://en.wikipedia.org/wiki/Product_topology" target="_blank">product topology</a> and the obvious base point


category_id: Met_c
type: products
description:
[countable case] In the finite case, take direct products with the metric $d(x,y) = \sup_i d_i(x_i,y_i)$, but other metrics such as $d(x,y) = \sum_i d_i(x_i,y_i)$ also work. In the countable case, one can assume $d_i \leq 1$ and then define $d(x,y) = \sum_i d_i(x,y) / 2^i$.


category_id: Sch
type: products
description:
[finite case] The idea is to use $\mathrm{Spec}(A) \times \mathrm{Spec}(B) = \mathrm{Spec}(A \otimes B)$ and then to glue affine pieces together. See EGA I, Chap. I, Thm. 3.2.1.


category_id: Man
type: products
description:
[finite case] direct products $X \times Y$ with the product topology and the charts $\mathbb{R}^{n + m} = \mathbb{R}^n \times \mathbb{R}^m \cong U \times V \hookrightarrow X \times Y$ for charts $\mathbb{R}^n \cong U \hookrightarrow X$ and $\mathbb{R}^m \cong V \hookrightarrow Y$


# special_morphisms

category_id: BN
type: isomorphisms
description: only the number $0$
reason:
The $0$ is the only natural number which has an additive inverse, since for $n > 0$ we have $n + m > 0$ for all $m$.


category_id: BOn
type: isomorphisms
description: only the ordinal $0$
reason:
The $0$ is the only ordinal which has an additive inverse, since for $\alpha > 0$ we have $\alpha + \beta > 0$ for all $\beta$.


category_id: Delta
type: isomorphisms
description: bijective order-preserving maps
reason:
This is easy. Notice that bijective order-preserving maps automatically also reflect the order (because we work with totally ordered sets).


category_id: Rel
type: isomorphisms
description: bijective functions
reason:
For the non-trivial direction, assume that $R : A \to B$ is a relation which has an inverse relation $S : B \to A$. For every $a \in A$ we have $(a,a) \in \mathrm{id}_A = S \circ R$, so there is some $b \in B$ with $(a,b) \in R$ (and $(b,a) \in S$). This shows that $R$ is left-total, and for right-total the argument is similar. By symmetry, this also holds for $S$. To show that $R$ is a function, assume $(a,b_1), (a,b_2) \in R$. Choose some $b_3 \in B$ with $(b_3,a) \in S$. It follows $(b_3,b_1) \in S \circ R = \mathrm{id}_A$, so $b_3 = b_1$. Similarly, $(b_3,b_2) \in S \circ R = \mathrm{id}_A$, so $b_3 = b_2$. This shows that $R$ is a function, i.e. left-unique. That $R$ is injective, i.e. right-unique, follows by symmetry. Finally, $R$ is surjective since it is right-total.


category_id: Ab
type: monomorphisms
description: injective homomorphisms
reason:
This holds in every finitary algebraic category: the forgetful functor to $\mathbf{Set}$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms.


category_id: Ab_fg
type: monomorphisms
description: injective homomorphisms
reason:
Let $f : A \to B$ be a monomorphism of finitely generated abelian groups. Let $a \in A$ be in the kernel of $a$. Then we may view $a$ as a morphism $a : \mathbb{Z} \to A$ with $f \circ a = 0$, and $\mathbb{Z}$ is finitely generated. Hence, $a = 0$.


category_id: Ban
type: monomorphisms
description: injective linear contractions
reason:
The unit ball functor $U : \mathbf{Ban} \to \mathbf{Set}$ is faithful and representable (by $\mathbb{C}$), hence reflects and preserves monomorphisms.


category_id: Cat
type: monomorphisms
description: faithful functors that are injective on objects
reason:
Faithful functors that are injective on objects are clearly monomorphisms. For the converse, assume that $F : \mathcal{C} \to \mathcal{D}$ is a monomorphic functor. Two objects $A,B \in \mathcal{C}$ with $F(A) = F(B)$ can be regarded as functors $A,B : 1 \to \mathcal{C}$ from the trivial category satisfying $F \circ A = F \circ B$, so that $A = B$. Now assume that $f,g : A \to B$ are morphisms in $\mathcal{C}$ with $F(f) = F(g)$. These can be regarded as functors $f,g : \{ 0 \to 1 \} \to \mathcal{C}$ from the walking morphism category satisfying $F \circ f = F \circ g$, so that $f = g$.


category_id: Delta
type: monomorphisms
description: injective order-preserving maps
reason:
For the non-trivial direction, the forgetful functor to $\mathbf{Set}$ is representable (by the terminal object), hence preserves monomorphisms.


category_id: FinAb
type: monomorphisms
description: injective homomorphisms
reason:
Let $f : A \to B$ be a monomorphism of finite abelian groups. Let $a \in A$ be in the kernel of $a$, say of order $n$. Then we may view $a$ as a morphism $a : \mathbb{Z}/n \to A$ with $f \circ a = 0$, and $\mathbb{Z}/n$ is finite. Hence, $a = 0$.


category_id: FinGrp
type: monomorphisms
description: injective homomorphisms
reason:
Let $f : A \to B$ be a monomorphism of finite groups. Let $a \in A$ be in the kernel of $a$, say of order $n$. Then we may view $a$ as a morphism $a : C_n \to A$ with $f \circ a = 1$ (the trivial homomorphism), and $C_n$ is finite. Hence, $a = 1$.


category_id: FreeAb
type: monomorphisms
description: injective homomorphisms
reason:
Let $f : A \to B$ be a monomorphism of free abelian groups. Let $a \in A$ be in the kernel of $a$. Then we may view $a$ as a morphism $a : \mathbb{Z} \to A$ with $f \circ a = 0$, and $\mathbb{Z}$ is free. Hence, $a = 0$.


category_id: FS
type: monomorphisms
description: bijective maps
reason:
Assume that $f : X \to Y$ is a monomorphism in this category. If $a,b \in X$ are such that $a \neq b$ but $f(a) = f(b)$, let $h : X \to X$ be the transposition that swaps $a$, $b$. Then $f \circ \mathrm{id}_X = f = f \circ h$, so by assumption $\mathrm{id}_X = h$, a contradiction. This shows that $f$ is bijective.


category_id: Rel
type: monomorphisms
reason: See <a href="https://math.stackexchange.com/questions/350716/" target="_blank">MSE/350716</a>.
description:
A relation $R : A \to B$ is a monomorphism iff the map $R_* : P(A) \to P(B)$ defined by $T \mapsto \{b \in B : \exists \, a \in T: (a,b) \in R \}$ is injective.


category_id: Set_f
type: monomorphisms
description: injective maps
reason:
For the non-trivial direction, the forgetful functor to $\mathbf{Set}$ is representable (by the singleton), hence preserves monomorphisms.


category_id: Set
type: monomorphisms
description: injective maps
reason:
For the non-trivial direction, if $f : X \to Y$ is a monomorphism of sets and $a,b \in X$ satisfy $f(a) = f(b)$, then $f \circ a = f \circ b$ when we regard $a,b$ as morphisms $a,b : 1 \to X$. Hence, $a = b$. More generally, if $\mathcal{C}$ is a category that admits a faithful and representable functor $U : \mathcal{C} \to \mathbf{Set}$, then the monomorphisms in $\mathcal{C}$ are precisely the morphisms whose $U$-image is injective.


category_id: Sh(X)
type: monomorphisms
description: morphisms of sheaves that are injective on every open subset
reason:
For the non-trivial direction, assume that $f : F \to G$ is a monomorphism of sheaves on $X$. Then the diagonal $F \to F \times_G F$ is an isomorphism. Since pullbacks of sheaves are constructed section-wise, it follows that the diagonal $F(U) \to F(U) \times_{G(U)} F(U)$ is an isomorphism for every open set $U \subseteq X$. But this means that $f(U) : F(U) \to G(U)$ is injective.


category_id: Sh(X,Ab)
type: monomorphisms
description: morphisms of abelian sheaves that are injective on every open subset
reason:
For the non-trivial direction, assume that $f : F \to G$ is a monomorphism of abelian sheaves on $X$. Then the diagonal $F \to F \times_G F$ is an isomorphism. Since pullbacks of abelian sheaves are constructed section-wise, it follows that the diagonal $F(U) \to F(U) \times_{G(U)} F(U)$ is an isomorphism for every open set $U \subseteq X$. But this means that $f(U) : F(U) \to G(U)$ is injective.


category_id: Sp
type: monomorphisms
description: pointwise injective natural transformations
reason:
If $a : F \to G$ is a monomorphism of species, then the diagonal morphism $F \to F \times_G F$ is an isomorphism, so that for every $x$ the diagonal morphism $F(x) \to F(x) \times_{G(x)} F(x)$ is an isomorphism, i.e., $a(x) : F(x) \to G(x)$ is a monomorphism. This argument works for every functor category where the target has fiber products.


category_id: sSet
type: monomorphisms
description: pointwise injective transformations
reason:
If $a : F \to G$ is a monomorphism of simplicial sets, then the diagonal morphism $F \to F \times_G F$ is an isomorphism, so that for every $n$ the diagonal morphism $F(n) \to F(n) \times_{G(n)} F(n)$ is an isomorphism, i.e., $a(n) : F(n) \to G(n)$ is a monomorphism. This argument works for every functor category where the target has fiber products.


category_id: Top*
type: monomorphisms
description: injective pointed continuous maps
reason:
For the non-trivial direction, the forgetful functor to $\mathbf{Set}$ is representable (by the discrete two-point space), hence preserves monomorphisms.


category_id: walking_splitting
type: monomorphisms
description: the identities and $i$
reason:
The morphism $i$ is even a split monomorphism. The morphism $p$ is not a monomorphism since $p \circ \mathrm{id}_1 = p \circ ip$. The morphism $ip$ is not a monomorphism since it would imply that $p$ is a monomorphism.


category_id: Z
type: monomorphisms
description: pointwise injective natural transformations
reason:
If $a : F \to G$ is a monomorphism of Z-functors, then the diagonal morphism $F \to F \times_G F$ is an isomorphism, so that for every $R$ the diagonal morphism $F(R) \to F(R) \times_{G(R)} F(R)$ is an isomorphism, i.e., $a(R) : F(R) \to G(R)$ is a monomorphism. This argument works for every functor category where the target has fiber products.


category_id: Ab
type: epimorphisms
description: surjective homomorphisms
reason:
For the non-trivial direction, if $f : A \to B$ is an epimorphism, then $p \circ f = 0$ for the projection  $p : B \to B/f(A)$ implies that $p = 0$, so that $B = f(A)$.


category_id: Ban
type: epimorphisms
description: linear contractions with dense image
reason:
Let $f : X \to Y$ be an epimorphism of Banach spaces. The subspace $U := \overline{f(X)} \subseteq Y$ is closed. It is well-known that the quotient $Y/U$ is also a Banach space with a projection $p : Y \to Y/U$. Since $p \circ f = 0 = 0 \circ f$, we infer $p = 0$, so that $U = Y$.


category_id: Cat
type: epimorphisms
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is an epimorphism iff $F$ is surjective on objects and for every morphism $s$ in $\mathcal{D}$ there is a zigzag over $U := F(\mathcal{C})$, meaning morphisms $u_1,\dotsc,u_{m+1} \in U$, $v_1,\dotsc,v_m \in U$, $x_1,\dotsc,x_m \in \mathcal{D}$ and $y_1,\dotsc,y_m \in \mathcal{D}$ such that $s = x_1 u_1$, $u_1 = v_1 y_1$, $x_{i-1} v_{i-1} = x_i u_i$, $u_i y_{i-1} = v_i y_i$, $x_m v_m = u_{m+1}$ and $u_{m+1} y_m = s$.

reason:
This is an extension of the <a href="https://en.wikipedia.org/wiki/Isbell's_zigzag_theorem" target="_blank">corresponding theorem for monoids</a> and proven in <a href="https://www.jstor.org/stable/2373286" target="_blank">Epimorphisms and Dominions, III</a> by John R. Isbell.


category_id: CMon
type: epimorphisms
description:
A morphism in $\mathbf{CMon}$ is an epimorphism iff it is an epimorphism in $\mathbf{Mon}$, which in turn can be characterized by <a href="https://en.wikipedia.org/wiki/Isbell's_zigzag_theorem" target="_blank">Isbell's zigzag theorem</a>.

reason:
If $f : M \to N$ is a homomorphism of commutative monoids which is an epimorphism in $\mathbf{Mon}$, then it is trivially also an epimorphism in $\mathbf{CMon}$. The converse requires a proof, which can be found at <a href="https://math.stackexchange.com/a/5133596/1650">MSE/5133488</a>.
	


category_id: CAlg(R)
type: epimorphisms
description: a homomorphism of algebras which is an epimorphism of commutative rings
reason:
The forgetful functor $\mathbf{CAlg}(R) \to \mathbf{Ring}$ is faithful and hence reflects epimorphisms, but it also preserves epimorphisms since it preserves pushouts (since $\mathbf{CAlg}(R) \cong R / \mathbf{Ring}$). For epimorphisms of commutative rings see their <a href="/category/CRing">detail page</a>.


category_id: CRing
type: epimorphisms
description:
A ring map $f : R \to S$ is an epimorphism iff $S$ equals the <i>dominion</i> of $f(R) \subseteq S$, meaning that for every $s \in S$ there is some matrix factorization $(s) = Y X Z$ with $X \in M_{n \times n}(R)$, $Y \in M_{1 \times n}(S)$, and $Z \in M_{n \times 1}(S)$.

reason:
See <a href="https://stacks.math.columbia.edu/tag/04VM" target="_blank">Stacks Project</a> ,or, for many more results, the seminar <a href="https://www.numdam.org/issues/SAC_1967-1968__2_/">Les épimorphismes d'anneaux</a>. See also <a href="https://mathoverflow.net/questions/109/" target="_blank">MO/109</a> for some results.


category_id: Delta
type: epimorphisms
description: surjective order-preserving maps
reason:
We can use the same <a href="/category/FinOrd">proof as for $\mathbf{FinOrd}$</a> since this has merely used the non-empty ordered set $\{0 < 1\}$.


category_id: FI
type: epimorphisms
description: bijective maps
reason:
Take an epimorphism $X \to Y$ in this category, w.l.o.g. the inclusion of a subset $X \subseteq Y$. This means that for two injective maps $f,g : Y \rightrightarrows T$ with $f|_X = g|_X$ we must have $f = g$. Let $T := Y + (Y \setminus X)$ (disjoint union), $f$ be the inclusion into the first summand, $g|_X$ be the inclusion into the first summand, and $g|_{Y \setminus X}$ be the inclusion into the second summand. Then $f$ and $g$ are injective with $f|_X = g|_X$, so that $f = g$. But this means $X = Y$.


category_id: FinGrp
type: epimorphisms
description: surjective homomorphisms
reason:
For the non-trivial direction, if $f : G \to H$ is an epimorphism, we may factor it as $G \to f(G) \to H$, and $f(G) \to H$ is still an epimorphism, but also an inclusion and hence a monomorphism. Since we already know that the category is mono-regular, $f(G) \to H$ must be an isomorphism.


category_id: FinOrd
type: epimorphisms
description: surjective order-preserving maps
reason:
The proof is similar to the one for $\mathbf{Set}$: If $f : X \to Y$ is an epimorphism of (finite) orders, in particular for all morphisms $g,h : Y \to \{0 < 1\}$ with $g \circ f = h \circ f$ we have $g = h$. This means for all upper sets $A,B \subseteq Y$ with $f^*(A) = f^*(B)$ we have $A = B$. If $y \in Y$, apply this to the intervals $A = Y_{\geq y}$ and $B = Y_{> y}$, which are different. Hence, there is some $x \in f^*(A) \setminus f^*(B)$, which means $f(x) \geq y$ but not $f(x) > y$, so that $f(x) = y$.


category_id: FreeAb
type: epimorphisms
description:
homomorphisms $f : A \to B$ with the property that $f(A)$ is not contained in a proper direct summand of $B$.

reason:
Let $f : A \to B$ be a morphism of free abelian groups such that $f(A)$ is not contained in a proper direct summand of $B$. Then $f$ is an epimorphism: If $t : B \to C$ is a morphism with $t \circ f = 0$, we want to show $t = 0$. Since the image of $t$ is free abelian (as a subgroup of $B$), we may assume that $t$ is surjective. Since $C$ is free, $t$ splits, so its kernel $K$ is a direct summand of $B$ (Splitting Lemma). It contains $f(A)$ because of $t \circ f = 0$. By assumption, $K = B$, so that $t = 0$. Conversely, assume that $f : A \to B$ is an epimorphism, and that $f(A)$ is contained in a direct summand $K$ of $B$. Let $L$ be a complement of $B$ (which is free abelian) and let $p : B \to L$ be the projection. Then $t \circ f = 0$, so $t = 0$. But then $L = 0$, which means $K = B$.


category_id: Grp
type: epimorphisms
description: surjective homomorphisms
reason:
For the non-trivial direction, if $f : G \to H$ is an epimorphism, we may factor it as $G \to f(G) \to H$, and $f(G) \to H$ is still an epimorphism, but also an inclusion and hence a monomorphism. Since we already know that the category is balanced, $f(G) \to H$ must be an isomorphism.


category_id: Haus
type: epimorphisms
description: continuous maps with dense image
reason:
It is clear that continuous maps with dense image are epimorphism, but the other direction is non-trivial. See <a href="https://math.stackexchange.com/questions/214045" target="_blank">MSE/214045</a> for a proof.


category_id: M-Set
type: epimorphisms
description: surjective $M$-maps
reason:
This holds in every functor category $[\mathcal{C},\mathbf{Set}]$, here applied to the case that $\mathcal{C}$ has just one object.


category_id: Man
type: epimorphisms
description: smooth maps with dense image
reason:
Let $f : X \to Y$ be a smooth map which is not dense. Choose some non-empty open subset $U \subseteq Y$ that is disjoint from $f(X)$. Take any $p \in U$. There is a smooth <i>bump function</i> $g : Y \to \mathbb{R}$ with $g(p) = 1$ and $\mathrm{supp}(g) \subseteq U$ (John Lee, Introduction to Smooth Manifolds, Prop. 2.25). Hence, $g \neq 0$ and $g \circ f = 0 = 0 \circ f$. Thus, $f$ is not an epimorphism.


category_id: Mon
type: epimorphisms
description:
A monoid map $f : T \to S$ is an epimorphism iff $S$ equals the <i>dominion</i> of $U := f(T) \subseteq S$, meaning that for every $s \in S$ there are $u_1,\dotsc,u_{m+1} \in U$, $v_1,\dotsc,v_m \in U$, $x_1,\dotsc,x_m \in S$ and $y_1,\dotsc,y_m \in S$ such that $s = x_1 u_1$, $u_1 = v_1 y_1$, $x_{i-1} v_{i-1} = x_i u_i$, $u_i y_{i-1} = v_i y_i$, $x_m v_m = u_{m+1}$ and $u_{m+1} y_m = s$.

reason:
This is <a href="https://en.wikipedia.org/wiki/Isbell's_zigzag_theorem" target="_blank">Isbell's zigzag Theorem</a>, see references there.


category_id: PMet
type: epimorphisms
description: surjective non-expansive maps
reason:
Let $f : X \to Y$ be a non-expansive map that is not surjective. Choose $y_0 \in Y \setminus f(X)$. We extend the pseudo-metric from $Y$ to $Z := Y \sqcup \{y'_0\}$ via $d(y,y'_0) := d(y,y_0)$, i.e., we make $y_0,y'_0$ indistinguishable. Let $g : Y \to Z$ be the inclusion and $h : Y \to Z$ be the map that composes $g$ with the swap between $y_0$ and $y'_0$. Both are isometric and satisfy $g \circ f = h \circ f$. Therefore, $f$ is not an epimorphism.


category_id: Pos
type: epimorphisms
description: surjective order-preserving functions
reason:
Clearly, surjective maps are epimorphisms. Conversely, assume that $f : P \to Q$ is an order-preserving map which is not surjective. Choose $q \in Q \setminus f(P)$. The order-preserving maps $Q \to \{0 < 1\}$ correspond to the upper sets in $Q$, and composing them with $f$ corresponds to taking their $f$-preimages. Consider the two upper sets $Q_{> q}$ and $Q_{\geq q}$. Their $f$-preimages are the same since $q \notin f(P)$. Therefore, $f$ is not an epimorphism.


category_id: Prost
type: epimorphisms
description: surjective order-preserving functions
reason:
Clearly, surjective maps are epimorphisms. The converse follows since, as mentioned, the forgetful functor $\mathbf{Prost} \to \mathbf{Set}$ has a right adjoint hence preserves epimorphisms.


category_id: R-Mod
type: epimorphisms
description: surjective $R$-linear maps
reason:
The forgetful functor to abelian groups is faithful and preserves colimits, hence reflects and preserves epimorphisms. Alternatively, use the same proof as for abelian groups.


category_id: Rel
type: epimorphisms
reason: See <a href="https://math.stackexchange.com/questions/350716/" target="_blank">MSE/350716</a>.
description:
A relation $R : A \to B$ is an epimorphism iff the map $R^* : P(B) \to P(A)$ defined by $S \mapsto \{a \in A : \exists \, b \in S: (a,b) \in R \}$ is injective.


category_id: Set_c
type: epimorphisms
description: surjective maps
reason:
For the non-trivial direction, if $f : X \to Y$ is an epimorphism of countable sets, then in particular $f^* : \hom(Y,2) \to \hom(X,2)$ is injective, which identifies with $f^* : P(Y) \to P(X)$. Then for all $y \in Y$ we have $f^*(\{y\}) \neq f^*(\varnothing) = \varnothing$, so that $y$ has a preimage.


category_id: Set_f
type: epimorphisms
description: surjective maps with finite fibers
reason:
For the non-trivial direction, if $f : X \to Y$ is an epimorphism in this category, then in particular $f^* : \hom(Y,2) \to \hom(X,2)$ is injective, which identifies with $f^* : E(Y) \to E(X)$, where $E(-)$ denotes the set of finite subsets. Then for all $y \in Y$ we have $f^*(\{y\}) \neq f^*(\varnothing) = \varnothing$, so that $y$ has a preimage.


category_id: Set
type: epimorphisms
description: surjective maps
reason:
For the non-trivial direction, if $f : X \to Y$ is an epimorphism of sets, then in particular $f^* : \hom(Y,2) \to \hom(X,2)$ is injective, which identifies with $f^* : P(Y) \to P(X)$. Then for all $y \in Y$ we have $f^*(\{y\}) \neq f^*(\varnothing) = \varnothing$, so that $y$ has a preimage.


category_id: Set*
type: epimorphisms
description: surjective pointed maps
reason:
If $f : X \to Y$ is an epimorphism of pointed sets, then the inclusion $i : f(X) \hookrightarrow Y$ is an epimorphism as well. But it is also a split monomorphism (we may map everything in $Y \setminus f(X)$ to the base point of $f(X)$, and do not touch the rest), hence an isomorphism.


category_id: Setne
type: epimorphisms
description: surjective maps
reason:
For the non-trivial direction, let $f : X \to Y$ be an epimorphism of non-empty sets. Then the inclusion $f(X) \hookrightarrow Y$ is an epimorphism as well, but also a split monomorphism (we can just map everything in $Y \setminus f(X)$ to a fixed element of $f(X)$). As such, it must be an isomorphism.


category_id: Sh(X)
type: epimorphisms
description:
morphisms of sheaves $f : F \to G$ that are "locally surjective": for every local section $g \in G(U)$ there is an open covering $U = \bigcup_{i \in I} U_i$ such that each $g|_{U_i} \in G(U_i)$ is contained in the image of $f(U_i) : F(U_i) \to G(U_i)$.

reason:
The one direction is easy. For the other one, assume that $f : F \to G$ is an epimorphism of sheaves. For every $x \in X$ the map on stalks $f_x : F_x \to G_x$ is an epimorphism because the stalk functor $\mathrm{Sh}(X) \to \mathbf{Set}$ admits a right adjoint: take skyscraper sheaves. For $x \in U$ then $g_x \in G_x$ has a preimage in $F_x$, say represented by some $f \in F(V_x)$ for some $x \in V_x \subseteq U$. By construction of the stalk $G_x$, there is some $x \in U_x \subseteq V_x$ with $f(U_x)(f|_{U_x}) = g|_{U_x}$. Hence, the sets $(U_x)$ provide the open covering.


category_id: Sh(X,Ab)
type: epimorphisms
description:
morphisms of abelian sheaves $f : F \to G$ that are "locally surjective": for every local section $g \in G(U)$ there is an open covering $U = \bigcup_{i \in I} U_i$ such that each $g|_{U_i} \in G(U_i)$ is contained in the image of $f(U_i) : F(U_i) \to G(U_i)$.

reason:
The one direction is easy. For the other one, assume that $f : F \to G$ is an epimorphism of abelian sheaves. For every $x \in X$ the homomorphism on stalks $f_x : F_x \to G_x$ is an epimorphism because the stalk functor  $\mathrm{Sh}(X,\mathbf{Ab}) \to \mathbf{Ab}$ admits a right adjoint: take skyscraper sheaves. For $x \in U$ then $g_x \in G_x$ has a preimage in $F_x$, say represented by some $f \in F(V_x)$ for some $x \in V_x \subseteq U$. By construction of the stalk $G_x$, there is some $x \in U_x \subseteq V_x$ with $f(U_x)(f|_{U_x}) = g|_{U_x}$. Hence, the sets $(U_x)$ provide the open covering.


category_id: Top
type: epimorphisms
description: surjective continuous maps
reason:
The proof works exactly as for the category of sets, where we endow $2$ with the indiscrete topology.


category_id: Top*
type: epimorphisms
description: surjective pointed continuous maps
reason:
For the non-trivial direction: The forgetful functor $\mathbf{Top}_* \to \mathbf{Top}$ preserves pushouts and hence also epimorphisms, and we already know the epimorphisms in $\mathbf{Top}$.


category_id: Vect
type: epimorphisms
description: surjective linear maps
reason:
The forgetful functor to abelian groups is faithful and preserves colimits, hence reflects and preserves epimorphisms. Alternatively, just use the same proof as for abelian groups.


category_id: walking_splitting
type: epimorphisms
description: the identities and $p$
reason:
The morphism $p$ is even a split monomorphism. The morphism $i$ is not an epimorphism since $\mathrm{id}_1 \circ i = ip \circ i$. The morphism $ip$ is not a epimorphism since it would imply that $i$ is an epimorphism.


category_id: BOn
type: regular monomorphisms
description: ordinals of the form $\alpha \cdot \omega$, where $\alpha$ is any ordinal
reason:
This results from the proof that equalizers exist, see <a href="https://math.stackexchange.com/questions/5029668" target="_blank">MSE/5029668</a>.


category_id: Ban
type: regular monomorphisms
description: closed embeddings
reason:
The non-trivial direction follows from the <a href="https://math.stackexchange.com/questions/319867" target="_blank">well-known fact</a> that for every closed subspace of a Banach space its quotient space is again a Banach space.


category_id: Fld
type: regular monomorphisms
reason: See <a href="https://math.stackexchange.com/questions/5129895" target="_blank">MSE/5129895</a>.
description:
A Galois extension is a regular monomorphism iff it is procyclic, and the general case can be reduced to this situation; see the reference for details.


category_id: Haus
type: regular monomorphisms
description: embeddings with closed image
reason:
The explicit construction of equalizers shows that they are embeddings, and they have a closed image because of the well-known lemma that for a Hausdorff space the diagonal $X \to X \times X$ has closed image. For the other non-trivial direction, see <a href="https://math.stackexchange.com/questions/214045/" target="_blank">MSE/214045</a>.


category_id: Meas
type: regular monomorphisms
description: embeddings
reason:
(This is the same proof as for $\mathbf{Top}$.) Equalizers are embeddings by their construction. Conversely, if $f : X \to Y$ is an embedding, then $f$ is the equalizer of the two characteristic maps $\chi_Y, \chi_{f(X)} : Y \rightrightarrows \{0,1\}$, where $\{0,1\}$ carries the trivial $\sigma$-algebra.


category_id: Met_c
type: regular monomorphisms
description: embeddings of closed subspaces
reason:
A reference is Example 7.58 (3) in <a href="https://ncatlab.org/nlab/show/Abstract+and+Concrete+Categories" target="_blank">Joy of Cats</a>, but a proof is missing there.


category_id: Pos
type: regular monomorphisms
description: embeddings
reason:
Every regular monomorphism is an embedding by the explicit construction of equalizers. For the converse, let $i : P \to Q$ be an embedding, which we may assume to be an inclusion. Consider the cokernel pair $C := Q \cup_P Q$. It exists because the category has pushouts, but we can (and need to) describe it more concretely: As a set, $C$ is a disjoint union of $P$ and two copies of $Q \setminus P$. The elements will be denoted by $p,q_1,q_2$. The $i$th inclusion map $Q \to C$ maps $p \in P$ to itself and $q \in Q \setminus P$ to $q_i$. The ordering is directly induced by $Q$: We have $p \leq p'$ in $C$ iff $p \leq p'$ in $P$, we have $p \leq q_i$ iff $p \leq q$ in $Q$, etc. One verifies that this indeed defines a partial order, and by construction the two maps $Q \rightrightarrows C$ are order-preserving and have equalizer $P$.


category_id: Prost
type: regular monomorphisms
description: embeddings
reason:
Every regular monomorphism is an embedding by the explicit construction of equalizers. For the converse, let $i : P \to Q$ be an embedding, which we may assume to be an inclusion. Consider the cokernel pair $C := Q \cup_P Q$. It exists because the category has pushouts, but we can (and need to) describe it more concretely: As a set, $C$ is a disjoint union of $P$ and two copies of $Q \setminus P$. The elements will be denoted by $p,q_1,q_2$. The $i$th inclusion map $Q \to C$ maps $p \in P$ to itself and $q \in Q \setminus P$ to $q_i$. The ordering is directly induced by $Q$: We have $p \leq p'$ in $C$ iff $p \leq p'$ in $P$, we have $p \leq q_i$ iff $p \leq q$ in $Q$, etc. One verifies that this indeed defines a preorder, and by construction the two maps $Q \rightrightarrows C$ are order-preserving and have equalizer $P$.


category_id: Top
type: regular monomorphisms
description: embeddings
reason:
Equalizers are embeddings by their construction. Conversely, if $f : X \to Y$ is an embedding, then $f$ is the equalizer of the two characteristic maps $\chi_Y, \chi_{f(X)} : Y \to \{0,1\}$, where $\{0,1\}$ carries the indiscrete topology.


category_id: Top*
type: regular monomorphisms
description: embeddings
reason:
Equalizers are embeddings by their construction. Conversely, if $f : X \to Y$ is an embedding, then $f$ is the equalizer of the two characteristic maps $\chi_Y, \chi_{f(X)} : Y \to \{0,1\}$, where $\{0,1\}$ carries the indiscrete topology and $1$ is the base point.


category_id: walking_fork
type: regular monomorphisms
description: the identities and $i$
reason:
First, $i$ is the equalizer of $f,g$. The morphism $f$ is an epimorphism (and a monomorphism), but no isomorphism, hence cannot be a regular monomorphism. The same holds for $g$.


category_id: Top
type: regular epimorphisms
description: surjective quotient maps
reason:
Regular epimorphisms are surjective quotient maps by the explicit construction of coequalizers. Conversely, if $q : X \to Y$ is a surjective quotient map, then one checks that $q$ is the coequalizer of its kernel pair $X \times_Y X \rightrightarrows X$: This is true for the underlying sets, and continuity of the induced morphism follows since $q$ is a quotient map.


category_id: Top*
type: regular epimorphisms
description: surjective pointed quotient maps
reason:
Regular epimorphisms are surjective pointed quotient maps by the explicit construction of coequalizers. Conversely, if $q : X \to Y$ is a surjective pointed quotient map, then one checks that $q$ is the coequalizer of its kernel pair $X \times_Y X \rightrightarrows X$: This is true for the underlying pointed sets, and continuity of the induced morphism follows since $q$ is a quotient map.


# functors

id: id_Set
name: identity functor on the category of sets
source: Set
target: Set
nlab_link: https://ncatlab.org/nlab/show/identity+functor
description:
Every category $\mathcal{C}$ has an identity functor $\mathrm{id}_{\mathcal{C}}$. Here, we specify that it is for the category of sets.


id: power_set_covariant
name: covariant power set functor
source: Set
target: Set
nlab_link: https://ncatlab.org/nlab/show/power+set
description:
This functor maps a set $X$ to its power set $P(X)$ and a map of sets $f : X \to Y$ to the induced image operator $f_* : P(X) \to P(Y)$.


id: power_set_contravariant
name: contravariant power set functor
source: Set_op
target: Set
nlab_link: https://ncatlab.org/nlab/show/power+set
description:
This functor maps a set $X$ to its power set $P(X)$ and a map of sets $f : X \to Y$ to the induced preimage operator $f^* : P(Y) \to P(X)$.


# functor_properties

id: full
relation: is
nlab_link: https://ncatlab.org/nlab/show/full+functor
dual_property_id: full
description:
A functor is full when it is surjective on Hom-sets: Every morphism $F(A) \to F(B)$ is induced by a morphism $A \to B$.


id: essentially surjective
relation: is
nlab_link: https://ncatlab.org/nlab/show/essentially+surjective+functor
dual_property_id: essentially surjective
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is essentially surjective when every object $Y \in \mathcal{D}$ is isomorphic to $F(X)$ for some $X \in \mathcal{C}$.


id: product-preserving
relation: is
dual_property_id: coproduct-preserving
description:
A functor $F$ preserves products when for every family of objects $(A_i)$ in the source whose product $\prod_i A_i$ exists, also the product $\prod_i F(A_i)$ exists in the target and such that the canonical morphism $F(\prod_i A_i) \to \prod_i F(A_i)$ is an isomorphism.


id: coproduct-preserving
relation: is
dual_property_id: product-preserving
description:
A functor $F$ preserves coproducts when for every family of objects $(A_i)$ in the source whose coproduct $\prod_i A_i$ exists, also the coproduct $\coprod_i F(A_i)$ exists in the target and such that the canonical morphism $\coprod_i F(A_i) \to F(\coprod_i A_i)$ is an isomorphism.


id: finite-product-preserving
relation: is
dual_property_id: finite-coproduct-preserving
description:
A functor $F$ preserves finite products when for every finite family of objects $(A_i)$ in the source whose product $\prod_i A_i$ exists, also the product $\prod_i F(A_i)$ exists in the target and such that the canonical morphism $F(\prod_i A_i) \to \prod_i F(A_i)$ is an isomorphism.


id: finite-coproduct-preserving
relation: is
dual_property_id: finite-product-preserving
description:
A functor $F$ preserves finite coproducts when for every family of objects $(A_i)$ in the source whose coproduct $\prod_i A_i$ exists, also the coproduct $\coprod_i F(A_i)$ exists in the target and such that the canonical morphism $\coprod_i F(A_i) \to F(\coprod_i A_i)$ is an isomorphism.


id: terminal-object-preserving
relation: is
dual_property_id: initial-object-preserving
description:
A functor $F$ preserves terminal objects when it maps every terminal object to a terminal object. It is not assumed that the source category has a terminal object.


id: initial-object-preserving
relation: is
dual_property_id: terminal-object-preserving
description:
A functor $F$ preserves initial objects when it maps every initial object to an initial object. It is not assumed that the source category has a initial object.


id: equalizer-preserving
relation: is
dual_property_id: coequalizer-preserving
description:
A functor $F$ preserves equalizers when for every parallel pair of morphisms $f,g : A \rightrightarrows B$ whose equalizer $i : E \to A$ exists, also $F(i) : F(E) \to F(A)$ is an equalizer of $F(f),F(g) : F(A) \rightrightarrows F(B)$.


id: coequalizer-preserving
relation: is
dual_property_id: equalizer-preserving
description:
A functor $F$ preserves coequalizers when for every parallel pair of morphisms $f,g : A \rightrightarrows B$ whose coequalizer $p : B \to Q$ exists, also $F(p) : F(B) \to F(Q)$ is an coequalizer of $F(f),F(g) : F(A) \rightrightarrows F(B)$.


id: left adjoint
relation: is a
nlab_link: https://ncatlab.org/nlab/show/left+adjoint
dual_property_id: right adjoint
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is a left adjoint when there is a functor $G : \mathcal{D} \to \mathcal{C}$ such that there are natural bijections $\hom(F(A),B) \cong \hom(A,G(B))$.


id: right adjoint
relation: is a
nlab_link: https://ncatlab.org/nlab/show/right+adjoint
dual_property_id: left adjoint
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is a right adjoint when there is a functor $G : \mathcal{D} \to \mathcal{C}$ such that there are natural bijections $\hom(G(A),B) \cong \hom(A,F(B))$.


id: monadic
relation: is
nlab_link: https://ncatlab.org/nlab/show/monadic+functor
dual_property_id: comonadic
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is monadic when there is a monad $T$ on $\mathcal{D}$ such that $F$ is equivalent to the forgetful functor $U^T : \mathbf{Alg}(T) \to \mathcal{D}$.


id: comonadic
relation: is
nlab_link: https://ncatlab.org/nlab/show/comonadic+functor
dual_property_id: monadic
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is comonadic when there is a comonad $T$ on $\mathcal{D}$ such that $F$ is equivalent to the forgetful functor $U^T : \mathbf{CoAlg}(T) \to \mathcal{D}$.


id: conservative
relation: is
nlab_link: https://ncatlab.org/nlab/show/conservative+functor
dual_property_id: conservative
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is conservative when it is isomorphic-reflecting: If $f$ is a morphism in $\mathcal{C}$ such that $F(f)$ is an isomorphism, then $f$ is an isomorphism.


id: representable
relation: is
nlab_link: https://ncatlab.org/nlab/show/representable+functor
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ is representable if $\mathcal{C}$ is locally small, $\mathcal{D} = \mathbf{Set}$, and there is an object $A \in \mathcal{C}$ with $F \cong \mathrm{Hom}(A,-)$.


id: monomorphism-preserving
relation: is
nlab_link: https://ncatlab.org/nlab/show/monomorphism
dual_property_id: epimorphism-preserving
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ preserves monomorphisms if every monomorphism in $\mathcal{C}$ is mapped to a monomorphism in $\mathcal{D}$.<br>
        This property is useful to rule out some adjunctions.


id: epimorphism-preserving
relation: is
nlab_link: https://ncatlab.org/nlab/show/epimorphism
dual_property_id: monomorphism-preserving
description:
A functor $F : \mathcal{C} \to \mathcal{D}$ preserves epimorphisms if every epimorphism in $\mathcal{C}$ is mapped to an epimorphism in $\mathcal{D}$.<br>
        This property is useful to rule out some adjunctions.


# functor_implications

id: equivalence_criterion
reason:
This is standard, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. IV, Theorem 4.1.


id: continuous_criterion
reason:
This follows from the construction of limits via equalizers and products, see  <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. V, Theorem 2.2.


id: continuous_criterion_filtered
reason:
This is because every limit can be written as a filtered limit of finite limits, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. IX, Theorem 1.1.


id: product_criterion_filtered
reason:
This is because every product can be written as a filtered limit of finite products, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. IX, Theorem 1.1.


id: left_exact_criterion
reason:
This follows from the construction of finite limits via equalizers and finite products, see  <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. V, Theorem 2.2.


id: right_adjoint_consequences
reason:
This is standard, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. V, Theorem 5.1.


id: saft
reason:
This is the Special Adjoint Functor Theorem. The proof can be found, for example, at the <a href="https://ncatlab.org/nlab/show/adjoint+functor+theorem" target="_blank">nLab</a>, or in <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. V, Theorem 8.2.


id: monadic_consequences
reason:
This is clear since for a monad $T$ the forgetful functor from the category of $T$-algebras has these properties.


id: representable_is_continuous
reason:
This is standard, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. V, Theorem 4.1.


id: left_exact_preserves_mono
reason:
This is because $f : X \to Y$ is a monomorphism if and only the square displaying $\mathrm{id}_Y \circ f = \mathrm{id}_Y \circ f$ is a pullback square.


id: faithful-via-equalizers
reason:
Let $f,g : X \rightrightarrows Y$ be two morphisms in the source category, and choose an equalizer $E \hookrightarrow X$. By assumption, $F(E) \to F(X)$ is the equalizer of $F(f),F(g) : F(X) \rightrightarrows F(Y)$. Thus, if $F(f) = F(g)$, then $F(E) \to F(X)$ is an isomorphism. Since $F$ is conservative, $E \to X$ is an isomorphism, which means $f = g$.


id: monadic_full_trivial
reason:
Assume that $T$ is a monad on $\mathcal{C}$ such that the forgetful functor $U : \mathbf{Alg}(T) \to \mathcal{C}$ is full. Then by general facts about adjunctions (see <a href="https://math.stackexchange.com/questions/1994963" target="_blank">MSE/1994963</a>) the unit $\eta : \mathrm{id}_{\mathcal{C}} \to T$ is an isomorphism of functors. Then it is an isomorphism of monads, so that $U$ is an equivalence.


# lemmas

id: thin_algebraic_categories
title: Algebraic categories are "never" thin
claim:
Let $\mathcal{C}$ be a <a href="/category-property/thin">thin</a> and <a href="/category-property/finitary_algebraic">finitary algebraic</a> category. Then $\mathcal{C} \simeq 1$ or $\mathcal{C} \simeq \{0 < 1\}$.

proof:
Let $F : \mathbf{Set} \to \mathcal{C}$ denote the free algebra functor. Every object $A \in \mathcal{C}$ admits a regular epimorphism $F(X) \twoheadrightarrow A$ for some set $X$. But since $\mathcal{C}$ is thin, every regular epimorphism must be an isomorphism. Also, $F(X)$ is a coproduct of copies of $F(1)$, which means it is either the initial object $0$ or $F(1)$ itself (since $\mathcal{C}$ is thin). If $F(1) \cong 0$, then every object is isomorphic to the initial object $0$, and hence $\mathcal{C}$ is trivial. If not, then $\mathcal{C}$ has exactly two objects up to isomorphism, $0$ and $F(1)$, there is a morphism $0 \to F(1)$, but no morphism $F(1) \to 0$. Since $\mathcal{C}$ is thin, we conclude $\mathcal{C} \simeq \{0 \to 1\}$.


id: hilberts_hotel
title: Hilbert's Hotel
claim:
Let $\mathcal{C}$ be a category with countable powers. Then for every object $X \in \mathcal{C}$ there is an object $P \in \mathcal{C}$ with $P \cong X \times P$ and which has a morphism $X \to P$.

proof:
Take $P := X^{\mathbb{N}}$. Since $\mathbb{N} \cong 1 + \mathbb{N}$ as sets, we have $P \cong X \times P$. The diagonal provides a morphism $X \to P$.


id: preadditive_structure_unique
title: Uniqueness of preadditive structures
claim:
Let $\mathcal{C}$ be a preadditive category (or more generally, a category enriched in commutative monoids) with finite products and finite coproducts. Then for all objects $X,Y$ the canonical morphism
    $$\alpha : X \oplus Y \to X \times Y$$
    is an isomorphism. Moreover, the preadditive structure is <i>unique</i>: If $f,g : A \rightrightarrows B$ are morphisms, their sum
    $$f+g : A \to B$$
    is the composite of $(f,g) : A \to B \times B$, the inverse $\alpha^{-1} : B \oplus B \to B \times B$, and the codiagonal $\nabla : B \oplus B \to B$.

proof:
The morphism $\alpha : X \oplus Y \to X \times Y$ is defined by the equations
    $$p_1 \circ \alpha \circ i_1 = \mathrm{id}_X, \quad p_2 \circ \alpha \circ i_2 = \mathrm{id}_Y,$$
    $$p_2 \circ \alpha \circ i_1 = 0,\quad p_1 \circ \alpha \circ i_2 = 0.$$
    It does not depend on the choice of preadditive structure since zero morphisms are unique. It is an isomorphism: Define
    $$\beta := i_1 \circ p_1  + i_2 \circ p_2 : X \times Y \to X \oplus Y.$$
    Then $\alpha \circ \beta = \mathrm{id}_{X \times Y}$ because
    $$p_1 \circ \alpha \circ \beta = p_1 \circ \alpha \circ i_1 \circ p_1 + p_1 \circ \alpha \circ i_2 \circ p_2 = \mathrm{id}_1 \circ p_1 + 0 \circ p_2 = p_1$$
    and likewise $p_2 \circ \alpha \circ \beta = p_2$. We also have $\beta \circ \alpha = \mathrm{id}_{X \oplus Y}$ with a very similar calculation that shows $\beta \circ \alpha \circ i_1 = i_1$ and $\beta \circ \alpha \circ i_2 = i_2$. Therefore, for morphisms $f,g : A \rightrightarrows B$ the composite $A \to B$ in the claim is equal to
    $$\begin{align*}
    \nabla \circ \beta \circ (f,g) & = \nabla \circ (i_1 \circ p_1  + i_2 \circ p_2) \circ (f,g) \\
    & = \nabla \circ i_1 \circ p_1 \circ (f,g) + \nabla \circ i_2 \circ p_2 \circ (f,g) \\
    & = p_1 \circ (f,g) + p_2 \circ (f,g) \\
    & = f + g.
    \end{align*}$$


id: generator_construction
title: Construction of Generators
claim:
In a category let $S$ be a generating set which is <a href="/category-property/strongly_connected">strongly connected</a> (between any two objects in $S$ there is a morphism). If the coproduct $U := \coprod_{G \in S} G$ exists, then it is a generator.

proof:
This is a straight forward generalization of <a href="/category-implication/generator_via_coproduct">this result</a>. We remark that the assumption about $S$ implies that each inclusion $G \to U$ has a left inverse. Now let $f,g : A \rightrightarrows B$ be two morphisms with $f h = g h$ for all $h : U \to A$. If $G \in S$, any morphism $G \to A$ extends to $U$ by our preliminary remark. Thus, $fh = gh$ holds for all $h : G \to A$ and $G \in S$. Since $S$ is a generating set, this implies $f = g$.


id: limits_in_factor_categories
title: Limits in factor categories
claim:
Let $\mathcal{C},\mathcal{D}$ be two categories. Assume that $\mathcal{D}$ is inhabited. If $\mathcal{C} \times \mathcal{D}$ has limits of a given shape, then $\mathcal{C}$ also has limits of this shape.

proof:
Let $U \in \mathcal{D}$ be a fixed object. Assume that $X : \mathcal{J} \to \mathcal{C}$ is a diagram. Assume that the diagram $(X,U) : \mathcal{J} \to \mathcal{C} \times \mathcal{D}$ (which is constant in the second variable) has a limit cone $((p_i, u_i) : (L,V) \to (X_i,U))$, thus consisting of morphisms $p_i : L \to X_i$ and $u_i : V \to U$. Clearly, $(p_i : L \to X_i)$ is a cone over $X$. We prove that it is universal:<br>
    Let $(f_i : T \to X_i)$ be a cone over $X$. Then $((f_i,\mathrm{id}_U) : (T,U) \to (X_i,U))$ is a cone over $(X,U)$. Hence, there is a unique morphism $(f,v) : (T,U) \to (L,V)$, consisting of $f : T \to L$ and $v : U \to V$, such that $(p_i,u_i) \circ (f,v) = (f_i,\mathrm{id}_U)$, i.e. $p_i \circ f = f_i$ and $u_i \circ v = \mathrm{id}_U$. If $g : T \to L$ is another morphism with $p_i \circ g = f_i$, then uniqueness applied to $(g,v)$ shows that $f = g$.


id: cogenerators_in_product_categories
title: Cogenerators in product categories
claim:
For a family of categories $(\mathcal{C}_i)_{i \in I}$, each having a cogenerator $Q_i$ which is weakly terminal, the object $(Q_i)_{i \in I}$ is a cogenerator in the product category $\prod_{i \in I} \mathcal{C}_i$.

proof:
Let $(f_i: A_i \to B_i)_{i \in I}$ and $(g_i: A_i \to B_i)_{i \in I}$ be two parallel morphisms in the product category which are coequalized by any morphism $(B_i \to Q_i)_{i \in I}$. Let $i_0 \in I$. We claim that $f_{i_0},g_{i_0} : A_{i_0} \rightrightarrows B_{i_0}$ are coequalized by all morphisms $B_{i_0} \to Q_{i_0}$, and hence are equal: Indeed, for all $i \neq i_0$ we may choose some morphism $B_i \to Q_i$ since $Q_i$ is weakly terminal. Thus, the morphism $B_{i_0} \to Q_{i_0}$ extends to a morphism $(B_i \to Q_i)_{i \in I}$ in the product category. By assumption, it coequalizes $(f_i)_{i \in I}$ and $(g_i)_{i \in I}$. By looking at the $i_0$-component are done.


id: special_sequential_colimits
title: Finite structures usually have no sequential colimits
claim:
Let $\mathcal{C}$ be a category with finite powers, including a terminal object $1$. Let $a : 1 \to X$ be a morphism. Assume that the sequence of morphisms $(X^n,a) : X^n \to X^{n+1}$ for $n \geq 0$ admits a colimit $(i_n : X^n \to C)$. Then for every $m \geq 0$ there is a split epimorphism $C \to X^m$. In particular, if $U : \mathcal{C} \to \mathbf{Set}$ is a functor preserving finite powers and $\mathrm{card}(U(X)) \geq 2$, then $U(C)$ is infinite.

proof:
Let $m \geq 0$ be fixed. For $n \geq 0$ we define a morphism $u_n : X^n \to X^m$ as follows: It is the projection on the first $m$ factors for $m \leq n$, and $(X^n,a^{m-n})$ for $m \geq n$ (for $m=n$ these agree). With generalized elements this says:
    $$u_n(x_1,\dotsc,x_n) = \begin{cases} (x_1,\dotsc,x_m) & m \leq n \\ (x_1,\dotsc,x_n,a,\dotsc,a) & m \geq n \end{cases}$$
    We claim that $u_n = u_{n+1} \circ (X^n,a)$, i.e.
    $$u_n(x_1,\dotsc,x_n) = u_{n+1}(x_1,\dotsc,x_n,a).$$
    If $m \leq n$ (hence, $m \leq n+1$), both sides are equal to $(x_1,\dotsc,x_m)$. If $m > n$, i.e. $m \geq n+1$, both sides are equal to $(x_1,\dotsc,x_n,a,\dotsc,a)$. This proves the claim.
    <br>Hence, there is a unique morphism $\varphi : C \to X^m$ such that $\varphi \circ i_n = u_n$ for all $n \geq 0$. Since $u_m$ is the identity, $\varphi$ is a split epimorphism.
    <br>If $U$ is a functor with the mentioned properties, $U(\varphi)$ is also a split epimorphism from $U(C)$ to $U(X^m) \cong U(X)^m$, and $U(X)^m$ has $\geq 2^m$ elements. This holds for all $m$, so that $U(C)$ is infinite.


id: missing_cogenerating_sets
title: Missing cogenerating sets
claim:
Let $\mathcal{C}$ be a category with a faithful functor $U: \mathcal{C} \to \mathbf{Set}$. Assume there exists a collection of objects $\mathcal{F} \subseteq \operatorname{Ob}(\mathcal{C})$ satisfying the following conditions:
    <ol>
        <li>For any $X \in \mathcal{F}$ and any non-terminal $Y \in \mathcal{C}$, for every morphism $f: X \to Y$ its underlying map $U(f) : U(X) \to U(Y)$ is injective.</li>
        <li>For every infinite cardinal number $\kappa$, there exists an object $X \in \mathcal{F}$ such that $\mathrm{card}(U(X)) \geq \kappa$ and such that $X$ has a non-identity endomorphism.</li>
    </ol>
    Then $\mathcal{C}$ does not have a cogenerating set.

proof:
Assume that there is a cogenerating set $S$. By assumption (2) there is an object $X \in \mathcal{F}$ such that $U(X)$ is larger than all the $U(Y)$ with $Y \in S$ (w.r.t. cardinalities) and which has a non-identity endomorphism $\sigma : X \to X$. Since $S$ cogenerates, there is a morphism $f : X \to Y$ with $Y \in S$ and $f \sigma \neq f$. For this, $Y$ must be non-terminal. By (1) the map $U(f) : U(X) \to U(Y)$ is injective. This is a contradiction.


id: missing_cogenerator
title: Missing cogenerator
claim:
Let $\mathcal{C}$ be a pointed category with a faithful functor $U: \mathcal{C} \to \mathbf{Set}$. Assume there exists a collection of non-zero objects $\mathcal{F} \subseteq \operatorname{Ob}(\mathcal{C})$ satisfying the following conditions:
    <ol>
        <li>For any $X \in \mathcal{F}$ and any $Y \in \mathcal{C}$, every non-zero morphism $f: X \to Y$ is injective on underlying sets.</li>
        <li>For every $Y \in \mathcal{C}$ there is some object $X \in \mathcal{F}$ such that $\mathrm{card}(U(X)) > \mathrm{card}(U(Y))$.</li>
    </ol>
    Then $\mathcal{C}$ does not have a cogenerator.

proof:
Assume that there is a cogenerator $Y$. By assumption (2) there is an object $X \in \mathcal{F}$ such that $U(X)$ is larger than $U(Y)$ (w.r.t. cardinalities). Since $0,\mathrm{id}_X : X \rightrightarrows X$ are distinct, there is a morphism $f : X \to Y$ with $f \neq 0$. But then $U(f) : U(X) \to U(Y)$ is injective by assumption (1), which contradicts our choice of $X$.


id: subobject_classifiers_coreflection
title: Coreflection of subobject classifiers
claim:
Let $\mathcal{D}$ be a category with a (regular) subobject classifier $\Omega$. Assume that $\mathcal{C} \to \mathcal{D}$ is a full subcategory such that (1) any (regular) $\mathcal{D}$-subobject of an object in $\mathcal{C}$ already lies in $\mathcal{C}$, (2) it is coreflective, i.e. there is a functor $R : \mathcal{D} \to \mathcal{C}$ right adjoint to the inclusion. Then $R(\Omega)$ is a (regular) subobject classifier in $\mathcal{C}$.

proof:
If $X \in \mathcal{C}$, then $\mathrm{Hom}(X,R(\Omega)) \cong \mathrm{Hom}(X,\Omega)$ is isomorphic to the collection of $\mathcal{D}$-subobjects of $X$, which by assumption coincide with the $\mathcal{C}$-subobjects of $X$.


id: exact_filtered_colimits_descend
title: Exact filtered colimits descend to nice subcategories
claim:
Let $G : \mathcal{C} \to \mathcal{D}$ be a fully faithful functor with a left adjoint $F : \mathcal{D} \to \mathcal{C}$ that preserves finite limits. Assume that $\mathcal{D}$ has exact filtered colimits and that $\mathcal{C}$ has finite limits. Then $\mathcal{C}$ has exact filtered colimits as well.

proof:
It is well-known (and easy to prove) that the colimit of a diagram $(X_j)$ in $\mathcal{C}$ is constructed as $F(\mathrm{colim}_j G(X_j))$, provided that colimit in $\mathcal{D}$ exists. In particular, $\mathcal{C}$ has filtered colimits. By assumption, it also has finite limits, and $G$ preserves these since it is a right adjoint. Now let $X : \mathcal{I} \times \mathcal{J} \to \mathcal{C}$ be a diagram, where $\mathcal{I}$ is finite and $\mathcal{J}$ is filtered. We compute:
    $$\begin{align*} \mathrm{colim}_j {\lim}_i X(i,j) & \cong F(\mathrm{colim}_j G({\lim}_i X(i,j))) \\
    & \cong F(\mathrm{colim}_j {\lim}_i G(X(i,j))) \\
    & \cong F({\lim}_i \mathrm{colim}_j G(X(i,j))) \\
    & \cong {\lim}_i F(\mathrm{colim}_j G(X(i,j))) \\
    & \cong {\lim}_i \mathrm{colim}_j X(i,j)
    \end{align*}$$


id: filtered-monos
title: Detection of filtered-colimit-stable monomorphisms
proof: Since $U$ is faithful, it reflects monomorphisms. From here the proof is straight forward.
claim:
Let $\mathcal{C}$ be a category with filtered colimits. Assume that $U : \mathcal{C} \to \mathcal{D}$ is faithful functor which preserves monomorphisms and filtered colimits. If monomorphisms in $\mathcal{D}$ are stable under filtered colimits, then the same is true for $\mathcal{C}$.<br><br>
    For the record, here is the dual statement: Let $\mathcal{C}$ be a category with cofiltered limits. Assume that $U : \mathcal{C} \to \mathcal{D}$ is faithful functor which preserves epimorphisms and cofiltered limits. If epimorphisms in $\mathcal{D}$ are stable under cofiltered limits, then the same is true for $\mathcal{C}$.
    


id: pushouts-of-monos-via-congruence-quotients
title: Construction of a pushout of monomorphisms as a quotient of a congruence
claim:
Let $\mathcal{C}$ be an extensive category with quotients of congruences.  Then $\mathcal{C}$ has pushouts of monomorphisms.

proof:
Let $f : S \hookrightarrow X$, $g : S \hookrightarrow Y$ be monomorphisms.  We construct a congruence on $X+Y$ via the maps $p_1, p_2 : X + Y + S + S \rightrightarrows X+Y$ which act as the identity on $X+Y$, $i_1 \circ f, i_2 \circ g$ on the first copy of $S$, and $i_2 \circ g, i_1 \circ f$ on the second copy of $S$, respectively.  To show that $p_1, p_2$ are jointly monomorphic, and again in proving transitivity of the congruence, we use extensivity to split the domain of the generalized elements of $X+Y+S+S$ so that without loss of generality we may assume each factors through one of the coproduct inclusions.  Now a quotient of the congruence must be a pushout of $f$ and $g$.


id: monic-sequential-colimits-via-congruence-quotients
title: Construction of a colimit of a sequence of monomorphisms as a quotient of a congruence
claim:
Let $\mathcal{C}$ be a countably extensive category with quotients of congruences. Then $\mathcal{C}$ has colimits of sequences of monomorphisms.

proof:
Suppose we have a sequence $X_1 \hookrightarrow X_2 \hookrightarrow \cdots$ with corresponding monomorphisms $f_{m,n} : X_m \hookrightarrow X_n$ for $m \le n$. Define $Y$ to be the coproduct of all $X_n$. Now for each $m\le n$, define $E_{m,n} := X_m$ with two maps $i_m, i_n \circ f_{m,n} : E_{m,n} \rightrightarrows Y$, and similarly for $m \ge n$ define $E_{m,n} := X_n$ with two maps $i_m \circ f_{n,m}, i_n : E_{m,n} \rightrightarrows Y$. Then the coproduct of all $E_{m,n}$, with the induced morphisms to $Y$, forms a congruence. Here to prove the maps are jointly monomorphic, and again in proving transitivity, we use extensivity to split the domain of the generalized elements of $\sum_{m,n=1}^\infty E_{m,n}$ so that without loss of generality we may assume each factors through one of the coproduct inclusions. Now a quotient of this congruence must be a colimit of the sequence.


# category_property_assignments

category_id: 0
property_id: multi-algebraic
reason:
The terminal category $\mathbf{1}$ becomes an FPC-sketch by selecting the unique empty cone and cocone. Then, a $\mathbf{Set}$-valued model of this sketch is a functor $\mathbf{1} \to \mathbf{Set}$ sending the unique object to a terminal and initial object, which never exists. Hence, $\mathbf{0}$ is the category of models of this FPC-sketch.


category_id: 2
property_id: multi-algebraic
reason:
There is an FPC-sketch whose $\mathbf{Set}$-model is precisely a pair $(X,Y)$ of sets such that the coproduct $X+Y$ is a singleton. Any $\mathbf{Set}$-model of such a sketch is isomorphic to either $(\varnothing, 1)$ or $(1, \varnothing)$, hence the category of models is equivalent to $\mathbf{2}$.


category_id: Ab
property_id: abelian
reason:
This is standard, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. VIII.


category_id: Ab
property_id: split abelian
reason:
The short exact sequence $0 \xrightarrow{} \mathbb{Z} \xrightarrow{p} \mathbb{Z} \xrightarrow{} \mathbb{Z}/p \xrightarrow{} 0$ does not split. 


category_id: Ab
property_id: CSP
reason:
The canonical homomorphism $\bigoplus_{n \geq 0} \mathbb{Z} \to \prod_{n \geq 0} \mathbb{Z}$ is not surjective, hence no epimorphism.


category_id: Ab_fg
property_id: essentially countable
reason:
Every finitely generated abelian group is isomorphic to a group of the form $\mathbb{Z}^n / U$, where $n \in \mathbb{N}$ and $U$ is a subgroup of $\mathbb{Z}^n$. Since $\mathbb{Z}^n$ is Noetherian as a $\mathbb{Z}$-module, $U$ is finitely generated, hence the category $\mathbf{Ab}_\mathrm{fg}$ has only countably many objects up to isomorphism. Furthermore, for any objects $A \cong \mathbb{Z}^n / U$ and $B \cong \mathbb{Z}^m / T$, the hom-set $\mathrm{Hom}(A,B)$ is countable. Indeed, precomposition with the quotient map yields an injection $\mathrm{Hom}(A,B) \hookrightarrow \mathrm{Hom}(\mathbb{Z}^n, B) \cong B^n$, and $B^n$ is countable.


category_id: Ab_fg
property_id: abelian
reason:
This follows from the fact for abelian groups and the fact that subgroups of finitely generated abelian groups are also finitely generated.


category_id: Ab_fg
property_id: ℵ₁-accessible
reason:
The inclusion $\mathbf{Ab}_{\mathrm{fg}} \hookrightarrow \mathbf{Ab}$ is closed under $\aleph_1$-filtered colimits by <a href="https://mathoverflow.net/questions/400763/">MO/400763</a>. In particular, $\mathbf{Ab}_{\mathrm{fg}}$ has $\aleph_1$-filtered colimits. Since $\mathbf{Ab}_{\mathrm{fg}}$ is essentially small, there is a set $G$ such that every f.g. abelian group is isomorphic to one in $G$. So trivially it is also a $\aleph_1$-filtered colimit of such objects (take the constant diagram). Finally, every object is $\mathbf{Ab}_{\mathrm{fg}} = \mathbf{Ab}_{\mathrm{fp}}$ is finitely presentable in $\mathbf{Ab}$ and hence also in $\mathbf{Ab}_{\mathrm{fg}}$, a fortiori $\aleph_1$-presentable.


category_id: Ab_fg
property_id: cogenerator
reason:
Let $Q$ be a finitely generated abelian group. By their well-known classification, we have $Q = F \oplus T$ for a free abelian group $F$ and a finite abelian group $T$. Let $p$ be a prime number which does not divide the order of $T$. Then $\hom(\mathbb{Z}/p, Q) = 0$, but $\mathbb{Z}/p \neq 0$. Therefore, $Q$ is no cogenerator.


category_id: Ab_fg
property_id: split abelian
reason:
The short exact sequence $0 \xrightarrow{} \mathbb{Z} \xrightarrow{p} \mathbb{Z} \xrightarrow{} \mathbb{Z}/p \xrightarrow{} 0$ does not split.


category_id: Ab_fg
property_id: countable powers
reason:
If countable powers exist, then by <a href="/lemma/hilberts_hotel">Hilbert's Hotel</a> there is some object $P$ with $P \cong \mathbb{Z} \times P$. Tensoring with $\mathbb{Q}$ yields an isomorphism of finite-dimensional vector spaces $P_{\mathbb{Q}} \cong \mathbb{Q} \times P_{\mathbb{Q}}$, which is impossible: the dimension $d$ of $P_{\mathbb{Q}}$ (i.e. the rank of $P$) would satisfy $d = 1+d$.


category_id: Ab_fg
property_id: countable copowers
reason:
If countable powers exist, then by <a href="/lemma/hilberts_hotel">Hilbert's Hotel</a> there is some object $P$ with $P \cong \mathbb{Z} \oplus P$. Tensoring with $\mathbb{Q}$ yields an isomorphism of finite-dimensional vector spaces $C_{\mathbb{Q}} \cong \mathbb{Q} \oplus C_{\mathbb{Q}}$, which is impossible: the dimension $d$ of $C_{\mathbb{Q}}$ (i.e. the rank of $C$) would satisfy $d = 1+d$.


category_id: Alg(R)
property_id: locally small
reason:
There is a forgetful functor $\mathbf{Alg}(R) \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: Alg(R)
property_id: Malcev
reason:
This follows in the same way as for groups, see also Example 2.2.5 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.


category_id: Alg(R)
property_id: balanced
reason:
Take a prime ideal $P \subseteq R$ and consider the $R$-algebra $A := R/P$ (which is an integral domain). Then the inclusion $A \hookrightarrow Q(A)$ is a counterexample.


category_id: Alg(R)
property_id: cogenerating set
reason:
We apply <a href="/lemma/missing_cogenerating_sets">this lemma</a> to the collection of $R$-algebras which are fields: If $F$ is an $R$-algebra that is also a field and $A$ is a non-trivial $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


category_id: Alg(R)
property_id: codistributive
reason:
If $\sqcup$ denotes the coproduct of $R$-algebras (see <a href="https://math.stackexchange.com/questions/625874" target="_blank">MSE/625874</a> for their description) and $A$ is an $R$-algebra, the canonical morphism $A \sqcup R^2 \to (A \sqcup R)^2 = A^2$ is usually no isomorphism. For example, for $A = R[X]$ the coproduct on the LHS is not commutative, it has the algebra presentation $\langle X,E : E^2=E \rangle$.


category_id: Alg(R)
property_id: semi-strongly connected
reason:
This is because already the full subcategory $\mathrm{CAlg}(R)$ of commutative algebras is not semi-strongly connected, see <a href="/category/CAlg(R)">its category page</a> for details.


category_id: Alg(R)
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{Alg}(R) \to \mathbf{Set}$ and the relation $S \subseteq U^2$ defined by $S(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $R[X]$ and $S$ by $R \langle X,Y \rangle / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.


category_id: Alg(R)
property_id: coregular
reason:
We just need to tweak the proof for $\mathbf{Ring}$. Since $R \neq 0$, there is an infinite field $K$ with a homomorphism $R \to K$. Since $K$ is infinite, we may choose some $\lambda \in K \setminus \{0,1\}$. Let $B := M_2(K)$ and $A := K \times K$. Then $A \to B$, $(x,y) \mapsto \mathrm{diag}(x,y)$ is a regular monomorphism: A direct calculation shows that a matrix is diagonal iff it commutes with $M := \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & \lambda \end{smallmatrix}\bigr)$, so that $A \to B$ is the equalizer of the identity $B \to B$ and the conjugation $B \to B$, $X \mapsto M X M^{-1}$. Consider the homomorphism $A \to K$, $(a,b) \mapsto a$. We claim that $K \to K \sqcup_A B$ is not a monomorphism, because in fact, the pushout $K \sqcup_A B$ is zero: Since $A \to K$ is surjective with kernel $0 \times K$, the pushout is $B/\langle 0 \times K \rangle$, which is $0$ because $B$ is simple (<a href="https://math.stackexchange.com/questions/22629" target="_blank">proof</a>) or via a direct calculation with elementary matrices.


category_id: Alg(R)
property_id: regular quotient object classifier
reason:
We may copy the proof for the <a href="/category/CAlg(R)">category of commutative algebras</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\mathbf{Alg}(R)$ would produce one in $\mathbf{CAlg}(R)$ by <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).


category_id: Alg(R)
property_id: cocartesian cofiltered limits
reason:
Consider the ring $A = R[X]$ and the sequence of rings $B_n = R[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $R[[Y]]$. Every element in the coproduct of rings $R[X] \sqcup R[[Y]]$ has a finite "free product" length. Now consider the elements
	$$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n.$$
	Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.


category_id: Alg(R)
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{CAlg}(R)$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the forgetful functor $\mathbf{CAlg}(R) \to \mathbf{Alg}(R)$. It preserves epimorphisms by <a href="https://math.stackexchange.com/questions/5133488" target="_blank">MSE/5133488</a>.


category_id: BN
property_id: pullbacks
reason:
For natural numbers $n,m$ we need to find a universal pair $p,q$ of natural numbers satisfying $n + p = m + q$. We may assume w.l.o.g. $n \leq m$. Then take $p = m-n$, $q = 0$.


category_id: BN
property_id: locally cartesian closed
reason:
The slice category $B\mathbb{N} / *$ is isomorphic to the poset $(\mathbb{N},\geq)$ (not to $(\mathbb{N},\leq)$). This category is thin and and semi-strongly connected, <a href="/category-implication/sequential_implies_lcc">hence</a> cartesian closed.


category_id: BN
property_id: sequential limits
reason:
Assume that the sequence $\cdots \xrightarrow{1}  \bullet \xrightarrow{1} \bullet \xrightarrow{1} \bullet$ has a limit. This is a (universal) sequence of natural numbers $n_0,n_1,\dotsc$ satisfying $n_i = n_{i+1} + 1$. But then $n_i = n_0 - i$, and in particular $n_{n_0 + 1} = - 1$, a contradiction.


category_id: BOn
property_id: left cancellative
reason:
It is well-known that ordinal addition satisfies $\alpha + \beta = \alpha + \gamma \implies \beta = \gamma$.


category_id: BOn
property_id: well-copowered
reason:
This follows from the description of epimorphisms as finite ordinals, see <a href="https://math.stackexchange.com/questions/5029605" target="_blank">MO/5029605</a>.


category_id: BOn
property_id: pullbacks
reason:
If $\alpha,\beta$ are ordinals, w.l.o.g. $\alpha \leq \beta$, choose $\delta$ with $\beta = \alpha + \delta$. Then the pair $\delta,0$ is a pullback of $\alpha,\beta$: We have $\alpha + \delta = \beta + 0$, and if $\epsilon_1,\,\epsilon_2$ is a pair of ordinals with $\alpha + \epsilon_1 = \beta + \epsilon_2$, we may cancel $\alpha$ to get $\epsilon_1 = \delta + \epsilon_2$, so that the pair factors as $\delta + \epsilon_2,\, 0 + \epsilon_2$.


category_id: BOn
property_id: locally cartesian closed
reason:
The slice category $B\mathbf{On} / *$ is isomorphic to the poset $(\mathbf{On},\geq)$ (not to $(\mathbf{On},\leq)$). This category is thin and and semi-strongly connected, <a href="/category-implication/sequential_implies_lcc">hence</a> cartesian closed.


category_id: BOn
property_id: right cancellative
reason:
Since $1 + \omega = 0 + \omega$, the ordinal $\omega$ is not an epimorphism. In fact, the epimorphisms are exactly the finite ordinals (see below).


category_id: BOn
property_id: balanced
reason:
Every finite ordinal is both a mono- and an epimorphism (see below), but only $0$ is an isomorphism.


category_id: BOn
property_id: binary powers
reason:
Assume that the unique object has a product with itself. This amounts to a pair of ordinals $\pi_1,\pi_2$ such that for every pair of ordinals $\alpha_1,\alpha_2$ there is a unique ordinal $\tau$ with $\pi_i + \tau = \alpha_i$ for $i = 1,2$. Applying this to $\alpha_i = 0$ we get $\pi_i = 0$. Now we get a contradiction for any distinct $\alpha_1,\alpha_2$.


category_id: BOn
property_id: sequential colimits
reason:
Assume that the sequence $\bullet \xrightarrow{1} \bullet \xrightarrow{1} \cdots$ has a colimit. This mounts to a (universal) sequence of ordinals $\alpha_n$ with $\alpha_n = \alpha_{n+1} + 1$. But then $\alpha_{n+1} < \alpha_n$, contradicting the fact that $\alpha_0$ is well-ordered.


category_id: BOn
property_id: pushouts
reason:
Assume that $1,\omega$ have a pushout. This is a (universal) pair of ordinals $\alpha,\beta$ with $\alpha + 1 = \beta + \omega$. But $\beta + \omega$ is a limit ordinal, while $\alpha + 1$ is not.


category_id: BOn
property_id: cofiltered-limit-stable epimorphisms
reason:
The epimorphisms are the finite ordinals (see below), but the limit of the sequential diagram $\cdots \xrightarrow{1} * \xrightarrow{1} *$ is the ordinal $\omega$ by <a href="https://math.stackexchange.com/questions/5129138">MSE/5129138</a>.


category_id: Ban
property_id: locally ℵ₁-presentable
reason:
Example 1.48 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.


category_id: Ban
property_id: cartesian filtered colimits
reason:
If $X$ is a Banach space and $(Y_i)$ is a filtered diagram of Banach spaces, the canonical map $\mathrm{colim}_i (X \times Y_i) \to X \times \mathrm{colim}_i Y_i$ is the completion of the canonical map in the category of normed vector spaces with non-expansive linear maps. Now the claim follows directly from the <a href="/category/Met">category of metric spaces</a> with non-expansive maps.


category_id: Ban
property_id: cocartesian cofiltered limits
reason:
If $X$ is a Banach space and $(Y_i)$ is a cofiltered diagram of Banach spaces, the canonical map $X \oplus \lim_i Y_i \to \lim_i (X \oplus Y_i)$ is an isomorphism: Since the forgetful functor $\mathbf{Ban} \to \mathbf{Vect}$ preserves finite coproducts and all limits, and $\mathbf{Vect}$ has the claimed property (see <a href="/category-implication/biproducts_cartesian_filtered_colimits">here</a>), the canonical map is bijective. It remains to show that it is isometric. For $(x,y) \in X \oplus \lim_i Y_i$ the norm in the domain is $|x| + \sup_i |y_i|$, and the norm in the codomain is $\sup_i (|x| + |y_i|)$, and these clearly agree.


category_id: Ban
property_id: balanced
reason:
The linear map $\mathbb{C} \to \mathbb{C}$, $x \mapsto x/2$ is a counterexample. It is bijective, hence a mono- and epimorphism, but not isometric and therefore no isomorphism.


category_id: Ban
property_id: filtered-colimit-stable monomorphisms
reason:
The proof is similar to <a href="/category/Met">$\mathbf{Met}$</a>. For $n \geq 1$ let $V_n$ be the Banach space with underlying vector space $\mathbb{C}$ and the norm $|x|_n := \frac{1}{n} |x|$. For $n \leq m$ the identity map provides a morphism $V_n \to V_m$, which is clearly a monomorphism (also an epimorphism by the way, but an isomorphism iff $n=m$). Let $V$ be the colimit of all $V_n$ in the category of semi-normed vector spaces. It is constructed as the colimit in the category of vector spaces with the semi-norm $|x| := \inf \{|x|_m : n \leq m \}$ for $x \in V_n$. So clearly, the semi-norm is zero. Hence, the colimit in the category of normed vector spaces is $0$. The colimit in the category of Banach spaces is its completion, also $0$. Thus, the monomorphisms $V_1 \to V_n$ become $V_1 \to 0$ in the colimit.


category_id: Ban
property_id: regular subobject classifier
reason:
If $\Omega$ is a regular subobject classifier, then by the classification of regular monomorphisms, $\mathrm{Hom}(X,\Omega)$ is isomorphic to the set of closed subspaces of $X$ for any Banach space $X$. For $X = \mathbb{C}$ this implies that there are exactly two vectors in $\Omega$ with norm $\leq 1$, which is absurd. (For $\Omega = 0$ there is just one, and for $\Omega \neq 0$ there are infinitely many.)


category_id: Ban
property_id: CSP
reason:
By using the concrete description of products and coproducts, for the constant family $X_n = \mathbb{C}$ the canonical morphism $\coprod_n X_n \to \prod_n X_n$ becomes the canonical inclusion map $\ell^1 \hookrightarrow \ell^\infty$. This is not an epimorphism (i.e., has no dense image) since the closure of the image is precisely $c_0$. So for example, $(1,1,\dotsc)$ is not contained in the closure of the image.


category_id: Ban
property_id: cofiltered-limit-stable epimorphisms
reason:
We show that epimorphisms are not stable under sequential limits. Let $X_n = Y_n = \mathbb{C}$ for all $n \geq 0$. The transition morphism $Y_{n+1} \to Y_n$ is the identity, and the transition morphism $X_{n+1} \to X_n$ is $x \mapsto x/2$. The morphisms $X_n \to Y_n$, $x \mapsto x/2^n$ are compatible with the transitions, and they are surjective, hence epimorphisms. Now we check $\lim_n X_n = 0$: An element $(x_n) \in \lim_n X_n$ is a family of complex numbers satisfying $x_n = x_{n+1}/2$ <i>and</i> $\sup_n |x_n| < \infty$. But then $x_n = 2^n x_0$ and this can only be bounded when $x_0=0$. Hence, $0 = \lim_n X_n \to \lim_n Y_n = \mathbb{C}$ is no epimorphism.


category_id: CAlg(R)
property_id: locally small
reason:
There is a forgetful functor $\mathbf{CAlg(R)} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: CAlg(R)
property_id: balanced
reason:
Take a prime ideal $P \subseteq R$ and consider the commutative $R$-algebra $A := R/P$ (which is an integral domain). Then the inclusion $A \hookrightarrow Q(A)$ is a counterexample.


category_id: CAlg(R)
property_id: cogenerating set
reason:
We apply <a href="/lemma/missing_cogenerating_sets">this lemma</a> to the collection of commutative $R$-algebras which are fields: If $F$ is a commutative $R$-algebra that is also a field and $A$ is a non-trivial commutative $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


category_id: CAlg(R)
property_id: countably codistributive
reason:
The canonical homomorphism $A \otimes_R R^{\mathbb{N}} \to A^{\mathbb{N}}$ is given by $a \otimes (r_n)_n \mapsto (r_n a)_n$ and does not have to be surjective: Since $R \neq 0$, there is a commutative $R$-algebra $K$ which is a field. Now take $A := K[X]$ and consider the sequence $(X^n)_{n} \in A^{\mathbb{N}}$.


category_id: CAlg(R)
property_id: semi-strongly connected
reason:
Choose a maximal ideal $\mathfrak{m}$ of $R$, so $K := R/\mathfrak{m}$ is a field. If $\mathbf{CAlg}(R)$ is semi-strongly connected, then also $\mathbf{CAlg}(K)$ is semi-strongly connected. This has been disproven in <a href="https://math.stackexchange.com/questions/5129689" target="_blank">MSE/5129689</a>.


category_id: CAlg(R)
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{CAlg}(R) \to \mathbf{Set}$ and the relation $S \subseteq U^2$ defined by $S(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $R[X]$ and $S$ by $R[X,Y] / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.


category_id: CAlg(R)
property_id: regular quotient object classifier
reason:
The strategy is similar to the one for $\mathbf{CRing}$: Assume that $P \to R$ is a regular quotient object classifier. If $J$ denotes the kernel of $P \to R$, every ideal $I \subseteq A$ of any commutative $R$-algebra has the form $I = \langle \varphi(J) \rangle$ for a unique homomorphism $\varphi : P \to A$. If $\sigma : A \to A$ is an automorphism with $\sigma(I)=I$, then uniqueness gives us $\sigma \circ \varphi = \varphi$, which means that $\varphi(J)$ lies in $A^{\sigma}$, the fixed algebra of $\sigma$. But then $I$ is generated by elements in $A^{\sigma} \cap I$. If $K$ is a residue field of $R$, this fails for $A = K[X,Y]$, $I = \langle X,Y \rangle$, $\sigma(X)=Y$, $\sigma(Y)=X$. The fixed algebra is the subalgebra of symmetric polynomials, which is $K[X+Y,XY]$. So $\langle X,Y \rangle$ is generated by symmetric polynomials without constant term, which implies $\langle X,Y \rangle \subseteq \langle X+Y,XY \rangle$ in $K[X,Y]$. But reducing an equation like $X = a(X,Y) \cdot (X+Y) + b(X,Y) \cdot (XY)$ modulo $\langle X^2,Y^2,XY \rangle$ yields a contradiction.


category_id: CAlg(R)
property_id: cofiltered-limit-stable epimorphisms
reason:
Let $K$ be a field over $R$. Consider the sequence of projections $\cdots \to K[X]/\langle X^2 \rangle \to K[X]/\langle X \rangle$ and the constant sequence $\cdots \to K[X] \to K[X]$. The surjective homomorphisms $K[X] \to K[X]/\langle X^n \rangle$ induce the inclusion $K[X] \hookrightarrow K[[X]]$ in the limit, where $K[[X]]$ is the algebra of formal power series. It is clearly not surjective, but this is not sufficient, we need to argue that it is not an epimorphism in $\mathbf{CAlg}(R)$, or equivalently, in $\mathbf{CRing}$. For a proof, see <a href="https://math.stackexchange.com/questions/2391187" target="_blank">MSE/2391187</a>.


category_id: CMon
property_id: preadditive
reason:
In categories with finite products and finite coproducts, the preadditive structure <a href="/lemma/preadditive_structure_unique">is unique</a> if it exists. In the case of $\mathbf{CMon}$, this is just the pointwise addition of maps. This is indeed an enrichment of $\mathbf{CMon}$ over itself, but not over $\mathbf{Ab}$, since for example $\mathrm{Hom}(\mathbb{N},\mathbb{N}) \cong \mathbb{N}$ (with respect to addition) is not a group.


category_id: CMon
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{CMon} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by the free monoid on a single generator and $R$ by the free commutative monoid on two generators $x,y$ subject to the relation $xy=x^2$. It is clear that $R$ is reflexive, but not symmetric.


category_id: CMon
property_id: coregular
reason:
We can show this analogously to the case of commutative rings <a href="https://math.stackexchange.com/a/3746890" target="_blank">MSE/3746890</a>. Consider the commutative monoid $\mathbb{N}^2$ and its submonoid $U\coloneqq\{(m,n)\mid m\ge n\}$ with the inclusion $i\colon U\hookrightarrow\mathbb{N}^2$. Then, the pushout of $i$ along itself is $\langle x,y,z : x+y=x+z \rangle$, and the equalizer of the cokernel pair of $i$ is $D\coloneqq\{(m,n)\mid m=0 \implies n=0 \}$. If the category $\mathbf{CMon}$ were coregular, the canonical inclusion $j\colon U\hookrightarrow D$ would have to be an epimorphism. However, it is not: let $I\coloneqq\{0,1\}$ be the two-element commutative monoid with $1+1=1$, and let $u,v\colon D \rightrightarrows I$ be the morphisms defined by $u^{-1}(0)=\{(0,0)\}$ and $v^{-1}(0)=\{(0,0),(1,2)\}$; then we have $u\circ j = v\circ j$.


category_id: CMon
property_id: regular quotient object classifier
reason:
If $P \in \mathbf{CMon}$ is a regular quotient object classifier, this means that every surjective homomorphism of commutative monoids $A \to B$ is the cokernel of a unique homomorphism $P \to A$. But there are many surjective homomorphisms which are no cokernels at all: Consider the Boolean monoid $(\{0,1\},\vee)$ with $1 \vee 1 = 1$ and the surjective homomorphism $f : (\mathbb{N},+) \to (\{0,1\},\vee)$ defined by $f(0)=0$ and $f(n)=1$ for $n \geq 1$. It has trivial kernel, but is no isomorphism, so it cannot be a cokernel.


category_id: CMon
property_id: CSP
reason:
First of all, epimorphisms in $\mathbf{CMon}$ are preserved and reflected by the forgetful functor to $\mathbf{Mon}$ (see below). Furthermore, if $M \to N$ is an epimorphism in $\mathbf{Mon}$ and $M$ is infinite, then $\mathrm{card}(N) \leq \mathrm{card}(M)$ (see <a href="https://mathoverflow.net/questions/510431/" target="_blank">MO/510431</a>). This implies that in $\mathbf{CMon}$ the canonical homomorphism $\bigoplus_{n \geq 0} \mathbb{N} \to \prod_{n \geq 0} \mathbb{N}$ is not an epimorphism because its domain is countable and its codomain is uncountable.


category_id: CRing
property_id: coextensive
reason:
[Sketch] A ring homomorphism $f : A \times B \to R$ yields the idempotent element $e := f(1,0) \in R$, so that $R \cong eR \times (1-e)R$. Then $f$ decomposes into the ring homomorphisms $f_A : A \to eR$, $f_A(a) := f(a,0)$ and $f_B : B \to (1-e)R$, $f_B(b) := f(0,b)$.


category_id: CRing
property_id: cogenerating set
reason:
We apply <a href="/lemma/missing_cogenerating_sets">this lemma</a> to the collection of fields: If $F$ is a field and $R$ is a non-trivial commutative ring, any ring homomorphism $F \to R$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


category_id: CRing
property_id: countably codistributive
reason:
The canonical homomorphism $\mathbb{Q} \otimes \mathbb{Z}^{\mathbb{N}} \to (\mathbb{Q} \otimes \mathbb{Z})^{\mathbb{N}} = \mathbb{Q}^{\mathbb{N}}$ is not an isomorphism: its image consists of those sequences of rational numbers whose denominators can be bounded.


category_id: CRing
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{CRing} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $\mathbb{Z}[X]$ and $R$ by $\mathbb{Z}[X,Y] / \langle XY-X^2 \rangle$. It is clear that $R$ is reflexive, but not symmetric.


category_id: CRing
property_id: regular quotient object classifier
reason:
Assume that $P \to \mathbb{Z}$ is a regular quotient object classifier. If $J$ denotes its kernel, this means that every ideal $I \subseteq A$ of any commutative ring has the form $I = \langle \varphi(J) \rangle$ for a unique homomorphism $\varphi : P \to A$. If $\sigma : A \to A$ is an automorphism with $\sigma(I)=I$, then uniqueness gives us $\sigma \circ \varphi = \varphi$, which means that $\varphi(J)$ lies in $A^{\sigma}$, the fixed ring of $\sigma$. But then $I$ is generated by elements in the fixed ring. This fails for $A = \mathbb{Z}[X]$, $I = \langle X \rangle$, $\sigma(X)=-X$. The fixed ring is $\mathbb{Z}[X^2]$, and if $I$ was generated by elements $f \in \mathbb{Z}[X^2] \cap I$, they would be multiples of $X^2$, but $X$ is not a multiple of $X^2$.


category_id: CRing
property_id: cofiltered-limit-stable epimorphisms
reason:
For a prime $p$ consider the sequence of projections $\cdots \to \mathbb{Z}/p^2 \to \mathbb{Z}/p$ and the constant sequence $\cdots \to \mathbb{Z} \to \mathbb{Z}$. The surjective homomorphisms $\mathbb{Z} \to \mathbb{Z}/p^n$ induce the homomorphism $\mathbb{Z} \to \mathbb{Z}_p$ in the limit, where $\mathbb{Z}_p$ is the ring of $p$-adic integers. It is not surjective since $\mathbb{Z}_p$ is uncountable, but this is not sufficient (at least, for this category): We need to use <a href="https://stacks.math.columbia.edu/tag/04W0" target="_blank">SP/04W0</a> to conclude that it is no epimorphism in $\mathbf{CRing}$.


category_id: Cat
property_id: locally small
reason:
There is a forgetful functor $\mathbf{Cat} \to \mathbf{Set} \times \mathbf{Set}$, $\mathcal{C} \mapsto (\mathrm{Ob}(\mathcal{C}),\mathrm{Mor}(\mathcal{C}))$, and $\mathbf{Set} \times \mathbf{Set}$ is locally small.


category_id: Cat
property_id: cartesian closed
reason:
See p. 98 in <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>.


category_id: Cat
property_id: generator
reason:
The interval category $\{0 \to 1\}$ is a generator: Assume that $F,G : \mathcal{C} \rightrightarrows \mathcal{D}$ are functors that agree when being precomposed with any functor from $\{0 \to 1\}$. This means that $F(f) = G(f)$ for all morphisms $f : X \to Y$ in $\mathcal{C}$. By comparing the domains and applying this to $f = \mathrm{id}_X$, we see that $F(X) = G(X)$ for all objects $X$. And we just saw that $F,G$ also agree on morphisms.


category_id: Cat
property_id: infinitary extensive
reason:
[Sketch] This is straight forward from the fact that $\mathbf{Set}$ is infinitary extensive: A functor $\mathcal{C} \to \coprod_i \mathcal{D}_i$ yields full subcategories $\mathcal{C}_i \subseteq \mathcal{C}$ (the preimages of $\mathcal{D}_i)$ with $\mathcal{C} = \coprod_i \mathcal{C}_i$.


category_id: Cat
property_id: balanced
reason:
Since we know that the <a href="/category/Mon">category of monoids</a> is not balanced, there is a monoid map $M \to N$ which is a monomorphism and an epimorphism which is not an isomorphism. Then $B(M) \to B(N)$ has the corresponding properties.


category_id: Cat
property_id: cogenerating set
reason:
Assume that $S$ is a cogenerating set in $\mathbf{Cat}$. Then one checks that the set of monoids $\{\mathrm{End}(X) : X \in \mathcal{C} \in S\}$ is a cogenerating set in the <a href="/category/Mon">category of monoids</a>, which we know does not exist.


category_id: Cat
property_id: regular
reason:
See Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a>.


category_id: Cat
property_id: co-Malcev
reason:
We can adapt the proof from <a href="/category/Mon">$\mathbf{Mon}$</a> as follows: Consider the functor $U : \mathbf{Cat} \to \mathbf{Set}^+$ sending a category $\mathcal{C}$ to the (large) set $\{(x,u) : x \in \mathrm{Ob}(\mathcal{C}) ,\, u \in \mathrm{End}(x) \}$. It is represented by $B \mathbb{N}$, the one-object category associated to the free monoid in one generator. Consider the relation $R \subseteq U^2$ consisting of those pairs $((x,u),(y,v))$ where $x = y$ and $uv = u^2$. This also representable, namely be the one-object category associated to the monoid with the presentation $\langle u,v : uv = u^2 \rangle$. Clearly, $R$ is reflexive, but not symmetric.


category_id: Cat
property_id: coregular
reason:
We already know that the <a href="/category/Mon">category of monoids</a> is not coregular, in fact there is a regular monomorphism $M \to N$ of monoids and a morphism $M \to K$ such that $K \to K \sqcup_M N$ is not a monomorphism. The delooping functor $B : \mathbf{Mon} \to \mathbf{Cat}$ has a left adjoint (<a href="https://math.stackexchange.com/questions/574745" target="_blank">MSE/574745</a>), hence it preserves regular monomorphisms. It also preserves pushouts (<a href="https://math.stackexchange.com/questions/5130854" target="_blank">MSE/5130854</a>), and it reflects monomorphisms since it is faithful. Therefore, $B(M) \to B(N)$ provides the desired counterexample of a non-stable regular monomorphism of categories.


category_id: Cat
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\mathbf{Set} \to \mathbf{Cat}$ that maps a set to its discrete category.


category_id: Delta
property_id: cogenerator
reason:
The ordered set $[1] = \{0 < 1\}$ is a cogenerator, even for <a href="/category/Pos">the category of posets</a>, see there for a proof.


category_id: Delta
property_id: coequalizers
reason:
Assume that $X \rightrightarrows Y$ are morphisms in $\mathbf{FinOrd} \setminus \{\varnothing\}$. Since <a href="/category/FinOrd">$\mathbf{FinOrd}$ has coequalizers</a>, we have a coequalizer $Y \to Q$. Since $Y$ is non-empty, $Q$ is non-empty as well, and clearly $Y \to Q$ is then also the coequalizer in $\mathbf{FinOrd} \setminus \{\varnothing\}$.


category_id: Delta
property_id: skeletal
reason:
If $f : [n] \to [m]$ is an isomorphism, then $n + 1 = m + 1$ by comparing the cardinalities, hence $n = m$.


category_id: Delta
property_id: core-thin
reason:
The category $\mathbf{FinOrd} \setminus \{\varnothing\}$ is core-thin because already $\mathbf{FinOrd}$ is core-thin (see its <a href="/category/FinOrd">category page</a>).


category_id: Delta
property_id: mono-regular
reason:
The <a href="/category/FinOrd">proof for $\mathbf{FinOrd}$</a> also works for $\mathbf{FinSet} \setminus \{\varnothing\}$.


category_id: Delta
property_id: cosifted
reason:
Let $X,Y \in \Delta$. We may pick $x \in X$, $y \in Y$. Then there is a "point span" $X \xleftarrow{x} [0] \xrightarrow{y} Y$. Every span $X \xleftarrow{f} Z \xrightarrow{g} Y$ is connected to such a point span: Pick $z \in Z$. This defines a morphism of spans:
    $$\begin{CD} X @<{f(z)}<< [0] @>{g(z)}>> Y \\ @| @VV{z}V @| \\ X @<<{f}< Z @>>{g}> Y \end{CD}$$
    It remains to show that all point spans are connected to each other. Assume $x_0,x_1 \in X$ and $y \in Y$, w.l.o.g. $x_0 \leq x_1$. Define the map $f : [1] \to X$ by $f(0) = x_0$, $f(1) = x_1$, and the map $g : [1] \to Y$ by $g(0)=g(1)=y$. They are order-preserving and fit into a zig-zag of spans:
    $$\begin{CD} X @<{x_0}<< [0] @>{y}>> Y \\ @| @V{0}VV @| \\ X @<{f}<< [1] @>{g}>> Y \\ @| @A{1}AA @| \\ X @<{x_1}<< [0] @>{y}>> Y \end{CD}$$
    This shows that the choice of $x \in X$ does not matter, and for $y \in Y$ the proof is the same.


category_id: Delta
property_id: coreflexive equalizers
reason:
The two maps $d^0,d^1 : [0] \rightrightarrows [1]$ have a common left inverse, the unique map $s^0 : [1] \to [0]$, but are not equalized by any morphism.


category_id: Delta
property_id: binary powers
reason:
The <a href="/category/FinOrd">proof for $\mathbf{FinOrd}$</a> also works for $\mathbf{FinSet} \setminus \{\varnothing\}$: The forgetful functor to $\mathbf{Set}$ is representable, hence preserves all limits. Thus, if the power $\{0 < 1\} \times \{0 < 1\}$ exists in $\mathbf{FinOrd} \setminus \{\varnothing\}$, we may assume its underlying set is the cartesian product and the projection morphisms are the usual projection maps. Moreover, these maps are order-preserving. Since the result must be a total order, we have $(0,1) \leq (1,0)$ or $(1,0) \leq (0,1)$. In the first case, apply $p_2$ to get $1 \leq 0$, a contradiction. In the second case, use $p_1$ to get a contradiction.


category_id: Delta
property_id: binary copowers
reason:
The <a href="/category/FinOrd">proof for $\mathbf{FinOrd}$</a> also works for $\mathbf{FinSet} \setminus \{\varnothing\}$: Assume that the copower $1+1$, i.e. the coproduct of two terminal objects exists, denoted $\{x\}$ and $\{y\}$. If $x \leq y$ holds in the coproduct, then the universal property would imply this relation for all pairs of elements in any non-empty finite order, which is absurd. Otherwise, $y \leq x$ holds in the coproduct, which yields the same contradiction.


category_id: Delta
property_id: sequential colimits
reason:
We can just copy the <a href="/category/FinOrd">proof for $\mathbf{FinOrd}$</a> to show that the sequence of inclusions $[0] \hookrightarrow [1] \hookrightarrow [2] \hookrightarrow \cdots$ has no colimit.


category_id: Delta
property_id: sequential limits
reason:
We can just copy the <a href="/category/FinOrd">proof for $\mathbf{FinOrd}$</a> to show that the sequence of truncations $\cdots \twoheadrightarrow [2] \twoheadrightarrow [1] \twoheadrightarrow [0]$ has no limit.


category_id: Delta
property_id: pushouts
reason:
Assume that the two inclusions $\{0 < 1\} \leftarrow \{0\} \rightarrow \{0 < 2\}$ have a pushout in $\mathbf{FinOrd} \setminus \{\varnothing\}$. This would be a universal non-empty finite ordered set $X$ with three elements $0,1,2$ satisfying $0 \leq 1$ and $0 \leq 2$. Assume w.l.o.g. $1 \leq 2$ (the case $2 \leq 1$ is similar). The universal property yields an order-preserving map $X \to \{a < b < c\}$ with $0 \mapsto a$, $1 \mapsto c$, $2 \mapsto b$. But then $c \leq b$, which is a contradiction.


category_id: FI
property_id: generator
reason:
The one-point set is a generator since it represents the forgetful functor $\mathbf{FI} \to \mathbf{Set}$.


category_id: FI
property_id: equalizers
reason:
We construct equalizers just like in $\mathbf{FinSet}$ and observe that the universal property still holds.


category_id: FI
property_id: wide pullbacks
reason:
We construct wide pullbacks just like in $\mathbf{Set}$, i.e., for a w.l.o.g. non-empty family of injective maps $f_i : X_i \to S$ we consider the subset $P \subseteq \prod_{i \in I} X_i$ of those tuples $x$ where $f_i(x_i) = f_j(x_j)$. Each projection $P \to X_i$ is injective, so in particular $P$ is finite, and $P \to X_i$ becomes a morphism in $\mathbf{FI}$. It is easy to check that the universal property still holds in $\mathbf{FI}$.


category_id: FI
property_id: mono-regular
reason:
If $f : X \to Y$ is an injective map of finite sets, it is the equalizer of the two injective maps $i_1,i_2 : Y \rightrightarrows Y \sqcup_X Y$, and $Y \sqcup_X Y$ is finite.


category_id: FI
property_id: semi-strongly connected
reason:
If $X,Y$ are two finite sets, we have $\mathrm{card}(X) \leq \mathrm{card}(Y)$ or $\mathrm{card}(Y) \leq \mathrm{card}(X)$. In the first case there will be an injection $X \to Y$, in the second case there will be an injection $Y \to X$.


category_id: FI
property_id: locally cartesian closed
reason:
IF $X$ is a finite set, the slice category $\mathbf{FI} / X$ is equivalent to the poset of subsets of $X$. This is cartesian closed because $A \cap B \subseteq C$ holds if and only if $B \subseteq (X \setminus A) \cup C$, where $A,B,C \subseteq X$.


category_id: FI
property_id: cofiltered-limit-stable epimorphisms
reason:
This is because every epimorphism is an isomorphism in this category (see below), and isomorphisms are always stable under any type of limit.


category_id: FI
property_id: cogenerator
reason:
Let $Q$ be finite set. When $Q$ is empty, it is clearly no cogenerator. Otherwise, $Q + 1$ has at least two elements, so that there are two different morphisms $1 \rightrightarrows Q + 1$. But there is no morphism $Q + 1 \to Q$ at all. Hence, $Q$ is no cogenerator.


category_id: FI
property_id: binary powers
reason:
Assume that two finite sets $X,Y$ have a product $P$ in this category. Elements of $P$ are the same as maps $1 \to P$, and they are automatically injective. Therefore, $P \cong \hom(1,P) \times \hom(1,X) \times \hom(1,Y) \cong X \times Y$, and the projections must agree as well. But they are usually not injective. In particular, the product $X \times X$ never exists when $X$ has $>1$ elements.


category_id: FI
property_id: sequential colimits
reason:
Let $X_n := \{1,\dotsc,n\}$. Assume the sequence of inclusion maps $X_n \hookrightarrow X_{n+1}$ has a colimit $(f_n : X_n \to X)$ in this category. But $f_n$ must be an injective map, so that $\mathrm{card}(X) \geq n$ for all $n$. Since $X$ is finite, this is a contradiction.


category_id: FI
property_id: core-thin
reason:
Its core is the <a href="/category/B">category of finite sets and bijections</a>, which we know is not thin.


category_id: FS
property_id: cogenerator
reason:
We prove that $\{0,1\}$ is a cogenerator: The surjective maps $X \to \{0,1\}$ correspond to the non-empty proper subsets of $X$. If $a,b \in X$ are elements that have the same image under each surjective map $X \to \{0,1\}$, it therefore means that they lie in the same non-empty proper subsets of $X$. This implies $a=b$: If $X = \{a\}$, this is trivial. Otherwise, use the subset $\{a\}$.


category_id: FS
property_id: wide pushouts
reason:
We construct wide pushouts as in $\mathbf{Set}$ and observe that the universal property still holds when we restrict to surjective maps. If $f_i : S \to X_i$ are surjective maps and $P$ is their wide pushout, then each $X_i \to P$ is surjective, so that in particular $P$ is finite.


category_id: FS
property_id: coequalizers
reason:
We construct coequalizers as in $\mathbf{FinSet}$ (or $\mathbf{Set}$) and observe that the universal property still holds when we restrict to surjective maps.


category_id: FS
property_id: epi-regular
reason:
If $f : X \to Y$ is a surjective map of finite sets, it is the coequalizer of the two projections $p_1, p_2 : X \times_Y X \rightrightarrows X$ in $\mathbf{FinSet}$, but also in $\mathbf{FS}$. Notice that $p_1,p_2$ are surjective. Even though $X \times_Y X$ is not a pullback in $\mathbf{FS}$, we can use this finite set here.


category_id: FS
property_id: filtered-colimit-stable monomorphisms
reason:
This is because every monomorphism is an isomorphism in this category (see below), and isomorphisms are always stable under any type of colimit.


category_id: FS
property_id: connected
reason:
If $f : \varnothing \to X$ is surjective, then $X = \varnothing$, and if $f : X \to \varnothing$ is any map, then also $X = \varnothing$. This shows that $\{ \varnothing \}$ is a connected component in this category. (The other connected component consists of all non-empty finite sets.)


category_id: FS
property_id: generator
reason:
Let $G$ be a finite set. There are at least two morphisms $G + 2 \rightrightarrows 2$, but there is no morphism $G \to G + 2$ at all. Hence, $G$ is not a generator.


category_id: FS
property_id: sequential limits
reason:
Let $X_n := \{1,\dotsc,n\}$. We define the truncation $p_n : X_{n+1} \to X_n$ by extending the identity of $X_n$ with $p_n(n+1) := n$. Assume the sequence of truncations $\cdots \to X_2 \to X_1$ has a limit $(f_n : X \to X_n)$ in this category. But $f_n$ is surjective, so that $\mathrm{card}(X) \geq n$ for all $n$. Since $X$ is finite, this is a contradiction.


category_id: FS
property_id: pullbacks
reason:
The connected component of non-empty sets has a terminal object, $1$, and it suffices to prove that it has no products. Let $X$ be a finite set with more than $1$ element. Assume that the product $P$ of $X$ with itself exists. The diagonal $X \to P$ is a split monomorphism, hence injective, but also surjective, i.e. an isomorphism. In other words, the two projections $P \rightrightarrows X$ are equal. The universal property of $P$ now implies that every two morphisms $Y \rightrightarrows X$ are equal, which is absurd.


category_id: FS
property_id: binary copowers
reason:
Assume that the copower $X := 2+2$ exists. Since we have a surjective map $2 \to X$, the set $X$ has at most $2$ elements. The codiagonal $X \to 2$ shows that $X$ has at least $2$ elements. Thus, $X \cong 2$. For all finite sets $Y$ we get a bijection $\mathrm{Hom}(2,Y) \cong \mathrm{Hom}(2,Y)^2$, in particular the cardinalities are the same. For $Y=2$ this gives the contradiction $2 = 4$.


category_id: FS
property_id: locally cocartesian coclosed
reason:
If $X$ is a finite set, the coslice category $X / \mathbf{FS}$ is thin and in fact equivalent to the lattice of equivalence relations on $X$. If $X$ has $\geq 3$ elements, it is not codistributive* and <a href="/category-implication/dual_distributive_criterion">hence</a> not cocartesian coclosed: For simplicity assume $X = \{a,b,c\}$. The bottom element $\bot$ corresponds to the partition $\{\{a\},\{b\},\{c\}\}$, the top element $\top$ to the partition $\{\{a,b,c\}\}$. Now consider the three equivalence relations $E_1,E_2,E_3$ corresponding to the three partitions
	$$\{\{a,b\},\{c\}\}, \, \{\{a,c\},\{b\}\}, \, \{\{b,c\},\{a\}\}.$$
	Then
	$$E_1 \vee (E_2 \wedge E_3) = E_1 \vee \bot = E_1,$$
	but
	$$(E_1 \vee E_2) \wedge (E_1 \vee E_3) = \top \wedge \top = \top.$$
	*For thin categories, the properties codistributive and distributive <a href="/category-implication/distributive_duality">are equivalent</a>.


category_id: FS
property_id: multi-initial object
reason:
If a multi-initial object exists, then the connected component consisting of non-empty finite sets has an initial object $X$. Then, any non-empty finite set cannot have a cardinality strictly greater than $X$,  which is a contradiction.


category_id: FinAb
property_id: self-dual
reason:
This is a simple special case of Pontryagin duality: The functor $\hom(-,\mathbb{Q}/\mathbb{Z})$ provides the equivalence.


category_id: FinAb
property_id: generator
reason:
If $A,B$ are finite abelian groups whose orders are coprime, then we know that $\hom(A,B)$ is trivial. But a generator would admit a non-trivial homomorphism to any other non-trivial finite abelian group.


category_id: FinAb
property_id: countable powers
reason:
If countable powers exist, then by <a href="/lemma/hilberts_hotel">Hilbert's Hotel</a> there is some object $P$ with $P \cong \mathbb{Z}/2 \times P$. If $P$ has $n$ elements, this means $n = 2n$, i.e. $n = 0$, a contradiction.


category_id: FinGrp
property_id: epi-regular
reason:
This holds since every epimorphism is surjective (see below), and a surjective homomorphism $A \to B$ is the coequalizer of its kernel pair $A \times_B A \rightrightarrows A$.


category_id: FinGrp
property_id: mono-regular
reason:
See Prop. 4.2 at the <a href="https://ncatlab.org/nlab/show/regular+monomorphism#Examples" target="_blank">nLab</a>. The proof also works for finite groups.


category_id: FinGrp
property_id: Malcev
reason:
A direct argument is possible, but this can also be derived from the observation that $\mathbf{FinGrp}$ is the category of group objects in $(\mathbf{FinSet},\times)$ and Example 2.2.16 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.


category_id: FinGrp
property_id: regular
reason:
The category is Malcev and hence finitely complete, and it has all coequalizers. The regular epimorphisms coincide with the surjective group homomorphisms (see below), hence are clearly stable under pullbacks.


category_id: FinGrp
property_id: conormal
reason:
Since epimorphisms are surjective (see below), this is the first isomorphism theorem for finite groups.


category_id: FinGrp
property_id: cogenerator
reason:
We apply <a href="/lemma/missing_cogenerator">this lemma</a> to the collection of finite simple groups: Any non-trivial homomorphism from a finite simple group to a finite group must be injective, and for every $n \in \mathbb{N}$ there is a finite simple group of size $\geq n$ (for example, the alternating group on $n+5$ elements).


category_id: FinGrp
property_id: binary copowers
reason:
Assume that $C_2 \sqcup C_2$ exists. This is a finite group, say of order $N$, with two involutions $u,v$ such that for every finite group $G$ with two involutions $a,b$ there is a unique homomorphism $\varphi : C_2 \sqcup C_2 \to G$ with $\varphi(u)=a$ and $\varphi(v)=2$. In particular, when $G$ is generated by $a,b$, then $\mathrm{ord}(G) \leq N$. But then the dihedral group $G := D_N$ of order $2N$ yields a contradiction.


category_id: FinGrp
property_id: generator
reason:
If $A,B$ are finite groups whose orders are coprime, then we know that $\hom(A,B)$ is trivial. But a generator would admit a non-trivial homomorphism to any other non-trivial finite group.


category_id: FinGrp
property_id: countable powers
reason:
If countable powers exist, then by <a href="/lemma/hilberts_hotel">Hilbert's Hotel</a> there is some object $P$ with $P \cong C_2 \times P$. If $P$ has $n$ elements, this means $n = 2n$, i.e. $n = 0$, a contradiction.


category_id: FinOrd
property_id: generator
reason:
The one-point finite ordered set is a generator since it represents the forgetful functor $\mathbf{FinOrd} \to \mathbf{Set}$.


category_id: FinOrd
property_id: cogenerator
reason:
The ordered set $\{0 < 1\}$ is a cogenerator, even for <a href="/category/Pos">the category of posets</a>, see there for a proof.


category_id: FinOrd
property_id: coequalizers
reason:
It suffices to construct quotients by equivalence relations. Let $\sim$ be an equivalence relation on $X$, where $(X,\leq)$ is a finite ordered set. Since $X$ is finite, by induction we may assume that $\sim$ is generated by a single relation $(a,b)$. If $a=b$, there is nothing to prove. If $a < b$ and $X = \{0,1,\dotsc,n-1\}$ with the usual order, the quotient is $\{0,1,\dotsc,a,b+1,\dotsc,n-1\}$ with the usual order.


category_id: FinOrd
property_id: mono-regular
reason:
Let $i : A \to B$ be a monomorphism of finite ordered sets. If $A$ is empty, then $i$ is clearly regular, so assume it is not. The map $i$ is injective (see below), hence order-reflecting. Define maps $u,v  : B \to A$ by $u(b) := \max \{a \in A : i(a) \leq b \}$ and $v(b) := \min \{a \in A : b \leq i(a) \}$. These are order-preserving and satisfy $u \circ i = v \circ i$, both sides are $\mathrm{id}_A$. Conversely, if $b \in B$ satisfies $u(b) = v(b) =: a$, then $i(a) \leq b$ and $b \leq i(a)$, hence $b = i(a)$. This shows that $i$ is the equalizer of $u,v$.


category_id: FinOrd
property_id: epi-regular
reason:
Let $f : A \to B$ be an epimorphism of finite ordered sets. It is surjective (see below). Define $u,v : B \to A$ by $u(b) := \min(f^{-1}(b))$ and $v(b) := \max(f^{-1}(b))$. One can easily check that $u,v$ are order-preserving maps with $f \circ u = f \circ v$ (both sides are $\mathrm{id}_B$). Let $h : A \to T$ be an order-preserving map with $h \circ u = h \circ v$. Then $h(a)$ only depends on $b := f(a)$: We have $u(b) \leq a \leq v(b)$, hence $h(u(b)) \leq h(a) \leq h(v(b)) = h(u(b))$. Therefore, there is a unique map $\tilde{h} : B \to T$ with $\tilde{h}(f(a)) = h(a)$, and one easily checks that it is order-preserving. This shows that $f$ is the coequalizer of $u,v$.


category_id: FinOrd
property_id: locally small
reason:
There is a forgetful functor $\mathbf{FinOrd} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: FinOrd
property_id: core-thin
reason:
Let $f : \{1 < \cdots < n \} \to \{1 < \cdots < n \}$ be an automorphism. Then $f(i)$ is the smallest element not contained in $\{f(j) : j < i\}$. From this one can deduce $f(i)=i$ by induction.


category_id: FinOrd
property_id: binary powers
reason:
The forgetful functor to $\mathbf{Set}$ is representable, hence preserves all limits. Thus, if the power $\{0 < 1\} \times \{0 < 1\}$ exists in $\mathbf{FinOrd}$, we may assume its underlying set is the cartesian product and the projection morphisms are the usual projection maps. Moreover, these maps are order-preserving. Since the result must be a total order, we have $(0,1) \leq (1,0)$ or $(1,0) \leq (0,1)$. In the first case, apply $p_2$ to get $1 \leq 0$, a contradiction. In the second case, use $p_1$ to get a contradiction.


category_id: FinOrd
property_id: binary copowers
reason:
Assume that the copower $1+1$, i.e. the coproduct of two terminal objects exists, denoted $\{x\}$ and $\{y\}$. If $x \leq y$ holds in the coproduct, then the universal property would imply this relation for all pairs of elements in any finite order, which is absurd. Otherwise, $y \leq x$ holds in the coproduct, which yields the same contradiction.


category_id: FinOrd
property_id: sequential limits
reason:
Consider the (non-empty) ordered set $[n] := \{0 < \cdots < n\}$ for $n \in \mathbb{N}$. The forgetful functor to $\mathbf{Set}$ is representable, hence preserves all limits. Thus, if the diagram of truncation maps $\cdots \twoheadrightarrow [2] \twoheadrightarrow [1] \twoheadrightarrow [0]$ has a limit in $\mathbf{FinOrd}$, its underlying set is isomorphic to the limit taken in $\mathbf{Set}$, which is $\mathbb{N} \cup \{\infty\}$. But this is not a finite set.


category_id: FinOrd
property_id: sequential colimits
reason:
Consider the (non-empty) ordered set $[n] := \{0 < \cdots < n\}$ for $n \in \mathbb{N}$. Assume the sequence of inclusion maps $[0] \hookrightarrow [1] \hookrightarrow [2] \hookrightarrow \cdots$ has a colimit $(f_n : [n] \to X)$ in $\mathbf{FinOrd}$. Let $n_0 \geq 0$ be fixed. I claim that $f_{n_0}$ is injective, which will then yield a contradiction by taking $n_0 \geq \mathrm{card}(X)$. For $n \geq 0$ define $g_n : [n] \to [n_0]$ as follows. For $n \leq n_0$ it is the inclusion, and for $n \geq n_0$ it is the surjection which keeps all elements of $[n_0]$ and maps all other elements to $n_0$. Observe that $g_n$ preserves the order and $g_{n+1} |_{[n]} = g_n$. Hence, there is a unique order-preserving map $g : X \to [n_0]$ with $g \circ f_n = g_n$ for all $n$. For $n = n_0$ this shows $g \circ f_{n_0} = \mathrm{id}_{[n_0]}$, and $f_{n_0}$ is injective.


category_id: FinSet
property_id: locally small
reason:
There is a forgetful functor $\mathbf{FinSet} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: FinSet
property_id: generator
reason:
The one-point set is a generator since it represents the forgetful functor $\mathbf{FinSet} \to \mathbf{Set}$.


category_id: FinSet
property_id: ℵ₁-accessible
reason:
The inclusion $\mathbf{FinSet} \hookrightarrow \mathbf{Set}$ is closed under ℵ₁-filtered colimits, that is, any ℵ₁-filtered colimit of finite sets is again finite. Since every finite set is ℵ₁-presentable in $\mathbf{Set}$, it is still ℵ₁-presentable in $\mathbf{FinSet}$. Therefore, $\mathbf{FinSet}$ is ℵ₁-accessible, where every object is ℵ₁-presentable.


category_id: FinSet
property_id: countable copowers
reason:
If countable copowers exist, then by <a href="/lemma/hilberts_hotel">Hilbert's Hotel</a> there is some object $C$ with $C \cong 1 \sqcup C$. If $C$ has $m \in \mathbb{N}$ elements, this implies the contradiction $m = 1 + m$.


category_id: FinSet
property_id: countable powers
reason:
If countable powers exist, then by <a href="/lemma/hilberts_hotel">Hilbert's Hotel</a> there is an object $P$ with $P \cong \{0,1\} \times P$ and a morphism $\{0,1\} \to P$. If $P$ has $m \in \mathbb{N}$ elements, this implies $m = 2m$ and hence $m = 0$, i.e. $P = \varnothing$. But then there cannot be a map $\{0,1\} \to P$.


category_id: FinSet
property_id: natural numbers object
reason:
If $(N,z,s)$ is a natural numbers object, then
	$$1 \xrightarrow{z} N \xleftarrow{s} N$$
	is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. But there is no finite set $N$ with $N \cong 1 + N$.


category_id: Fld
property_id: well-copowered
reason:
Epimorphisms are the purely inseparable field extensions. If $K \to L$ is purely inseparable, then for all $x \in L$ there is some $n \in \mathbb{N}$ with $x^n \in L$. An element of $K$ has at most $n$ $n$th-roots. So we can bound the size of $L$.


category_id: Fld
property_id: multi-algebraic
reason:
See Eg. 4.3(1) in <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01]</a>.


category_id: Fld
property_id: connected
reason:
A field of characteristic $0$ cannot be connected with a field of characteristic $p > 0$. in fact, the connected components of $\mathbf{Fld}$ are the subcategories $\mathbf{Fld}_p$ of fields of characteristic $p$, where $p$ is a prime or $0$.


category_id: Fld
property_id: pushouts
reason:
By <a href="https://math.stackexchange.com/questions/359352/" target="_blank">MSE/359352</a>, the pushout $E \sqcup_K F$ of two field homomorphisms $E \leftarrow K \rightarrow F$ exists if and only if the tensor product $E \otimes_K F$ has a "fieldification": this means that the nilradical is a prime ideal whose quotient ring is a field. This is quite rare: Consider $E = K(X)$, $F = K(Y)$. Then $E \otimes_K F$ is isomorphic to $K[X,Y] (K[X]-\{0\})^{-1} (K[Y]-\{0\})^{-1}$, which is an integral domain but not a field: for example, $X-Y$ has no inverse.


category_id: Fld
property_id: balanced
reason:
Every non-trivial purely inseparable field extension, such as $\mathbb{F}_p(X^p) \to \mathbb{F}_p(X)$, provides a counterexample by the descriptions of special morphisms below.


category_id: Fld
property_id: generator
reason:
Assume that $G$ is a generator, say of characteristic $p$. Then for all $q \neq p$ all homomorphisms between two fields of characteristic $q$ would be equal, which is absurd.


category_id: Fld
property_id: cogenerating set
reason:
We apply <a href="/lemma/missing_cogenerating_sets">this lemma</a> to the collection of fields: Any homomorphism of fields is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


category_id: Fld
property_id: locally cartesian closed
reason:
Assume that $K$ is a field such that $\mathbf{Fld} / K$ is cartesian closed. This slice category is equivalent to the poset of subfields of $K$. This poset is a lattice, and our assumption implies that it is distributive (see <a href="/category-implication/distributive_criterion">here</a>). But this is quite rare: Consider $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. By Galois theory, the lattice of subfields is isomorphic to the diamond lattice $M_3$ which is not distributive. Specifically, $(\mathbb{Q}(\sqrt{2}) \wedge \mathbb{Q}(\sqrt{6})) \vee (\mathbb{Q}(\sqrt{3}) \wedge \mathbb{Q}(\sqrt{6})) = \mathbb{Q} \vee \mathbb{Q} = \mathbb{Q}$, while $(\mathbb{Q}(\sqrt{2}) \vee \mathbb{Q}(\sqrt{3})) \wedge \mathbb{Q}(\sqrt{6}) = \mathbb{Q}(\sqrt{2},\sqrt{3}) \wedge \mathbb{Q}(\sqrt{6}) = \mathbb{Q}(\sqrt{6})$.


category_id: Fld
property_id: core-thin
reason:
If this category was core-thin, Galois theory would not exist. Specifically, the conjugation $\mathbb{C} \to \mathbb{C}$, $z \mapsto \overline{z}$ is a non-trivial automorphism.


category_id: Fld
property_id: binary powers
reason:
Assume that the product $P := \mathbb{Q}(\sqrt{2}) \times \mathbb{Q}(\sqrt{2})$ exists. This field is isomorphic to a subfield of $\mathbb{Q}(\sqrt{2})$, hence $P \cong \mathbb{Q}$ or $P \cong \mathbb{Q}(\sqrt{2})$. In the first case, the two projections $P \rightrightarrows \mathbb{Q}(\sqrt{2})$ must be equal, which means that every two homomorphisms $K \rightrightarrows \mathbb{Q}(\sqrt{2})$ are equal, which is absurd (take $K = \mathbb{Q}(\sqrt{2})$ and its two automorphisms). In the second case, the projections induce for every field $K$ a bijection $\mathrm{Hom}(K,\mathbb{Q}(\sqrt{2})) \cong \mathrm{Hom}(K,\mathbb{Q}(\sqrt{2}))^2$, which however fails for $K = \mathbb{Q}(\sqrt{2})$: the left hand side has $2$ elements, the right hand side has $4$ elements. A more general result about products in $\mathbf{Fld}$ can be found at <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.


category_id: Fld
property_id: multi-terminal object
reason:
Every field has a non-trivial extension, for instance, the rational function field over itself in one variable. Hence, a multi-terminal object never exists.


category_id: Fld
property_id: cofiltered-limit-stable epimorphisms
reason:
Inside of $\mathbb{F}_p(X)$ consider the descending sequence of subfields
	$$\mathbb{F}_p(X) \supseteq \mathbb{F}_p(X^p) \supseteq \mathbb{F}_p(X^{p^2}) \supseteq \cdots,$$
	whose intersection is $\mathbb{F}_p$. Each $\mathbb{F}_p(X^{p^n}) \hookrightarrow \mathbb{F}_p(X)$ is purely inseparable, hence an epimorphism, but in the limit we get $\mathbb{F}_p \hookrightarrow \mathbb{F}_p(X)$, which is not even algebraic.


category_id: FreeAb
property_id: locally small
reason:
There is a forgetful functor $\mathbf{FreeAb} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: FreeAb
property_id: regular
reason:
This follows formally from the fact that $\mathbf{Ab}$ is regular and $\mathbf{FreeAb}$ is closed under subobjects and finite products: By Prop. 2.5 in the <a href="https://ncatlab.org/nlab/show/regular+category">nlab</a> it suffices to prove that there are pullback-stable (reg epi, mono)-factorizations. Every homomorphism $f  : A \to B$ in $\mathbf{FreeAb}$ factors as $f = i \circ p : A \twoheadrightarrow C \hookrightarrow B$, where $C$ is a subgroup, hence free, and $A \to C$ is surjective. Clearly, surjective homomorphisms are pullback-stable. It remains to show that they coincide with the regular epimorphisms.<br>
	(1) If $f : A \to B$ is surjective, it is the coequalizer of $A \times_B A \rightrightarrows A$ in $\mathbf{Ab}$. Since $A \times_B A$ is free abelian, $f$ is also an coequalizer in $\mathbf{FreeAb}$.<br>
	(2) If $f : A \to B$ is a regular epimorphism in $\mathbf{FreeAb}$, consider the factorization $f = i \circ p$ as above. Since $f$ is an extremal epimorphism, $i$ must be an isomorphism, so that $f$ is surjective.


category_id: FreeAb
property_id: countable powers
reason:
Assume that the power $P = \prod_{n \geq 0} \mathbb{Z}$ exists in this category. The projections $P \to \mathbb{Z}$ induce an isomorphism of abelian groups $\mathrm{Hom}(\mathbb{Z}, P) \to \prod_n \mathrm{Hom}(\mathbb{Z}, \mathbb{Z})$, hence $P \cong \mathbb{Z}^{\mathbb{N}}$. But it is <a href="https://groupprops.subwiki.org/wiki/Baer-Specker_group_is_not_free_abelian" target="_blank">well-known</a> that $\mathbb{Z}^{\mathbb{N}}$ is not free abelian.


category_id: Grp
property_id: mono-regular
reason:
See Prop. 4.2 at the <a href="https://ncatlab.org/nlab/show/regular+monomorphism#Examples" target="_blank">nLab</a>.


category_id: Grp
property_id: epi-regular
reason:
This holds since every epimorphism is surjective, and a surjective homomorphism $A \to B$ is the coequalizer of its kernel pair $A \times_B A \rightrightarrows A$.


category_id: Grp
property_id: Malcev
reason:
See Example 2.2.4 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.


category_id: Grp
property_id: cogenerator
reason:
We apply <a href="/lemma/missing_cogenerator">this lemma</a> to the collection of simple groups: Any non-trivial homomorphism from a simple group to a group must be injective, and for every infinite cardinal $\kappa$ there is a simple group of size $\geq \kappa$ (for example, the alternating group on $\kappa$ elements).


category_id: Grp
property_id: coregular
reason:
This is because injective group homomorphisms are not stable under pushouts, see e.g. <a href="https://math.stackexchange.com/questions/601463/" target="_blank">MSE/601463</a> or <a href="https://math.stackexchange.com/questions/5088032" target="_blank">MSE/5088032</a>.


category_id: Grp
property_id: counital
reason:
The canonical morphism $F_2 = \mathbb{Z} \sqcup \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ is not a monomorphism since $F_2$ is not abelian.


category_id: Grp
property_id: regular quotient object classifier
reason:
Assume that $\mathbf{Grp}$ has a (regular) quotient object classifier, i.e. a group $P$ such that every surjective homomorphism $G \to H$ is the cokernel of a unique homomorphism $\varphi : P \to G$. Equivalently, every normal subgroup $N \subseteq G$ is $\langle \langle \varphi(P) \rangle \rangle$ for a unique homomorphism $\varphi : P \to G$, where $\langle \langle - \rangle \rangle$ denotes the normal closure. If $c_g : G \to G$ denotes the conjugation with $g \in G$, then the images of $\varphi$ and $c_g \circ \varphi$ have the same normal closures, so the homomorphisms must be equal. In other words, $\varphi$ factors through the center $Z(G)$. But then every normal subgroup of $G$, in particular $G$ itself, would be contained in $Z(G)$, which is wrong for every non-abelian group $G$.


category_id: Grp
property_id: cocartesian cofiltered limits
reason:
For cofiltered diagrams of groups $(H_i)$ and a group $G$ the canonical homomorphism
	$$\textstyle \alpha : G \sqcup \lim_i H_i \to \lim_i (G \sqcup H_i)$$
	is injective, but often fails to be surjective because the components of an element in the image have bounded <i>free product length</i> (the number of factors appearing in the reduced form). Specifically, consider the free groups $G = \langle y \rangle$ and $H_n = \langle x_1,\dotsc,x_n \rangle$ for $n \in \mathbb{N}$ with the truncation maps $H_{n+1} \to H_n$, $x_{n+1} \mapsto 1$. Define
	$$p_n := x_1 \, y \, x_2 \, y \, \cdots \, x_{n-1} \, y \, x_n \, y^{-(n-1)} \in G \sqcup H_n.$$
	If we substitute $x_{n+1}=1$ in $p_{n+1}$, we get $p_n$. Thus, we have $p = (p_n) \in \lim_n (G \sqcup H_n)$. This element does not lie in the image of $\alpha$ since the free product length of $p_n$ (which is well-defined) is $2n$, which is unbounded.


category_id: Grp
property_id: CSP
reason:
The canonical homomorphism $\coprod_{n \geq 0} \mathbb{Z} \to \prod_{n \geq 0} \mathbb{Z}$ is not surjective because its domain is countable and its codomain is uncountable. Hence it is no epimorphism.


category_id: Grp
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Ab}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the forgetful functor $\mathbf{Ab} \to \mathbf{Grp}$ which indeed preserves epimorphisms.


category_id: Haus
property_id: equalizers
reason:
This follows from the corresponding fact for $\mathbf{Top}$ since subspaces of Hausdorff spaces are again Hausdorff.


category_id: Haus
property_id: products
reason:
This follows from the corresponding fact for $\mathbf{Top}$ since products of Hausdorff spaces are again Hausdorff.


category_id: Haus
property_id: cocomplete
reason:
This follows since $\mathbf{Haus}$ is a reflective subcategory of $\mathbf{Top}$, which is cocomplete. For the reflector, see e.g. the <a href="https://ncatlab.org/nlab/show/Hausdorff+space#HausdorffReflection" target="_blank">nLab</a>. Explicitly, we construct the colimit of Hausdorff spaces by applying the reflector to the colimit of the underlying topological spaces.


category_id: Haus
property_id: infinitary extensive
reason:
This follows exactly as for $\mathbf{Top}$ since Hausdorff spaces are closed under taking subspaces and coproducts in $\mathbf{Top}$.


category_id: Haus
property_id: well-copowered
reason:
Every epimorphism has dense image, so it suffices to prove that if a Hausdorff space $X$ has a dense subset $D \subseteq X$, we can bound the cardinality of $X$ in terms of the cardinality of $D$. For $x \in X$ let $\alpha(x)$ be the set of all $U \cap D$, where $U$ is an open neighborhood of $x$. This defines a map $\alpha : X \to P(P(D))$, and we claim that it is injective: this is because $\{x\}$ is the intersection of all $\overline{U} = \overline{U \cap D}$, where $U$ runs through the open neighborhoods of $x$.


category_id: Haus
property_id: generator
reason:
The one-point space is a generator since it represents the forgetful functor $\mathbf{Haus} \to \mathbf{Set}$.


category_id: Haus
property_id: cartesian filtered colimits
reason:
It is shown in <a href="https://math.stackexchange.com/questions/1255678">MSE/1255678</a> that $\mathbb{Q} \times - : \mathbf{Top} \to \mathbf{Top}$ does not preserve sequential colimits (so that it cannot be a left adjoint). The same example also works in $\mathbf{Haus}$: Surely $\mathbb{Q}$ is Hausdorff, $X_n$ is Hausdorff, as is their colimit $X$, and the colimit (taken in $\mathbf{Top}$) of the $X_n \times \mathbb{Q}$ admits a bijective continuous map to a Hausdorff space, therefore is also Hausdorff, meaning it is also the colimit taken in $\mathbf{Haus}$.


category_id: Haus
property_id: filtered-colimit-stable monomorphisms
reason:
The proof is similar to <a href="/category/Met">$\mathbf{Met}$</a>. For $n \geq 1$ let $X_n$ be the pushout of $[-1/n,+1/n] \hookrightarrow \mathbb{R}$ with itself. That is, $X_n$ is the union of two lines $\mathbb{R} \times \{1\}$ and $\mathbb{R} \times \{2\}$ where we identify $(x,1) \equiv (x,2)$ when $|x| \leq 1/n$. Then $X_n$ is Hausdorff, and there is a canonical surjective continuous map $X_n \to X_{n+1}$. The colimit in $\mathbf{Top}$ is the union of two lines where we identify $(x,1) \equiv (x,2)$ when $|x| \leq 1/n$ for some $n$, i.e. when $x \neq 0$. This is the line with the double origin, which is not Hausdorff. Its Hausdorff reflection is the line $\mathbb{R}$ where all points of both lines are identified, and it provides the colimit in $\mathbf{Haus}$. Now, the injective continuous maps $\{1,2\} \to X_n$, $i \mapsto (0,i)$ (where $\{1,2\}$ is discrete) become the constant map $0 : \{1,2\} \to \mathbb{R}$ in the colimit, which is no monomorphism.


category_id: Haus
property_id: balanced
reason:
The inclusion $\mathbb{Q} \hookrightarrow \mathbb{R}$ is a counterexample; it is an epimorphism since $\mathbb{Q}$ is dense in $\mathbb{R}$.


category_id: Haus
property_id: Malcev
reason:
This is clear since $\mathbf{Set}$ is not Malcev and can be interpreted as the subcategory of discrete spaces (which are Hausdorff).


category_id: Haus
property_id: regular subobject classifier
reason:
Assume that there is a regular subobject classifier $\Omega$. By the classification of regular monomorphisms, we would have an isomorphism between $\mathrm{Hom}(X,\Omega)$ and the set of closed subsets of $X$ for any Hausdorff space $X$. If we take $X = 1$ we see that $\Omega$ has two points. Since $\Omega$ is Hausdorff, $\Omega \cong 1 + 1$ must be discrete. But then $\mathrm{Hom}(X,\Omega)$ is isomorphic to the set of all clopen subsets of $X$, of which there are usually far fewer than closed subsets (consider $X = [0,1]$).


category_id: Haus
property_id: cofiltered-limit-stable epimorphisms
reason:
Our <a href="/category/Set">counterexample for $\mathbf{Set}$</a> (using infinite intersections) also works in $\mathbf{Haus}$ by using discrete topologies. We could also apply a variant of (the dual of) <a href="/lemma/filtered-monos">this lemma</a> to the discrete topology functor $\mathbf{Set} \to \mathbf{Haus}$, which does not preserve all cofiltered limits, but does preserve intersections.


category_id: LRS
property_id: locally small
reason:
For two ringed spaces $(X,O_X), (Y,O_Y)$, the collection of morphisms $X \to Y$ of ringed spaces is the collection $\prod_{f \in \hom(X,Y)} \hom(O_Y,f_* O_X)$, which is a set since $\hom(X,Y)$ is a set and each $\hom(O_Y,f_* O_X)$ is a set.


category_id: LRS
property_id: cocomplete
reason:
See Demazure-Gabriel's "Groupes algébriques", I. §1. 1.6. Specifically, the forgetful functor to ringed spaces preserves colimits, and colimits of ringed spaces are built from colimits of topological spaces and limits of commutative rings, see <a href="https://math.stackexchange.com/questions/1646202" target="_blank">MSE/1646202</a>.


category_id: LRS
property_id: well-powered
reason:
Let $f : X \to Y$ be a monomorphism of locally ringed spaces. I claim that $f$ is injective, from which the claim will follow. The diagonal $\Delta : X \to X \times_Y X$ is an isomorphism. By the explicit construction of fiber products, $X \times_Y X$ consists of triples $(x_1,x_2,\mathfrak{p})$ where $x_1,x_2 \in X$, $y := f(x_1) = f(x_2)$ and $\mathfrak{p}$ is a prime ideal in $k(x_1) \otimes_{k(y)} k(x_2)$. The map $\Delta$ is given by $\Delta(x) = \bigl(x,x,\ker(k(x) \otimes_{k(f(x))} k(x) \to k(x)\bigr)$, and it is bijective. This clearly implies that $f$ is injective (and that each tensor product $k(x) \otimes_{k(f(x))} k(x)$ has a unique prime ideal, namely the kernel mentioned).


category_id: LRS
property_id: well-copowered
reason:
It is enough to prove that every epimorphism of locally ringed spaces is surjective. The forgetful functor $\mathbf{LRS} \to \mathbf{RS}$ has a right adjoint (see <a href="https://arxiv.org/abs/1103.2139" target="_blank">Localization of ringed spaces</a> by W. Gillam), so it preserves epimorphisms. The forgetful functor $\mathbf{RS} \to \mathbf{Top}$ also has a right adjoint, namely $X \mapsto (X,\underline{\mathbb{Z}})$, so it also preserves epimorphisms.


category_id: LRS
property_id: infinitary extensive
reason:
[Sketch] Since $\mathbf{Top}$ is infinitary extensive, a morphism $f : Y \to \coprod_i X_i =: X$ yields a decomposition $Y = \coprod_i Y_i$ (as topological spaces) with continuous maps $f_i : Y_i \to X_i$. Endow the open subset $Y_i \subseteq Y$ with the restricted sheaf. Then each $f_i$ becomes a morphism of locally ringed spaces, and $Y = \coprod_i Y_i$ holds as locally ringed spaces.


category_id: LRS
property_id: balanced
reason:
The canonical morphism $\mathrm{Spec}(\mathbb{Z}/2 \times \mathbb{Z}[1/2]) \longrightarrow \mathrm{Spec}(\mathbb{Z})$ is a mono- and an epimorphism, but no isomorphism.


category_id: LRS
property_id: semi-strongly connected
reason:
This is because already the full subcategory of affine schemes is not semi-strongly connected, see <a href="/category/CRing">the entry for $\mathbf{CRing}$</a>. Specifically, there is no morphism between $\mathrm{Spec}(\mathbb{F}_2)$ and $\mathrm{Spec}(\mathbb{F}_3)$.


category_id: LRS
property_id: co-Malcev
reason:
We can adjust the proof for <a href="/category/Top">$\mathbf{Top}$</a> (see <a href="https://mathoverflow.net/questions/509548" target="_blank">MO/509548</a>) as follows: Let $k$ be a field, $X$ be a singleton and $Y = \{u,v\}$ be the Sierpinski space where $\{u\}$ is open, but $\{v\}$ is not. Endow both with the sheaf of locally constant functions to $k$. Thus, $\mathcal{O}_X(X) = k$, $\mathcal{O}_Y(Y) = \mathcal{O}_Y(\{u\}) = k$. There is a canonical morphism $p : X + X \to Y$. It is a coreflexive corelation that is not cosymmetric.


category_id: M-Set
property_id: locally small
reason:
There is a forgetful functor $M{-}\mathbf{Set} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: M-Set
property_id: Malcev
reason:
Endow the set $\mathbb{N}$ with the trivial $M$-action, and consider the subset $\{(a,b) : a \leq b \}$ of $\mathbb{N}^2$.


category_id: M-Set
property_id: cofiltered-limit-stable epimorphisms
reason:
We know that $\mathbf{Set}$ does not have this property. Now use the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> applied to the functor $\mathbf{Set} \to M{-}\mathbf{Set}$ that equips a set with the trivial $M$-action.


category_id: Man
property_id: finite products
reason:
In short, this follows from the corresponding statement for topological spaces and $\mathbb{R}^n \times \mathbb{R}^m \cong \mathbb{R}^{n+m}$.


category_id: Man
property_id: well-copowered
reason:
This follows from the fact that epimorphisms have dense image (see below), which bounds the size of the codomain.


category_id: Man
property_id: generator
reason:
The $0$-dimensional one-point manifold is a generator since it represents the forgetful functor $\mathbf{Top} \to \mathbf{Set}$.


category_id: Man
property_id: cogenerator
reason:
The manifold $\mathbb{R}$ is a cogenerator, since for every smooth manifold $M$ and points $p \neq q$ in $M$ there is a smooth function $f : M \to \mathbb{R}$ with $f(p) = 1$ and $f(q) = 0$ (John Lee, Introduction to Smooth Manifolds, Prop. 2.25).


category_id: Man
property_id: extensive
reason:
[Sketch] Since $\mathbf{Top}$ is infinitary extensive, a continuous map $f : M \to \coprod_i N_i$ corresponds to a decomposition $M = \coprod_i M_i$ (as topological spaces) with continuous maps $f_i : M_i \to N_i$. Endow the open subset $M_i \subseteq M$ with the smooth structure inherited from $M$. Now remark that $f$ is smooth iff each $f_i$ is smooth.


category_id: Man
property_id: countably distributive
reason:
To construct countable coproducts, take the usual disjoint union of spaces, which is clearly locally Euclidean and Hausdorff, and it is second countable since we are using only countable many spaces. (Without that condition, all coproducts would exist.) Now we need to check that the canonical smooth map $\coprod_i X \times Y_i \to X \times \coprod_i Y_i$ is a diffeomorphism (for countable families). It is a homeomorphism since $\mathbf{Top}$ is infinitary distributive. The inverse $X \times \coprod_i Y_i \to \coprod_i X \times Y_i$ is smooth since the domain is covered by the open subsets $X \times Y_i$ on which the map is clearly smooth.


category_id: Man
property_id: Cauchy complete
reason:
See Theorem 2.1 at the <a href="https://ncatlab.org/nlab/show/smooth+manifold" target="_blank">nLab</a>.


category_id: Man
property_id: essentially small
reason:
This is a consequence of the <a href="https://en.wikipedia.org/wiki/Whitney_embedding_theorem" target="_blank">Whitney embedding theorem</a>. But there is also a more direct proof: Since a manifold is second-countable, it is Lindelöf (<a href="https://topospaces.subwiki.org/wiki/Second-countable_implies_Lindelof" target="_blank">proof</a>). In particular, there is a countable atlas. It is then completely determined by countable many open subsets of Euclidean spaces and the transition maps.


category_id: Man
property_id: coquotients of cocongruences
reason:
Let $p : X + X \twoheadrightarrow E$ be a cocongruence with coreflexivity morphism $r : E \to X$, so that $r \circ p : X + X \to X$ is the codiagonal. Since $p$ is an epimorphism, it has dense image (see below). We first claim that in fact $p$ also has closed image and therefore is surjective.
	Because $r \circ (p \circ i_1) : X \to X$ is the identity, the image of $p \circ i_1$ is the equalizer of $\mathrm{id}_E$ and $(p \circ i_1) \circ r$, hence closed. Likewise, the image of $p \circ i_2$ is closed. Thus, the image of $p$, which is the union of these images, is closed.<br>
	Now, since the pushforward maps of tangent spaces compose to the identity, we see that $p$ must be a local immersion and $r$ must be a submersion.  Also, since the fibers of $r$ have one or two points each, we see that the dimension of $E$ must locally be the same as the dimension of $X$.  This implies that in fact $p$ and $r$ are local diffeomorphisms.  Therefore, the cardinality of the fiber of $r$ is locally constant.  Thus, if $U$ is the subset of $X$ where $r$ has fiber of a single point, with the subspace topology, then $U$ is a clopen submanifold of $X$ which serves as the equalizer of $p \circ i_1$ and $p \circ i_2$.


category_id: Man
property_id: countable powers
reason:
The power $\mathbb{R}^{\mathbb{N}}$ does not exist. More generally, let $(M_n)_{n \geq 0}$ be a sequence of smooth manifolds of positive dimension whose product $(\pi_n : P \to M_n)_{n \geq 0}$ exists. This product cone in $\mathbf{Man}$ yields a product cone in $\mathbf{Set}$ since the forgetful functor $\mathbf{Man} \to \mathbf{Set}$ is representable, hence preserves all limits. Choose points $x_n \in M_n$ with $T_{x_n}(M_n) \neq 0$. Choose the point $x \in P$ with $\pi_n(x) = x_n$. Consider the linear map $T_x(P) \to \prod_{n \geq 0} T_{x_n}(M_n)$ induced by the derivatives $d_x(\pi_n) : T_x(P) \to T_{x_n}(M_n)$. Since $T_x(P)$ is finite-dimensional and $\prod_{n \geq 0} T_{x_n}(M_n)$ is not, it cannot be surjective. But actually, it is: Choose tangent vectors $v_n \in T_{x_n}(M_n)$. Choose smooth curves $\gamma_n : \mathbb{R} \to M_n$ with $\gamma_n(0)=x_n$ and ${\gamma_n}'(0) = v_n$. By the universal property there is a unique smooth curve $\gamma : \mathbb{R} \to P$ with $\pi_n \gamma = \gamma_n$. In particular, $\gamma(0) = x$. The chain rule now implies that $\gamma'(0) \in T_x(P)$ is a preimage of $(v_n)$ – a contradiction.


category_id: Man
property_id: copowers
reason:
If $I$ is uncountable, then the copower $I \otimes \mathbb{R}$ does not exist. More generally, if $(X_i)_{i \in I}$ is a family of non-empty manifolds whose coproduct $M$ exists, then using maps $M \to \{0,1\}$ we find that the $M_i$ map to disjoint connected components of $M$. Since $M$ only has countably many components, $I$ must be countable.
	See also the answer at <a href="https://math.stackexchange.com/questions/5083641">MSE/5083641</a>.


category_id: Man
property_id: pushouts
reason:
The pushout of $\mathbb{R} \leftarrow \{0\} \rightarrow \mathbb{R}$ does not exist, see <a href="https://mathoverflow.net/questions/19116">MO/19116</a>.


category_id: Man
property_id: pullbacks
reason:
See <a href="https://math.stackexchange.com/questions/5129579/" target="_blank">MSE/5129579</a> or <a href="https://math.stackexchange.com/questions/322485" target="_blank">MSE/322485</a>.


category_id: Man
property_id: balanced
reason:
The irrational winding of a torus $\mathbb{R} \to S^1 \times S^1$, $t \mapsto (e^{i t}, \, e^{i \alpha t})$, where $\alpha \in \mathbb{R} \setminus \mathbb{Q}$, provides a counterexample. It is injective and has a dense image. Hence, it is a mono- and an epimorphism. But it is not surjective, hence no isomorphism.


category_id: Man
property_id: sequential colimits
reason:
If $\mathbf{Man}$ had sequential colimits, then by <a href="/lemma/special_sequential_colimits">this lemma</a> there would be a manifold $M$ that admits a split epimorphism $M \to \mathbb{R}^n$ for every $n$. But then $M$ will have an infinite-dimensional tangent space, which is a contradiction.


category_id: Man
property_id: ℵ₁-accessible
reason:
We already know that the <a href="/category/Set_c">category of countable sets</a> $\mathbf{Set}_\mathrm{c}$ does not have $\aleph_1$-filtered colimits. The functor $\pi_0: \mathbf{Man} \to \mathbf{Set}_\mathrm{c}$ is well-defined (because manifolds are second-countable), and it admits a fully faithful right adjoint (regarding a countable set as a discrete manifold). Therefore, $\mathbf{Man}$ does not have $\aleph_1$-filtered colimits.


category_id: Man
property_id: quotients of congruences
reason:
If $\mathbf{Man}$ had quotients of congruences, then by <a href="/lemma/pushouts-of-monos-via-congruence-quotients">this lemma</a>, it would have a pushout of $\mathbb{R} \leftarrow \{ 0 \} \rightarrow \mathbb{R}$.  This contradicts <a href="https://mathoverflow.net/questions/19116">MO/19916</a>.


category_id: Meas
property_id: complete
reason:
Take the limit of the underlying sets and take the smallest $\sigma$-algebra making all projections measurable.


category_id: Meas
property_id: cocomplete
reason:
Take the colimit of the underlying sets and take the largest $\sigma$-algebra making all inclusions measurable. That is, a set is measurable iff its preimage under each inclusion is measurable.


category_id: Meas
property_id: infinitary extensive
reason:
[Sketch] Since $\mathbf{Set}$ is infinitary extensive, a map $f : Y \to \coprod_i X_i =: X$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow the measurable subset $Y_i \subseteq Y$ with the restricted $\sigma$-algebra. If $f$ is measurable, each $f_i$ is measurable, and $Y = \coprod_i Y_i$ holds as measurable spaces.


category_id: Meas
property_id: generator
reason:
The one-point measurable space (with the unique $\sigma$-algebra) is a generator since it represents the forgetful functor $\mathbf{Meas} \to \mathbf{Set}$.


category_id: Meas
property_id: cogenerator
reason:
Take the two-element set $2$ endowed with the trivial $\sigma$-algebra (where only $\varnothing$ and $2$ are measurable), and use that $2$ is a cogenerator for $\mathbf{Set}$.


category_id: Meas
property_id: filtered-colimit-stable monomorphisms
reason:
This follows from <a href="/lemma/filtered-monos">this lemma</a> applied to the forgetful functor to $\mathbf{Set}$.


category_id: Meas
property_id: regular subobject classifier
reason:
The set $\{0,1\}$ with the trivial $\sigma$-algebra is a regular subobject classifier since measurable maps $X \to \{0,1\}$ correspond to subsets of $X$.


category_id: Meas
property_id: balanced
reason:
Take a set $X$ with two different $\sigma$-algebras $\mathcal{A} \subset \mathcal{B}$ (for example, $\mathcal{A} = \{\varnothing,X\}$ and $\mathcal{B} = P(X)$ when $X$ has at least $2$ elements), then the identity map $(X,\mathcal{B}) \to (X,\mathcal{A})$ provides a counterexample.


category_id: Meas
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\mathbf{Set} \to \mathbf{Meas}$ which equips a set with the trivial $\sigma$-algebra.


category_id: Met
property_id: binary products
reason:
The product of two metric spaces $(X,d)$, $(Y,d)$ is $(X \times Y,d)$ with $d((x_1,y_1),(x_2,x_2)) := \sup(d(x_1,x_2),d(y_1,y_2))$.


category_id: Met
property_id: coequalizers
reason:
This is because the <a href="/category/PMet">category of pseudo-metric spaces</a> has coequalizers and $\mathbf{Met} \hookrightarrow \mathbf{PMet}$ has a left adjoint, mapping a pseudo-metric space $X$ to $X /{\sim}$ where $x \sim y \iff d(x,y)=0$. Concretely, we take the coequalizer in the category of pseudo-metric spaces and then identify points with distance zero.


category_id: Met
property_id: directed colimits
reason:
This is because the <a href="/category/PMet">category of pseudo-metric spaces</a> has directed colimits and $\mathbf{Met} \hookrightarrow \mathbf{PMet}$ has a left adjoint, mapping a pseudo-metric space $X$ to $X /{\sim}$ where $x \sim y \iff d(x,y)=0$. Concretely, we take the directed colimit in the category of pseudo-metric spaces and then identify points with distance zero.


category_id: Met
property_id: cartesian filtered colimits
reason:
The canonical map $\mathrm{colim}_i (X \times Y_i) \to X \times \mathrm{colim}_i Y_i$ is an isomorphism for directed diagrams $(Y_i)$: It is surjective by the concrete description of directed colimits. It is isometric because of the elementary observation $\inf_i \max(r, s_i) = \max(r, \inf_i s_i)$ for $r, s_i \in \mathbb{R}$, where $i \leq j \implies s_i \geq s_j$.


category_id: Met
property_id: generator
reason:
The one-point metric space is a generator since it represents the forgetful functor $\mathbf{Met} \to \mathbf{Set}$.


category_id: Met
property_id: cogenerator
reason:
We claim that $\mathbb{R}$ with the usual metric is a cogenerator. Let $a,b \in X$ be two points of a metric space such that $f(a)=f(b)$ for all non-expansive maps $f : X \to \mathbb{R}$. This applies in particular to $f(x) := d(a,x)$ and shows that $0=d(a,a)=d(a,b)$, so that $a=b$.


category_id: Met
property_id: well-copowered
reason:
If $f : X \to Y$ is an epimorphism, then $f(X)$ is dense in $Y$ (see below). Hence, there is an injective map $Y \to X^{\mathbb{N}}$, which bounds the size of $Y$.


category_id: Met
property_id: countable powers
reason:
Assume that the power $P = \mathbb{R}^{\mathbb{N}}$ exists, where $\mathbb{R}$ has the usual metric. Since the forgetful functor $\mathbf{Met} \to \mathbf{Set}$ is representable, it preserves limits, powers in particular. Thus, we may assume that $P$ is the set of sequences of numbers and that the projections $p_n : P \to \mathbb{R}$ are given by $p_n(x) = x_n$. Now consider the sequences $x = (n)_n$ and $y = (0)_n$. Since each $p_n$ is non-expansive, we get $d(x,y) \geq d(p_n(x),p_n(y)) = d(n,0) = n$. But then $d(x,y) = \infty$, a contradiction.


category_id: Met
property_id: binary copowers
reason:
The coproduct of two non-empty metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>. For example, the copower $\mathbb{R} \sqcup \mathbb{R}$ does not exist. We only get coproducts when allowing $\infty$ as a distance, as in $\mathbf{Met}_{\infty}$.


category_id: Met
property_id: balanced
reason:
The inclusion $\mathbb{Q} \hookrightarrow \mathbb{R}$ is a counterexample; it is an epimorphism since $\mathbb{Q}$ is dense in $\mathbb{R}$. Alternatively, consider the identity map $(X,2d) \to (X,d)$ for any non-trivial metric space $(X,d)$.


category_id: Met
property_id: cartesian closed
reason:
This is proven in <a href="https://math.stackexchange.com/questions/5131457" target="_blank">MSE/5131457</a>.


category_id: Met
property_id: filtered-colimit-stable monomorphisms
reason:
The following example is taken from Remark 2.7 in <a href="https://arxiv.org/abs/2006.01399" target="_blank">Approximate injectivity and smallness in metric-enriched categories</a> by Adamek-Rosicky: For $n \geq 1$ let $X_n$ denote the metric space with underlying set $\{0,1\}$ in which $0,1$ have distance $1/n$. We have bijective non-expansive maps $X_n \to X_{n+1}$, $x \mapsto x$. The colimit of this sequence in $\mathbf{PMet}$ is $\{0,1\}$ where $0,1$ have distance $0$, so the colimit in $\mathbf{Met}$ collapses to $\{0\}$. Therefore, the colimit of the monomorphisms $X_0 \to X_n$, $x \mapsto x$ is the non-injective map $X_0 \to \{0\}$.


category_id: Met
property_id: natural numbers object
reason:
If $(N,z,s)$ is a natural numbers object in $\mathbf{Met}$, then
	$$1 \xrightarrow{z} N \xleftarrow{s} N$$
	is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. Since there is a map $1 \to N$, we have $N \neq \varnothing$. However, the coproduct of two non-empty metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>.


category_id: Met_c
property_id: infinitary extensive
reason:
This follows from the existence of coproducts and from the fact that $\mathbf{Top}$ is infinitary extensive.


category_id: Met_c
property_id: generator
reason:
The one-point metric space is a generator since it represents the forgetful functor $\mathbf{Met}_c \to \mathbf{Set}$.


category_id: Met_c
property_id: cogenerator
reason:
The same proof as for $\mathbf{Met}$ shows that $\mathbb{R}$ with the usual metric is a cogenerator.


category_id: Met_c
property_id: countable products
reason:
For finite products, we take the cartesian product with, say, the sup-metric. The product of countably many metric spaces $(X_i,d_i)_{i \geq 0}$ is given by the cartesian product $\prod_{i \geq 0} X_i$ with the metric $d(x,y) := \sum_{i \geq 0} d_i(x_i,y_i)/(1 + d_i(x_i,y_i))$. See Engelking's book <i>General Topology</i>.


category_id: Met_c
property_id: powers
reason:
See <a href="https://math.stackexchange.com/questions/139168" target="_blank">MSE/139168</a> for a proof that uncountable powers do not exist.


category_id: Met_c
property_id: regular subobject classifier
reason:
We recycle the proof from the <a href="/category/Haus">category of Hausdorff spaces</a>: Assume that there is a regular subobject classifier $\Omega$. By the classification of regular monomorphisms, we would have an isomorphism between $\mathrm{Hom}(X,\Omega)$ and the set of closed subsets of $X$ for any metric space $X$. If we take $X = 1$ we see that $\Omega$ has two points. Since $\Omega$ is Hausdorff, $\Omega \cong 1 + 1$ must be discrete. But then $\mathrm{Hom}(X,\Omega)$ is isomorphic to the set of all clopen subsets of $X$, of which there are usually far fewer than closed subsets (consider $X = [0,1]$).


category_id: Met_c
property_id: quotients of congruences
reason:
If $\mathbf{Met}_c$ had quotients of congruences, then by <a href="/lemma/monic-sequential-colimits-via-congruence-quotients">this lemma</a> it would have sequential colimits of sequences of monomorphisms. This contradicts <a href="https://mathoverflow.net/questions/510316" target="_blank">MO/510316</a>.


category_id: Met_oo
property_id: locally small
reason:
There is a forgetful functor $\mathbf{Met}_{\infty} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: Met_oo
property_id: generator
reason:
The singleton metric space $1$ is a generator, since morphisms $1 \to X$ correspond to the elements of $X$.


category_id: Met_oo
property_id: cartesian filtered colimits
reason:
We can use the same proof as for the <a href="/category/Met">category of metric spaces</a> since the equation $\inf_i \max(r, s_i) = \max(r, \inf_i s_i)$ also holds for for $r, s_i \in \mathbb{R} \cup \{\infty\}$.


category_id: Met_oo
property_id: infinitary extensive
reason:
[Sketch] Since $\mathbf{Set}$ is infinitary extensive, a map $f : Y \to \coprod_i X_i =: X$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow $Y_i$ with the restricted metric. If $f$ is non-expansive, each $f_i$ is non-expansive, and for $x_i \in Y_i$ and $i \neq j$ we have $d_Y(x_i,x_j) \geq d_X(f(x_i),f(x_j)) = \infty$, so that $Y = \coprod_i Y_i$ holds as metric spaces.


category_id: Met_oo
property_id: cogenerator
reason:
The proof is similar to $\mathbf{Met}$, a cogenerator is given by $\mathbb{R} \cup \{\infty\}$ with the metric in which $d(a,\infty)=\infty$ for $a \in \mathbb{R}$. Then one checks that the maps $d(a,-) : X \to \mathbb{R} \cup \{\infty\}$ are non-expansive and finishes as for $\mathbf{Met}$.


category_id: Met_oo
property_id: balanced
reason:
The inclusion $\mathbb{Q} \hookrightarrow \mathbb{R}$ provides a counterexample. Alternatively, consider the identity map $(X,2d) \to (X,d)$ for any non-trivial metric space $(X,d)$.


category_id: Met_oo
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{Met}_{\infty} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(X) := \{(a,b) \in U(X)^2 : d(x,y) \leq 1 \}$. Both are representable: $U$ by the singleton metric space and $R$ by the metric space $\{ 0,1 \}$ where $d(0,1) := 1$. It is clear that $R$ is reflexive, but not transitive.


category_id: Met_oo
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\mathbf{Set} \to \mathbf{Met}_{\infty}$ that equips a set with the discrete topology.


category_id: Mon
property_id: unital
reason:
If a submonoid of $X \times Y$ contains $X \times \{1\}$ and $\{1\} \times Y$, then for all $x \in X$ and $y \in Y$ it also contains $(x,1) \cdot (1,y) = (x,y)$.


category_id: Mon
property_id: cogenerator
reason:
We apply <a href="/lemma/missing_cogenerator">this lemma</a> to the collection of simple groups: Any non-trivial homomorphism $G \to M$ from a simple group $G$ to a monoid $M$ must be injective (as it corestricts to a homomorphism of groups $G \to M^{\times}$), and for every infinite cardinal $\kappa$ there is a simple group of size $\geq \kappa$ (for example, the alternating group on $\kappa$ elements).


category_id: Mon
property_id: counital
reason:
The canonical morphism $\mathbb{N} \sqcup \mathbb{N} \to \mathbb{N} \times \mathbb{N}$ is not a monomorphism since $\mathbb{N} \sqcup \mathbb{N}$ is not commutative.


category_id: Mon
property_id: coregular
reason:
Consider the monoid $M := \langle x_0, x_1, s : x_0 s = x_1 s = 1 \rangle$. Notice that every element in $M$ has a unique expression as $s^k \cdot u$ with $k \in \mathbb{N}$ and $u \in \langle x_0,x_1 \rangle_M$. Moreover, the canonical homomorphism $\iota : \langle x_0, x_1 \rangle \to M$ (from the free monoid) is injective. We will prove that it is a regular monomorphism, which however is not pushout-stable. Consider $N := \langle x_0, x_1, s_0, s_1 : x_i s_j = 1 \rangle$ and define $f_i : M \to N$ for $i=0,1$ by $f_i(x_j) = x_j$ and $f_i(s) = s_i$. Then $\iota$ is the equalizer of $f_0,f_1$. Now consider $g : \langle x_0,x_1 \rangle \to \langle y_0 \rangle$ defined by $g(x_0) = y_0$, $g(x_1) = 1$. The pushout of $\iota$ with $g$ is given by $\langle x_0, x_1, s, y_0 : x_0 s = x_1 s = 1 , \, x_0 = y_0, \, x_1 = 1 \rangle$, which simplifies to $\langle x_0, s : x_0 s = s = 1 \rangle$, which is trivial.


category_id: Mon
property_id: regular subobject classifier
reason:
Assume that $\Omega$ is a regular subobject classifier. Since the trivial monoid is a zero object, every regular submonoid $U \subseteq M$ of any monoid $M$ would have the form $\{m \in M : h(m) = 1 \}$ for some homomorphism $M \to \Omega$. Now take any monoid $M$ with zero that has two different homomorphisms with zero $f,g : M \rightrightarrows N$ (for example, let $M = N = \{0\} \cup \{x^n : n \geq 0\}$ be the free monoid with zero on one generator, $f(x) = 0$,and $g(x) = x$). Take their equalizer $U \subseteq M$, and choose a homomorphism $h : M \to \Omega$ with $U = \{m \in M : h(m) = 1\}$. Since $0 \in U$, we have $h(0)=1$. But then for all $m \in M$ we have $h(m) = h(m) h(0) = h(m 0) = h(0) = 1$, i.e. $U = M$, which yields the contradiction $f = g$.


category_id: Mon
property_id: regular quotient object classifier
reason:
We can just copy the proof for the <a href="/category/CMon">category of commutative monoids</a>. Alternatively, we may use <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).


category_id: Mon
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Grp}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the forgetful functor $\mathbf{Grp} \to \mathbf{Mon}$. It preserves epimorphisms since it has a right adjoint, the unit group functor.


category_id: Mon
property_id: cocartesian cofiltered limits
reason:
We know that the <a href="/category/Grp">category of groups</a> fails to satisfy this property. The same counterexample works here since the inclusion $\mathbf{Grp} \hookrightarrow \mathbf{Mon}$ preserves limits and colimits (it has a left and a right adjoint) and is conservative. A similar counterexample is given by the free monoids $N_n = \langle x_1,\dotsc,x_n \rangle$ and the Boolean monoid $M = \langle e : e^2=e \rangle$ with the maps $N_{n+1} \to N_n$, $x_{n+1} \mapsto 1$. Then the element $(x_1 e \cdots x_n e) \in \lim_n (M \sqcup N_n)$ does not come from $M \sqcup \lim_n N_n$ because its components have unbounded free product length.


category_id: Mon
property_id: CSP
reason:
If $M \to N$ is an epimorphism in $\mathbf{Mon}$ and $M$ is infinite, then $\mathrm{card}(N) \leq \mathrm{card}(M)$ (see <a href="https://mathoverflow.net/questions/510431/" target="_blank">MO/510431</a>). This implies that in $\mathbf{Mon}$ the canonical homomorphism $\coprod_{n \geq 0} \mathbb{N} \to \prod_{n \geq 0} \mathbb{N}$ is not an epimorphism because its domain is countable and its codomain is uncountable.


category_id: N_oo
property_id: locally strongly finitely presentable
reason:
Every natural number is strongly finitely presentable, and $\infty$ is the colimit of all $n < \infty$.


category_id: On
property_id: binary products
reason:
For ordinal numbers $\alpha,\beta$ their minimum exists, it is $\alpha$ when $\alpha \leq \beta$ and $\beta$ otherwise.


category_id: On
property_id: well-powered
reason:
This is because for every ordinal $\alpha$ the collection of ordinals $\leq \alpha$ is a set (namely, $\alpha + 1$).


category_id: On
property_id: semi-strongly connected
reason:
It is well-known that for ordinals $\alpha,\beta$ we have $\alpha \leq \beta$ or $\beta \leq \alpha$.


category_id: PMet
property_id: binary products
reason:
The product of two pseudo-metric spaces $(X,d)$, $(Y,d)$ is $(X \times Y,d)$ with $d((x_1,y_1),(x_2,x_2)) := \sup(d(x_1,x_2),d(y_1,y_2))$.


category_id: PMet
property_id: directed colimits
reason:
Given a directed diagram $(X_i)$ of pseudo-metric spaces, take the directed colimit $X$ of the underlying sets with the following pseudo-metric: If $x,y \in X$, let $d(x,y)$ be infimum of all $d(x_i,y_i)$, where $x_i,y_i \in X_i$ are some preimages of $x,y$ in some $X_i$. The definition ensures that each $X_i \to X$ is non-expansive, and the universal property is easy to check.


category_id: PMet
property_id: exact filtered colimits
reason:
Let $\mathcal{I}$ be a finite category and $\mathcal{J}$ be a small filtered category, w.l.o.g. a directed poset. Let $X : \mathcal{I} \times \mathcal{J} \to \mathbf{PMet}$ be a diagram. We need to show that the canonical map $\mathrm{colim}_{j \in \mathcal{J}} \lim_{i \in \mathcal{I}} X(i,j) \to \lim_{i \in \mathcal{I}} \mathrm{colim}_{j \in \mathcal{J}}  X(i,j)$ is an isomorphism. It is bijective since the forgetful functor to $\mathbf{Set}$ preserves finite limits and filtered colimits and since $\mathbf{Set}$ has exact filtered colimits. That the map is isometric can easily be reduced to the following lemma: If $d_{i,j} \in \mathbb{R}_{\geq 0}$ are numbers for $i \in \mathcal{I}$, $j \in \mathcal{J}$ with $j \leq k \implies d_{i,k} \leq d_{i,j}$, then $\inf_j \sup_i d_{i,j} = \sup_i \inf_j d_{i,j}$. This can be proven directly. Alternatively, use that the thin category $(\mathbb{R}_{\geq 0} \cup \{\infty\},\leq)$ is isomorphic to $([0,1],\leq)$, and we already know that <a href="/category/real_interval">it has exact filtered colimits</a>.


category_id: PMet
property_id: generator
reason:
The one-point (pseudo-)metric space is a generator since it represents the forgetful functor $\mathbf{PMet} \to \mathbf{Set}$.


category_id: PMet
property_id: cogenerator
reason:
The set $\{0,1\}$ equipped with the pseudo-metric $d(0,1)=0$ is a cogenerator since every map into is automatically non-expansive and since $\{0,1\}$ is a cogenerator in $\mathbf{Set}$.


category_id: PMet
property_id: balanced
reason:
Let $d : \mathbb{R} \times \mathbb{R} \to \mathbb{R}_{\geq 0}$ be the usual Euclidean metric on $\mathbb{R}$ and $0 : \mathbb{R} \times \mathbb{R} \to \mathbb{R}_{\geq 0}$ be the zero pseudo-metric. Then the identity map $(\mathbb{R},d) \to (\mathbb{R},0)$ provides a counterexample.


category_id: PMet
property_id: countable powers
reason:
Assume that the power $P = \mathbb{R}^{\mathbb{N}}$ exists, where $\mathbb{R}$ has the usual (pseudo-)metric. Since the forgetful functor $\mathbf{PMet} \to \mathbf{Set}$ is representable, it preserves limits, powers in particular. Thus, we may assume that $P$ is the set of sequences of numbers and that the projections $p_n : P \to \mathbb{R}$ are given by $p_n(x) = x_n$. Now consider the sequences $x = (n)_n$ and $y = (0)_n$. Since each $p_n$ is non-expansive, we get $d(x,y) \geq d(p_n(x),p_n(y)) = d(n,0) = n$. But then $d(x,y) = \infty$, a contradiction.


category_id: PMet
property_id: binary copowers
reason:
The coproduct of two non-empty pseudo-metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a> (the proof also works for pseudo-metric spaces). For example, the copower $\mathbb{R} \sqcup \mathbb{R}$ does not exist. We only get coproducts when allowing $\infty$ as a distance.


category_id: PMet
property_id: natural numbers object
reason:
If $(N,z,s)$ is a natural numbers object in $\mathbf{PMet}$, then
	$$1 \xrightarrow{z} N \xleftarrow{s} N$$
	is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. Since there is a map $1 \to N$, we have $N \neq \varnothing$. However, the coproduct of two non-empty pseudo-metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>.


category_id: Pos
property_id: locally finitely presentable
reason:
See <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Example 1.10.


category_id: Pos
property_id: generator
reason:
The singleton poset $1$ is a generator, since morphisms $1 \to P$ correspond to the elements of $P$.


category_id: Pos
property_id: cartesian closed
reason:
For posets $P,Q$ we endow $\hom(P,Q)$ with the partial order in which $f \leq g$ holds iff $f(p) \leq g(p)$ for all $p \in P$. The universal evaluation map is $\hom(P,Q) \times P \to Q$, $(f,p) \mapsto f(p)$, it is order-preserving, and it satisfies the universal property.


category_id: Pos
property_id: cogenerator
reason:
We prove that the poset $\{0 < 1\}$ is a cogenerator: Let $P$ be a poset and $a,b \in P$ be two elements such that $f(a) = f(b)$ for all order-preserving maps $f : P \to \{0 < 1 \}$. This means that $a$ and $b$ lie in the same upper sets. In particular, $b$ lies in the upper set generated by $a$, meaning $a \leq b$, and similarly we deduce $b \leq a$. Thus, $a = b$.


category_id: Pos
property_id: infinitary extensive
reason:
[Sketch] Since $\mathbf{Set}$ is infinitary extensive, a map $f : P \to \coprod_i Q_i$ corresponds to a decomposition $P = \coprod_i P_i$ (as sets) with maps $f_i : P_i \to Q_i$. Endow $P_i$ with the induced order. If $f$ is order-preserving, the elements in different $P_i$ cannot be comparable (since their $f$-images are not comparable), so that $P = \coprod_i P_i$ as posets, and each $f_i$ is order-preserving.


category_id: Pos
property_id: balanced
reason:
The inclusion $\{0,1\} \to \{0 < 1\}$ provides a counterexample (where in the domain there is no relation between $0$ and $1$).


category_id: Pos
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{Pos} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(A) := \{(a,b) \in U(A)^2 : a \leq b\}$. Both are representable: $U$ by the singleton poset and $R$ by $\{0 \leq 1 \}$. It is clear that $R$ is reflexive, but not symmetric.


category_id: Pos
property_id: regular subobject classifier
reason:
Assume that there is a regular subobject classifier $\Omega$, so that (by the classification of regular monomorphisms) $\mathrm{Hom}(P,\Omega)$ is isomorphic to the set of subsets of $P$. By taking $P = \{0\}$ we see that $\Omega$ has $2$ elements, so that either $\Omega \cong \{0,1\}$ (discrete) or $\Omega \cong \{0 < 1\}$. By taking $P = \{0 < 1\}$ we see that $\Omega$ has four pairs $(x,y)$ with $x \leq y$. But $\{0,1\}$ has only two and $\{0 < 1\}$ has only three such pairs.


category_id: Pos
property_id: cofiltered-limit-stable epimorphisms
reason:
Pick any poset $X$ which has a decreasing sequence of non-empty sets $X = X_0 \supseteq X_1 \supseteq \cdots$ with empty intersection, such as $X_n = \mathbb{N}_{\geq n}$ with the natural ordering. The unique map $X_n \to 1$ is surjective, but their limit $\varnothing \to 1$ is not surjective.


category_id: Prost
property_id: locally finitely presentable
reason:
The same proof as for $\mathbf{Pos}$ works, cf. <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Example 1.10.


category_id: Prost
property_id: generator
reason:
The singleton proset $1$ is a generator, since morphisms $1 \to P$ correspond to the elements of $P$.


category_id: Prost
property_id: cartesian closed
reason:
For prosets $P,Q$ we endow $\hom(P,Q)$ with the preorder in which $f \leq g$ holds iff $f(p) \leq g(p)$ for all $p \in P$. The universal evaluation map is $\hom(P,Q) \times P \to Q$, $(f,p) \mapsto f(p)$, it is order-preserving, and it satisfies the universal property.


category_id: Prost
property_id: cogenerator
reason:
Endow the set $\{ 0,1 \}$ with the preorder $0 \leq 1$, $1 \leq 0$ (which is not a partial order). Then every map $P \to \{0,1\}$ is order-preserving. Now the claim follows since the set $\{ 0,1 \}$ is a cogenerator in $\mathbf{Set}$.


category_id: Prost
property_id: infinitary extensive
reason:
[Sketch] Since $\mathbf{Set}$ is infinitary extensive, a map $f : P \to \coprod_i Q_i$ corresponds to a decomposition $P = \coprod_i P_i$ (as sets) with maps $f_i : P_i \to Q_i$. Endow $P_i$ with the induced order. If $f$ is order-preserving, the elements in different $P_i$ cannot be comparable (since their $f$-images are not comparable), so that $P = \coprod_i P_i$ as prosets, and each $f_i$ is order-preserving.


category_id: Prost
property_id: regular subobject classifier
reason:
The set $\{0,1\}$ with the chaotic preorder $(0 \leq 1$, $1 \leq 0)$ is a regular subobject classifier since order-preserving maps $P \to \{0,1\}$ correspond to subsets of $P$.


category_id: Prost
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{Prost} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(A) := \{(a,b) \in U(A)^2 : a \leq b\}$. Both are representable: $U$ by the singleton preordered set and $R$ by $\{0 \leq 1 \}$. It is clear that $R$ is reflexive, but not symmetric.


category_id: Prost
property_id: cofiltered-limit-stable epimorphisms
reason:
We know that $\mathbf{Set}$ does not have this property. Now use the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> applied to the functor $\mathbf{Set} \to \mathbf{Prost}$ that equips a set with the chaotic preorder.


category_id: R-Mod
property_id: locally small
reason:
There is a forgetful functor $R{-}\mathbf{Mod} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: R-Mod
property_id: finitary algebraic
reason:
Take the algebraic theory of an $R$-module (given by the algebraic theory of an abelian group and for each $r \in R$ a unary operation).


category_id: R-Mod
property_id: split abelian
reason:
By assumption, $R$ is not semisimple, so that there is a non-projective $R$-module, which yields a non-split sequence.


category_id: R-Mod
property_id: CSP
reason:
The canonical homomorphism $\bigoplus_{n \geq 0} R \to \prod_{n \geq 0} R$ is not surjective, hence no epimorphism.


category_id: R-Mod_div
property_id: split abelian
reason:
It is a standard fact that the category of $R$-modules is abelian for any ring $R$. If $R$ is a division ring, then by linear algebra every $R$-module has a basis, hence is projective, so that every short exact sequence splits.


category_id: Rel
property_id: self-dual
reason:
There is an isomorphism $\mathbf{Rel} \to \mathbf{Rel}^{\mathrm{op}}$ that is the identity on objects (sets) and maps a relation $R \subseteq X \times Y$ to the opposite relation $R^o \subseteq Y \times X$ defined by $R^o := \{(y,x) : (x,y) \in R \}$.


category_id: Rel
property_id: coproducts
reason:
It is an easy exercise to deduce this from the corresponding fact for sets and that sets form a distributive category.


category_id: Rel
property_id: well-powered
reason:
Any relation $R : A \to B$ induces an injective map $P(A) \to P(B)$, so that in particular there is an injective map $A \to P(B)$.


category_id: Rel
property_id: biproducts
reason:
This is a consequence of the description of coproducts and products, both are disjoint unions (even for infinite families).


category_id: Rel
property_id: kernels
reason:
Let $R : A \to B$ be a relation. Define $K = \bigl\{a \in A : \neg \exists b \in B ((a,b) \in R) \bigr\}$. We have an inclusion map $I : K \to A$, which can also be regarded as relation, and $R \circ I = \empty$ is the empty relation, i.e. the zero morphism. It is easy to check the universal property.


category_id: Rel
property_id: preadditive
reason:
In categories with finite products and finite coproducts, the preadditive structure <a href="/lemma/preadditive_structure_unique">is unique</a> if it exists. In the case of $\mathbf{Rel}$, where both products and coproducts are just disjoint unions, this operation is just the set-theoretic union $f+g = f \cup g$ of relations. This clearly has no inverses.


category_id: Rel
property_id: normal
reason:
The construction of equalizers in $\mathbf{Rel}$ shows that they are injective functions, but <a href="https://math.stackexchange.com/questions/350716" target="_blank">MSE/350716</a> shows that monomorphisms in $\mathbf{Rel}$ don't have to be functions.


category_id: Ring
property_id: disjoint finite products
reason:
To show that $A \sqcup_{A \times B} B$ is trivial, let $R$ be a ring which admits homomorphisms $f : A \to R$, $g : B \to R$ with $f(p_1(a,b))=g(p_2(a,b))$ for all $(a,b) \in A \times B$, i.e. $f(a)=g(b)$. Applying this to $a=0$, $b=1$ yields $1=0$ in $R$.


category_id: Ring
property_id: cogenerating set
reason:
We apply <a href="/lemma/missing_cogenerating_sets">this lemma</a> to the collection of fields: If $F$ is a field and $R$ is a non-trivial ring, any ring homomorphism $F \to R$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


category_id: Ring
property_id: codistributive
reason:
If $\sqcup$ denotes the coproduct of rings (see <a href="https://math.stackexchange.com/questions/625874" target="_blank">MSE/625874</a> for their description) and $R$ is a ring, the canonical morphism $R \sqcup \mathbb{Z}^2 \to (R \sqcup \mathbb{Z})^2 = R^2$ is usually no isomorphism. For example, for $R = \mathbb{Z}[X]$ the coproduct on the LHS is not commutative, it has the ring presentation $\langle X,E : E^2=E \rangle$.


category_id: Ring
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{Ring} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $\mathbb{Z}[X]$ and $S$ by $\mathbb{Z} \langle X,Y \rangle / \langle XY-X^2 \rangle$. It is clear that $R$ is reflexive, but not symmetric.


category_id: Ring
property_id: coregular
reason:
Let $B := M_2(\mathbb{Q})$ and $A := \mathbb{Q}^2$. Then $A \to B$, $(x,y) \mapsto \mathrm{diag}(x,y)$ is a regular monomorphism: A direct calculation shows that a matrix is diagonal iff it commutes with $M := \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 2 \end{smallmatrix}\bigr)$, so that $A \to B$ is the equalizer of the identity $B \to B$ and the conjugation $B \to B$, $X \mapsto M X M^{-1}$. Consider the homomorphism $A \to K$, $(a,b) \mapsto a$. We claim that $K \to K \sqcup_A B$ is not a monomorphism, because in fact, the pushout $K \sqcup_A B$ is zero: Since $A \to K$ is surjective with kernel $0 \times K$, the pushout is $B/\langle 0 \times K \rangle$, which is $0$ because $B$ is simple (<a href="https://math.stackexchange.com/questions/22629" target="_blank">proof</a>) or via a direct calculation with elementary matrices.


category_id: Ring
property_id: regular quotient object classifier
reason:
We may copy the proof for the <a href="/category/CRing">category of commutative rings</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\mathbf{Ring}$ would produce one in $\mathbf{CRing}$ by <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).


category_id: Ring
property_id: cocartesian cofiltered limits
reason:
Consider the ring $A = \mathbb{Z}[X]$ and the sequence of rings $B_n = \mathbb{Z}[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $\mathbb{Z}[[Y]]$. Every element in the coproduct of rings $\mathbb{Z}[X] \sqcup \mathbb{Z}[[Y]]$ has a finite "free product" length. Now consider the elements
	$$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n.$$
	Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.


category_id: Ring
property_id: cofiltered-limit-stable epimorphisms
reason:
We know that $\mathbf{CRing}$ does not have this property. Now use the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> applied to the forgetful functor $\mathbf{CRing} \to \mathbf{Ring}$. It preserves epimorphisms by <a href="https://math.stackexchange.com/questions/5133488" target="_blank">MSE/5133488</a>.


category_id: Rng
property_id: balanced
reason:
The inclusion $\mathbb{Z} \hookrightarrow \mathbb{Q}$ is a counterexample. (The proof can be reduced to the unital case.)


category_id: Rng
property_id: cogenerator
reason:
We apply <a href="/lemma/missing_cogenerator">this lemma</a> to the collection of fields: Any non-zero rng homomorphism from a field to a rng must be injective, and for every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$.


category_id: Rng
property_id: counital
reason:
If $\mathbb{Z}\langle X_1,\dotsc,X_n \rangle_0$ denotes the free rng on $n$ generators (non-commutative polynomials without constant term), then the canonical homomorphism $\mathbb{Z}\langle X,Y \rangle_0 = \mathbb{Z}\langle X \rangle_0 \sqcup \mathbb{Z}\langle Y \rangle_0 \to \mathbb{Z}\langle X \rangle_0 \times \mathbb{Z}\langle Y \rangle_0$ is not a monomorphism since $\mathbb{Z}\langle X,Y \rangle_0$ is not commutative.


category_id: Rng
property_id: regular subobject classifier
reason:
Assume that $\mathbf{Rng}$ has a subobject classifier $\Omega$. Since $0$ is a zero object, every regular subobject $R \subseteq S$ would be the kernel of some homomorphism $S \to \Omega$. In particular, it would be an ideal. Now take any pair of homomorphisms $f,g : S \rightrightarrows T$ in $\mathbf{Ring}$. Their equalizer $R \subseteq S$ is also the equalizer in $\mathbf{Rng}$, and it contains $1 \in S$. If it was an ideal, then $R = S$, and hence $f = g$, which is absurd.


category_id: Rng
property_id: coregular
reason:
We can copy the proof for the <a href="/category/Ring">category of rings</a>. In short, the inclusion of diagonal matrices $\mathbb{Q}^2 \hookrightarrow M_2(\mathbb{Q})$ is a regular monomorphism, but becomes zero after taking the pushout with $p_1 : \mathbb{Q}^2 \twoheadrightarrow \mathbb{Q}$ because $M_2(\mathbb{Q})$ is simple.


category_id: Rng
property_id: regular quotient object classifier
reason:
Assume that $\mathbf{Rng}$ has a regular quotient object classifier $P$. Consider the functor $N : \mathbf{Ab} \to \mathbf{Rng}$ that equips an abelian group with zero multiplication. It is fully faithful and has a left adjoint mapping a rng $R$ to the abelian group $R/R^2$. If $R$ is a rng with zero multiplication and $R \to S$ is a surjective homomorphism, then $S$ has zero multiplication. Therefore, the assumptions of <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized) apply and we conclude that $P/P^2$ is a regular quotient object classifier of $\mathbf{Ab}$. But we already know that <a href="/category/Ab">this category</a> has no such object (in fact, the only additive categories with such an object are trivial by <a href="https://math.stackexchange.com/questions/4086192" target="_blank">MSE/4086192</a>).


category_id: Rng
property_id: cocartesian cofiltered limits
reason:
Consider the ring $A = \mathbb{Z}[X]$ and the sequence of rings $B_n = \mathbb{Z}[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $\mathbb{Z}[[Y]]$ (both in $\mathbf{Ring}$ and $\mathbf{Rng}$). Every element in the coproduct of rngs $\mathbb{Z}[X] \sqcup \mathbb{Z}[[Y]]$ has a finite "free product" length. Now consider the elements
	$$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) - 1 \in A \sqcup B_n.$$
	Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.


category_id: Rng
property_id: CSP
reason:
Assume that $\coprod_n \mathbb{Z} \to \prod_n \mathbb{Z}$ is an epimorphism in $\mathbf{Rng}$. Then $((\coprod_n \mathbb{Z})^+)^{\mathrm{ab}} \to \prod_n \mathbb{Z}$ would be an epimorphism in $\mathbf{CRing}$, where $(-)^+$ denotes the unitalization and $(-)^{\mathrm{ab}}$ the abelianization. But if $R \to S$ is an epimorphism of commutative rings, then $\mathrm{card}(S) \leq \mathrm{card}(R)$ by <a href="https://stacks.math.columbia.edu/tag/04W0" target="_blank">SP/04W0</a>. Since $((\coprod_n \mathbb{Z})^+)^{\mathrm{ab}}$ is countable and $\prod_n \mathbb{Z}$ is not, we get a contradiction.


category_id: Rng
property_id: cofiltered-limit-stable epimorphisms
reason:
We know that $\mathbf{Ring}$ does not have this property. Now use the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> applied to the forgetful functor $\mathbf{Ring} \to \mathbf{Rng}$. We only need to verify that it preserves epimorphisms: Let $f : R \to S$ be an epimorphism in $\mathbf{Ring}$ and let $g,h : S \rightrightarrows T$ be two homomorphisms of rngs with $gf = hf$. The element $e = g(1) = h(1) \in T$ is idempotent, and $g,h$ become homomorphisms of rings $S \rightrightarrows eTe$. Hence, $g=h$.


category_id: Sch
property_id: pullbacks
reason:
This is the well-known construction of the fiber product of schemes, see e.g. EGA I, Chap. I, Thm. 3.2.1.


category_id: Sch
property_id: infinitary extensive
reason:
One uses the same proof as for locally ringed spaces, using that open subspaces of schemes are also schemes.


category_id: Sch
property_id: pushouts
reason:
The span $\mathbb{A}^1 \leftarrow \mathrm{Spec}(k(t)) \rightarrow \mathbb{A}^1$ has no pushout, see <a href="https://mathoverflow.net/questions/9961" target="_blank">MO/9961</a>.


category_id: Sch
property_id: countable powers
reason:
While all diagrams of affine schemes have a limit in the category of schemes, one can show that an infinite product of quasi-compact non-empty schemes only exists when almost all of them are affine, see <a href="https://mathoverflow.net/questions/65506" target="_blank">MO/65506</a>. Thus, for example the power $(\mathbb{P}^1)^{\mathbb{N}}$ does not exist in $\mathbf{Sch}$.


category_id: Sch
property_id: Malcev
reason:
Consider the subscheme $V(x-y) \cup V(y) \subseteq \mathbb{A}^2$. It contains the diagonal but it is not symmetric.


category_id: Sch
property_id: generating set
reason:
If $S$ is a generating set of schemes, then the set of affine open subsets of the schemes in $S$ would also be a generating set. This is then also a generating set in the category of affine schemes, corresponding to a cogenerating set in the <a href="/category/CRing">category of commutative rings</a>, which we know does not exist (see there for the proof).


category_id: Sch
property_id: quotients of congruences
reason:
If $\mathbf{Sch}$ had quotients of congruences, then by <a href="/lemma/pushouts-of-monos-via-congruence-quotients">this lemma</a> it would also have pushouts of monomorphisms, contradicting the fact that the span $\mathbb{A}^1 \leftarrow \mathrm{Spec}(k(t)) \rightarrow \mathbb{A}^1$ has no pushout - see <a href="https://mathoverflow.net/questions/9961" target="_blank">MO/9961</a>.


category_id: Set
property_id: cofiltered-limit-stable epimorphisms
reason:
Pick any decreasing sequence of non-empty sets $X_0 \supseteq X_1 \supseteq \cdots$ with empty intersection, such as $X_n = \mathbb{N}_{\geq n}$. The unique map $X_n \to 1$ is surjective, but their limit $\varnothing \to 1$ is not surjective.


category_id: Set_c
property_id: locally small
reason:
The collection of maps between two (countable) sets $X,Y$ is a subset of $X \times Y$ and therefore a set.


category_id: Set_c
property_id: equalizers
reason:
We can use the same construction as in $\mathbf{Set}$ since subsets of countable sets are again countable.


category_id: Set_c
property_id: countably distributive
reason:
By elementary set theory, a countable (disjoint) union of countable sets is again countable, and a finite direct product of countable sets is countable. Hence, countable coproducts and finite products exists in $\mathbf{Set}_\mathrm{c}$. The distributivity morphism is an isomorphism since this is the case in $\mathbf{Set}$ and the forgetful functor $\mathbf{Set}_\mathrm{c} \to \mathbf{Set}$ preserves finite products and countable coproducts.


category_id: Set_c
property_id: mono-regular
reason:
If $f : X \to Y$ is a monomorphism, i.e. an injective map, it is an equalizer of the maps $\chi_Y, \chi_{f(X)} : Y \to \{0,1\}$ in $\mathbf{Set}$ and hence also in $\mathbf{Set}_\mathrm{c}$.


category_id: Set_c
property_id: epi-regular
reason:
If $X \to Y$ is an epimorphism in $\mathbf{Set}_\mathrm{c}$, i.e. a surjective map, it is coequalizer of the two maps $X \times_Y X \rightrightarrows X$ in $\mathbf{Set}$ and hence also in $\mathbf{Set}_\mathrm{c}$.


category_id: Set_c
property_id: co-Malcev
reason:
For finite colimits we can use the same construction as in $\mathbf{Set}$ since quotients and finite unions of countable sets are again countable. The co-Malcev property now can be easily deduced from the corresponding fact for $\mathbf{Set}$.


category_id: Set_c
property_id: subobject classifier
reason:
This is because $\{0,1\}$ is a subobject classifier in $\mathbf{Set}$, which is countable, and the monomorphisms coincide.


category_id: Set_c
property_id: regular
reason:
From the other properties we know that the category is finitely complete and that it has coequalizers. The regular epimorphisms are stable under pullback since this holds in $\mathbf{Set}$ and both regular epimorphisms (they are surjective maps) and pullbacks coincide.


category_id: Set_c
property_id: coregular
reason:
From the other properties we know that the category is finitely cocomplete and that it has equalizers. The regular monomorphisms are stable under pushout since this holds in $\mathbf{Set}$ and both regular monomorphisms (they are injective maps) and pushouts coincide.


category_id: Set_c
property_id: countable powers
reason:
Since the forgetful functor $\mathbf{Set}_\mathrm{c} \to \mathbf{Set}$ is representable, it preserves (countable) products. Therefore, if the power $\{0,1\}^{\mathbb{N}}$ exists in $\mathbf{Set}_\mathrm{c}$, it must be the ordinary cartesian product, which however is uncountable.


category_id: Set_c
property_id: ℵ₁-accessible
reason:
In fact, $\mathbf{Set}_\mathrm{c}$ does not have $\aleph_1$-filtered colimits: Fix an uncountable set $X$, let $P_\mathrm{c}(X)$ be the poset of countable subsets of $X$, which is $\aleph_1$-filtered, and consider the functor $P_\mathrm{c}(X) \to \mathbf{Set}_\mathrm{c}$ taking a subset $Y \subseteq X$ to $Y$. The colimit of this diagram in $\mathbf{Set}$ is given by $X$ itself, so if $X_c$ were a colimit in $\mathbf{Set}_\mathrm{c}$, then $\mathrm{Hom}(X_c, \{0,1\}) \cong \mathrm{Hom}(X, \{0,1\})$. But the former has cardinality at most $2^{\aleph_0}$ and the latter has cardinality $2^{\mathrm{card}(X)}$, so we have obtained a contradiction if we pick $X$ large enough (e.g. $\mathrm{card}(X)=2^{\aleph_0}$).


category_id: Set_f
property_id: locally small
reason:
There is a forgetful functor $\mathbf{Set}_\mathrm{f} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: Set_f
property_id: generator
reason:
The singleton set (which is <i>not</i> terminal) is a generator as it represents the forgetful functor $\mathbf{Set}_\mathrm{f} \to \mathbf{Set}$.


category_id: Set_f
property_id: cogenerator
reason:
The set $\{0,1\}$ is a cogenerator in $\mathbf{Set}_\mathrm{f}$: Assume that $f,g : X \rightrightarrows Y$ are two finite-to-one maps such that $h \circ f = h \circ g$ for all finite-to-one maps $h : Y \to \{0,1\}$. This exactly means $f^*(A)=g^*(A)$ for all finite subsets $A \subseteq Y$. Applying this to $A = \{f(x)\}$ for $x \in X$ we get $x \in f^*(\{f(x)\}) = g^*(\{f(x)\})$, so that $g(x) = f(x)$.


category_id: Set_f
property_id: extensive
reason:
We first show that finite coproducts exist. The empty set is clearly initial. The disjoint union $X+Y$ of two sets $X,Y$ with the inclusion maps $X \rightarrow X+Y \leftarrow Y$ is a coproduct: The inclusions are injective, hence finite-to-one. If $f : X \to T$, $g : Y \to T$ are finite-to-one maps, the induced map $(f;g) : X + Y \to T$ is finite-to-one since the fiber of $t \in T$ is $f^*(\{t\}) + g^*(\{t\})$, which is finite. Hence, finite coproducts exist. A map $A \to X + Y$ yields a decomposition $A = A_X + A_Y$ with maps $A_X \to X$, $A_Y \to Y$ (since $\mathbf{Set}$ is extensive). Here, $A \to X + Y$ is finite-to-one iff $A_X \to X$ and $A_Y \to Y$ are finite-to-one.


category_id: Set_f
property_id: equalizers
reason:
Equalizers can be constructed as in $\mathbf{Set}$ because of the following trivial observation: if $f : X \to Y$ is a finite-to-one map and $E \subseteq Y$ is a subset with $f(X) \subseteq E$, then the induced map $f^E : X \to E$ is also finite-to-one.


category_id: Set_f
property_id: quotients of congruences
reason:
A congruence on a set $X$ in $\mathbf{Set}_\mathrm{f}$ is the same as an equivalence relation $R$ on $X$ whose equivalence classes are finite. In that case, the usual quotient map $p : X \to X/R$ is finite-to-one. Moreover, if $h : X/R \to Y$ is a map such that $h \circ p : X \to Y$ is finite-to-one, then $h$ is finite-to-one as well because $h^*(\{y\}) \subseteq p^*((h \circ p)^*(\{y\}))$ for all $y \in Y$. Therefore, $p$ is also the quotient in $\mathbf{Set}_\mathrm{f}$.


category_id: Set_f
property_id: locally cartesian closed
reason:
If $X$ is a set, the equivalence $\mathbf{Set}/X \simeq \mathbf{Set}^X$, $f \mapsto (f^*(\{x\}))_{x \in X}$ restricts to an equivalence $\mathbf{Set}_\mathrm{f} / X \simeq \mathbf{FinSet}^X$. This category is cartesian closed since <a href="/category/FinSet">the category $\mathbf{FinSet}$</a> is cartesian closed and products of cartesian closed categories are cartesian closed.


category_id: Set_f
property_id: mono-regular
reason:
If $f : X \to Y$ is a monomorphism, i.e. an injective map, it is an equalizer of the two maps $i_1, i_2 : Y \rightrightarrows Y \sqcup_X Y$ in $\mathbf{Set}$. But $i_1,i_2$ are injective, hence finite-to-one, so that $f$ is also an equalizer in $\mathbf{Set}_\mathrm{f}$.


category_id: Set_f
property_id: epi-regular
reason:
If $f : X \to Y$ is an epimorphism in $\mathbf{Set}_\mathrm{f}$, i.e. a surjective finite-to-one map, it is a coequalizer of the two maps $p_1, p_2 : X \times_Y Y \rightrightarrows Y$ in $\mathbf{Set}$. These maps are finite-to-one since $p_i^*(\{y\}) \cong f^*(\{y\})$ for $i=1,2$, and their coequalizer is also $f$ in $\mathbf{Set}_\mathrm{f}$: It suffices to observe that if $h : Y \to T$ is a map such that $h \circ f$ is finite-to-one, then $h$ is finite-to-one as well. In fact, surjectivity of $f$ implies $h^*(\{t\}) = f_*((h \circ f)^*(\{t\}))$ for $t \in T$.


category_id: Set_f
property_id: semi-strongly connected
reason:
From set theory it is known that for all sets $X,Y$ there is an injective map $X \to Y$ or an injective map $Y \to X$, and injective maps are finite-to-one.


category_id: Set_f
property_id: ℵ₁-accessible
reason:
We first show that $\aleph_1$-directed colimits exist and are preserved by the forgetful functor to $\mathbf{Set}$. Let $(X_i)_{i \in I}$ be a diagram in $\mathbf{Set}_\mathrm{f}$ indexed by a $\aleph_1$-directed poset $I$. Let $(u_i : X_i \to X)$ be the colimit in $\mathbf{Set}$. Each map $u_i$ is finite-to-one: Otherwise, some fiber $u_i^*(\{x\}) \subseteq X_i$ contains infinitely many elements $a_1,a_2,\dotsc$. For every $n \geq 1$ we find $i_n \geq i$ such that $a_1,a_n$ have the same image in $X_{i_n}$. Let $i_\infty \in I$ be an upper bound of all $i_n$. Then all $a_n$ have the same image in $X_{i_\infty}$. This is a contradiction since $X_i \to X_{i_\infty}$ is finite-to-one. Now if $f : X \to Y$ is a map such that each $f \circ u_i$ is finite-to-one, then $f$ is finite-to-one as well: If a fiber $f^*(\{y\})$ contains infinitely many distinct elements $x_1,x_2,\dotsc$, there is some index $i$ such that all have a preimage in $X_i$, but then $(f \circ u_i)^*(\{y\})$ would be infinite. This concludes the proof that $\aleph_1$-directed colimits exist.<br>
	In $\mathbf{Set}$, every set $X$ is the $\aleph_1$-directed colimit of its countable subsets. This remains true in $\mathbf{Set}_\mathrm{f}$ because a map $X \to Y$ is finite-to-one as long as each restriction to a countable subset of $X$ is finite-to-one. Moreover, every countable set is $\aleph_1$-presentable in $\mathbf{Set}$, but also in $\mathbf{Set}_\mathrm{f}$.


category_id: Set_f
property_id: binary powers
reason:
More generally, if $X,Y$ are two non-empty sets such that $X \times Y$ exists in $\mathbf{Set}_\mathrm{f}$, then both $X$ and $Y$ must be finite. In fact, the forgetful functor to $\mathbf{Set}$ is representable, so it must preserve products. This means we can assume $X \times Y$ is the usual cartesian product with the usual projections. Since $p_1 : X \times Y \to X$ is finite-to-one and $X$ is non-empty, $Y$ is finite. By symmetric, also $X$ is finite. (Conversely, if $X$ and $Y$ are finite, or one of them is empty, then indeed $X \times Y$ exists.)


category_id: Set_f
property_id: countable copowers
reason:
Assume that the copower $X := \mathbb{N} \otimes 1$ exists, where $1$ is the singleton set. This is a set with a map $i : \mathbb{N} \to X$ (not necessarily finite-to-one) such that for every other such map $j : \mathbb{N} \to Y$ there is a unique finite-to-one map $f : X \to Y$ with $f \circ i = j$. Applying this to $j : \mathbb{N} \to 1$, we see that $X$ is finite. Applying the universal property to maps $j : \mathbb{N} \to \{0,1\}$, we see that for every subset $E \subseteq \mathbb{N}$ there is a unique finite subset $F \subseteq X$ with $i^*(F) = E$. But finiteness of $F$ is automatic, so $i^* : P(X) \to P(\mathbb{N})$ is bijective. But then $P(\mathbb{N})$ is finite, which is absurd.


category_id: Set_f
property_id: filtered
reason:
Consider the maps $f,g : \mathbb{N} \rightrightarrows \mathbb{N}$ defined by $f(x)=x$ and $g(x)=2x$. They are injective, hence finite-to-one. If a map $h : \mathbb{N} \to X$ coequalizes them, we have $h(x)=h(2x)$, in particular $h(1)=h(2^n)$ for all $n \in \mathbb{N}$. Thus, $h$ is not finite-to-one.


category_id: Set_f
property_id: sequential limits
reason:
Consider the set $[n] := \{0,\dotsc,n\}$ for $n \in \mathbb{N}$. The forgetful functor to $\mathbf{Set}$ is representable (by the singleton set), hence preserves all limits. Thus, if the diagram of truncation maps $\cdots \twoheadrightarrow [2] \twoheadrightarrow [1] \twoheadrightarrow [0]$ has a limit in $\mathbf{Set}_\mathrm{f}$, its underlying set is isomorphic to the limit taken in $\mathbf{Set}$, which is $\mathbb{N} \cup \{\infty\}$. But there is no finite-to-one map $\mathbb{N} \cup \{\infty\} \to [0]$.


category_id: Set_f
property_id: multi-cocomplete
reason:
We will prove that the family of singleton sets $(1)_{n \in \mathbb{N}}$ has no multi-coproduct, generalizing the proof that the family does not have a coproduct given above. A cocone is just a map of sets $f : \mathbb{N} \to X$. A morphism from $f : \mathbb{N} \to X$ to $g : \mathbb{N} \to Y$ is a finite-to-one map $h : X \to Y$ with $g = h \circ f$. This describes the category of cocones, and we need to show that it has no multi-initial object. To this end, we claim that the connected component of the unique map $! : \mathbb{N} \to 1$ consists precisely of the maps $f : \mathbb{N} \to X$ with finite image. Once that is established, we can recycle the proof for missing coproducts since there we have only used finite codomains.<br>
	Let $g = h \circ f$ be as above. If $\mathrm{im}(f)$ is finite, then $\mathrm{im}(g) = h_*(\mathrm{im}(f))$ is finite as well. Conversely, if $\mathrm{im}(g)$ is finite, then $\mathrm{im}(f) \subseteq \bigcup_{y \in \mathrm{im}(g)} h^*(\{y\})$ is finite as well. This shows that the connected component of $!$ is contained in the collection of maps with finite image. Conversely, if $f$ has finite image, then there is a morphism from the corestriction $f' : \mathbb{N} \to \mathrm{im}(f)$ to $f$, and also a morphism from $f'$ to $!$. This proves the remaining inclusion.


category_id: Set*
property_id: epi-regular
reason:
Every epimorphism is surjective for this category, and in an algebraic category every surjective homomorphism is a regular epimorphism.


category_id: Set*
property_id: coregular
reason:
From the other properties we know that (co-)limits exist and that monomorphisms coincide with injective pointed maps. So it suffices to prove that these maps are stable under pushouts. This follows from the corresponding fact for the <a href="/category/Set">category of sets</a> and the observation that the forgetful functor $\mathbf{Set}_* \to \mathbf{Set}$ preserves pushouts.


category_id: Set*
property_id: co-Malcev
reason:
Malcev categories are closed under slice categories by Prop. 2.2.14 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>. It follows that co-Malcev categories are closed under coslice categories, and $\mathbf{Set}_*$ is a coslice category of $\mathbf{Set}$, which is co-Malcev since every elementary topos is co-Malcev.


category_id: Set*
property_id: cocartesian cofiltered limits
reason:
Let $X$ be a pointed set and $(Y_i)$ be a filtered diagram of pointed sets. Base points will be denoted by $0$. The canonical map $X \vee \lim_i Y_i \to \lim_i (X \vee Y_i)$ is injective since the wedge sum naturally embeds into the product and the natural map $X \vee \prod_i Y_i \to \prod_i (X \times Y_i)$ is injective. Now let $z = (z_i) \in \lim_i (X \vee Y_i)$.
	<br>Case 1: There is some index $i$ with $z_i \in X \setminus \{0\}$. We claim $z_j \in X$ for any index $j$ and $z_j = z_i$ in $X$, so that $z$ has a preimage in $X$. To see this, choose an index $k \geq i,j$. Since $X \vee Y_i \to X \vee Y_k$ maps $z_i \mapsto z_k$ and is the identity on $X$, we see that $z_k \in X$ and $z_k = z_i$ in $X$. Since $X \vee Y_j \to X \vee Y_k$ maps $z_j \mapsto z_k$, we see that $z_j \notin Y_j$, since otherwise $z_k \in Y_k \cap X = \{0\}$. Hence, $z_j \in X \setminus \{0\}$, and then $z_j = z_k = z_i$.
	<br>Case 2: We have $z_i \in Y_i$ for all $i$. Then clearly $(z_i) \in \lim_i Y_i$ is a preimage.


category_id: Set*
property_id: CIP
reason:
The coproduct (wedge sum) of a family of pointed sets $(X_i)_{i \in I}$ can be realized as the subset of $\prod_{i \in I} X_i$ consisting of those tuples $x$ such that $x_i = 0$ for all but (at most) one index.


category_id: Set*
property_id: unital
reason:
The joint image of $X \to X \times Y \leftarrow Y$ is just $\{(x,0) : x \in X\} \cup \{(0,y) : y \in Y\}$ (where $0$ denotes the base point), which is clearly a proper subset of $X \times Y$ when both $X,Y$ are non-trivial.


category_id: Set*
property_id: CSP
reason:
The image of $X \vee Y$ in $X \times Y$ is just $\{(x,0) : x \in X\} \cup \{(0,y) : y \in Y\}$ (where $0$ denotes the base point), which is clearly a proper subset when both $X,Y$ are non-trivial.


category_id: Set*
property_id: conormal
reason:
Every cokernel is "injective away from the base point". Formally, if $p : A \to B$ is a cokernel in $\mathbf{Set}_*$, it has the property that $p(x)=p(y) \neq 0$ implies $x=y$ (where $0$ denotes the base point). Clearly this is not satisfied for every surjective pointed map, consider $(\mathbb{N},0) \to (\{0,1\},0)$ defined by $0 \mapsto 0$ and $x \mapsto 1$ for $x > 0$.


category_id: Set*
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\mathbf{Set} \to \mathbf{Set}_*$ that freely adds a base point.


category_id: Setne
property_id: locally small
reason:
There is a forgetful functor $\mathbf{Set}_{\neq \varnothing} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.


category_id: Setne
property_id: products
reason:
Take the product of non-empty sets inside of $\mathbf{Set}$ and observe that it is non-empty by the axiom of choice.


category_id: Setne
property_id: wide pushouts
reason:
The wide pushout taken in $\mathbf{Set}$ is clearly non-empty, and it still satisfies the universal property.


category_id: Setne
property_id: coequalizers
reason:
The coequalizer taken in $\mathbf{Set}$ is clearly non-empty, and it still satisfies the universal property.


category_id: Setne
property_id: cartesian closed
reason:
This follows as for $\mathbf{Set}$, since for non-empty sets $X,Y$ there is at least one function $X \to Y$.


category_id: Setne
property_id: finitely accessible
reason:
Since the inclusion $\mathbf{Set}_{\neq \varnothing} \hookrightarrow \mathbf{Set}$ is closed under non-empty colimits, it is also closed under filtered colimits. Therefore, non-empty finite sets are still finitely presentable in $\mathbf{Set}_{\neq \varnothing}$, and every non-empty set is written as a filtered colimit of them.


category_id: Setne
property_id: generalized variety
reason:
Since the inclusion $\mathbf{Set}_{\neq \varnothing} \hookrightarrow \mathbf{Set}$ is closed under non-empty colimits, it is also closed under sifted colimits. Therefore, non-empty finite sets are still strongly finitely presentable in $\mathbf{Set}_{\neq \varnothing}$, and every non-empty set is written as a sifted colimit of them.


category_id: Setne
property_id: multi-complete
reason:
Let $D$ be a diagram in $\mathbf{Set}_{\neq \varnothing}$, and let $L$ be a limit of $D$ in $\mathbf{Set}$. If $L$ is non-empty, it gives a limit in $\mathbf{Set}_{\neq \varnothing}$ as well. If $L$ is the empty set, there is no cone over $D$ in $\mathbf{Set}_{\neq \varnothing}$; hence the empty set of cones gives a multi-limit of $D$ in $\mathbf{Set}_{\neq \varnothing}$.


category_id: Setne
property_id: natural numbers object
reason:
Any natural numbers object in $\mathbf{Set}$, such as $(\mathbb{N},0,n \mapsto n+1)$, is clearly also one in $\mathbf{Set}_{\neq \varnothing}$.


category_id: Setne
property_id: sequential limits
reason:
Assume that the sequence of inclusions $\cdots \to \mathbb{N}_{\geq 2} \to \mathbb{N}_{\geq 1} \to \mathbb{N}_{\geq 0} = \mathbb{N}$ as a limit $X$, consisting of maps $X \to \mathbb{N}_{\geq n}$. Since $X$ is non-empty, there is a map $1 \to X$. This corresponds to a family of compatible maps $ 1 \to \mathbb{N}_{\geq n}$, i.e. to compatible elements in $\mathbb{N}_{\geq n}$. But the set $\bigcap_{n \geq 0} \mathbb{N}_{\geq n}$ is empty.


category_id: Setne
property_id: cofiltered
reason:
The two maps $\{0\} \rightrightarrows \{0,1\}$ are equalized by no map $X \to \{0\}$ in this category.


category_id: Setne
property_id: coquotients of cocongruences
reason:
The two maps $\{0\} \rightrightarrows \{0,1\}$ form a cocongruence on $\{0\}$ &mdash; namely the cofull cocongruence on $\{0\}$ &mdash; but they do not have an equalizer.


category_id: SetxSet
property_id: generator
reason:
Assume that $(A,B)$ is a generator. Of course, $A$ and $B$ cannot be both empty. Assume w.l.o.g. that $A$ is non-empty. Then there is no morphism $(A,B) \to (0,1)$, but there are two different morphisms $(0,1) \rightrightarrows (0,2)$.


category_id: SetxSet
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the diagonal functor $\mathbf{Set} \to \mathbf{Set} \times \mathbf{Set}$.


category_id: Sh(X)
property_id: Malcev
reason:
Consider the subsheaf of $\underline{\mathbb{N}} \times \underline{\mathbb{N}}$ consisting of locally constant functions $(f,g) : X \to \mathbb{N} \times \mathbb{N}$ with $f \leq g$ pointwise. This is reflexive, but not symmetric.


category_id: Sh(X)
property_id: cofiltered-limit-stable epimorphisms
reason:
Our <a href="/category/Set">counterexample for $\mathbf{Set}$</a> (using infinite intersections) also works in $\mathbf{Sh}(X)$ by using constant sheaves. We could also apply a variant of (the dual of) <a href="/lemma/filtered-monos">this lemma</a> to the constant sheaf functor $\mathbf{Set} \to \mathbf{Sh}(X)$, which does not preserve all cofiltered limits, but does preserve intersections.


category_id: Sh(X,Ab)
property_id: Grothendieck abelian
reason:
This is standard, see for example Theorem 18.1.6. in <a href="https://ncatlab.org/nlab/show/Categories+and+Sheaves" target="_blank">Kashiwara-Schapira</a>.


category_id: Sh(X,Ab)
property_id: split abelian
reason:
Choose a point $x \in X$. The functor $x_* : \mathbf{Ab} \to \mathrm{Sh}(X,\mathbf{Ab})$ (skyscraper sheaf) is exact, and its left adjoint $x^* : \mathrm{Sh}(X,\mathbf{Ab}) \to \mathbf{Ab}$ (stalk) satisfies $x^* x_* \cong \mathrm{id}_{\mathbf{Ab}}$. Now, since $\mathbf{Ab}$ is not split abelian (see <a href="/category/Ab">here</a>), there is a short exact sequence of abelian groups $0 \to A \to B \to C \to 0$ that does not split. Then $0 \to x_* A \to x_* B \to x_* C \to 0$ is also exact, but it does not split: Otherwise it would also be split after applying $x^*$, which however gives the original sequence in $\mathbf{Ab}$.


category_id: Sp
property_id: elementary topos
reason:
The category is equivalent to $\prod_{n \geq 0} \Sigma_n{-}\mathbf{FinSet}$ (where $\Sigma_n$ denotes the symmetric group of order $n$), and each $\Sigma_n{-}\mathbf{FinSet}$ is an elementary topos since $\mathbf{FinSet}$ is and $\Sigma_n$ is a finite group, cf. <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part B, Corollary 2.3.18.


category_id: Sp
property_id: cogenerator
reason:
This follows from $\mathbf{Sp} \simeq \prod_{n \geq 0} \Sigma_n{-}\mathbf{FinSet}$, <a href="/lemma/cogenerators_in_product_categories">this lemma</a>, and the fact that if $G$ is a (finite) group, the power set $P(G)$ with the evident $G$-action is a weakly terminal cogenerator in $G{-}\mathbf{Set}$ (resp. $G{-}\mathbf{FinSet}$). For the proof, notice that $\varnothing,G \in P(G)$ are fixed points, yielding two $G$-maps $1 \rightrightarrows P(G)$. In particular, $P(G)$ is weakly terminal. If $X$ is a $G$-set with distinct points $x,y$, we construct a $G$-map $f : X \to P(G)$ that separates $x,y$: First, $X$ is a coproduct of orbits. If $x,y$ lie in different orbits, let $f|_{Gx}$ be constant $\varnothing$, $f|_{Gy}$ be constant $G$, and, say, $f$ be constant $\varnothing$ on all other orbits. If $x,y$ lie in the same orbit, say $y = g_0 x$, define $f|_{Gx} : Gx \to P(G)$ by $f(x) = G_x$ (stabilizer), which is well-defined, and choose $f$ to be $\varnothing$ on all other orbits. Then $f(y) = g_0 G_x \neq G_x = f(x)$.


category_id: Sp
property_id: locally small
reason:
Disclaimer: This result and its proof are not relevant for category theory and are also depending on the concrete model of set theory. That this category is locally essentially small is only what matters. Now, consider the terminal species $F=G=1$. Then $\mathrm{Hom}(F,G)$ has just a single element, namely the natural transformation $\alpha$ that sends every finite set $X$ to the unique map $\alpha_X : 1 \to 1$. Formally, $\alpha$ is a map, modelled as a set of ordered pairs $(X,\mathrm{id}_1)$, where $X$ is a finite set. Hence, $\alpha$ is not a set (since finite sets do not form a set), and therefore $\mathrm{Hom}(F,G) = \{\alpha\}$ is also not a set.


category_id: Sp
property_id: countable powers
reason:
If $\mathbf{Sp} \simeq \mathbf{FinSet} \times \prod_{n > 0} \Sigma_n{-}\mathbf{FinSet}$ has countable powers, then $\mathbf{FinSet}$ has countable powers as well by <a href="/lemma/limits_in_factor_categories">this lemma</a>, which we already know is false (see <a href="/category/FinSet">here</a>).


category_id: Sp
property_id: countable copowers
reason:
If $\mathbf{Sp} \simeq \mathbf{FinSet} \times \prod_{n > 0} \Sigma_n{-}\mathbf{FinSet}$ has countable copowers, then $\mathbf{FinSet}$ has countable copowers as well by <a href="/lemma/limits_in_factor_categories">this lemma</a> (dualized), which we already know is false (see <a href="/category/FinSet">here</a>).


category_id: Sp
property_id: Malcev
reason:
Since $\mathbf{FinSet}$ is not Malcev, there is some finite set $X$ with a non-symmetric reflexive relation $R \subseteq X^2$. Now consider these as constant functors $\mathbb{B} \to \mathbf{FinSet}$.


category_id: Sp
property_id: semi-strongly connected
reason:
For $n \geq 0$ let $E_n$ be the combinatorial species of $n$-sets: $E_n(A) = \{A\}$ when $A$ has cardinality $n$, otherwise $E_n(A) = \varnothing$. Then there is no morphism between $E_n$ and $E_m$ unless $n = m$.


category_id: Sp
property_id: generator
reason:
Assume that a generator $G$ exists. For $n \geq 0$ let $F_n$ be the combinatorial species of sets of cardinality $\neq n$: $F_n(A) = \varnothing$ when $A$ has cardinality $n$, otherwise $F_n(A) = \{A\}$. There are two different morphisms $F_n \rightrightarrows F_n \sqcup F_n$. Hence, there must be at least one morphism $G \to F_n$. If $A$ has cardinality $n$, this implies $G(A) = \varnothing$. Since this holds for all $n$, $G$ is the initial object. But this is clearly no generator (it would mean that the category is thin).


category_id: Sp
property_id: essentially countable
reason:
Any function $f\colon\mathbb{N} \to \mathbb{N}$ can be regarded as a combinatorial species with trivial actions, and distinct functions yield non-isomorphic species.


category_id: Sp
property_id: natural numbers object
reason:
If $(N,z,s)$ is a natural numbers object, then
	$$1 \xrightarrow{z} N \xleftarrow{s} N$$
	is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. But there is no combinatorial species $N$ with $N \cong 1 + N$, since evaluating this at, say, $\varnothing$, would yield a finite set $N$ with this property.


category_id: Top
property_id: complete
reason:
Take the limit of the underlying sets and endow it with the coarsest topology making all projections continuous.


category_id: Top
property_id: cocomplete
reason:
Take the colimit of the underlying sets and endow it with the finest topology making all inclusions continuous.


category_id: Top
property_id: infinitary extensive
reason:
[Sketch] Since $\mathbf{Set}$ is infinitary extensive, a map $f : Y \to \coprod_i X_i$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow $Y_i$ with the subspace topology. If $f$ is continuous, each $Y_i = f^{-1}(X_i)$ is open in $Y$, so that $Y = \coprod_i Y_i$ holds as topological spaces, and each $f_i$ is continuous.


category_id: Top
property_id: generator
reason:
The one-point space is a generator since it represents the forgetful functor $\mathbf{Top} \to \mathbf{Set}$.


category_id: Top
property_id: regular subobject classifier
reason:
The indiscrete two-point space $\{0,1\}$ is a regular subobject classifier since continuous maps $X \to \{0,1\}$ correspond to subsets of $X$.


category_id: Top
property_id: coregular
reason:
The category has all limits and colimits, and the regular monomorphisms are the subspace inclusions. Thus, it suffices to prove that subspace inclusions are stable under pushouts. For a proof see e.g. Lemma 3.6 at the <a href="https://ncatlab.org/nlab/show/subspace+topology#pushout" target="_blank">nLab</a>.


category_id: Top
property_id: cartesian filtered colimits
reason:
The functor $\mathbb{Q} \times - : \mathbf{Top} \to \mathbf{Top}$ does not preserve colimits, see <a href="https://math.stackexchange.com/questions/2969372" target="_blank">MSE/2969372</a>.


category_id: Top
property_id: locally presentable
reason:
In fact, it does not have any small dense subcategory by <a href="https://math.stackexchange.com/questions/4097315/" target="_blank">MSE/4097315</a>. For a related result, see <a href="https://mathoverflow.net/questions/288648" target="_blank">MO/288648</a>.


category_id: Top
property_id: balanced
reason:
If $X$ is a set, consider the discrete space $X_d$ on $X$ and the indiscrete space $X_i$ on $X$. The identity map $X \to X$ lifts to a continuous map $X_d \to X_i$, which is bijective and therefore both a mono- and an epimorphism, but it is not an isomorphism unless $X$ has at most one element.


category_id: Top
property_id: Malcev
reason:
This is clear since $\mathbf{Set}$ is not Malcev and can be interpreted as the subcategory of discrete spaces.


category_id: Top
property_id: co-Malcev
reason:
See <a href="https://mathoverflow.net/questions/509548" target="_blank">MO/509548</a>. We can also phrase the proof as follows: Consider the forgetful functor $U : \mathbf{Top} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(X) := \{(x,y) \in U(X)^2 : x \in \overline{\{y\}} \}$. Both are representable: $U$ by the singleton and $R$ by the Sierpinski space. It is clear that $R$ is reflexive, but not symmetric.


category_id: Top
property_id: coaccessible
reason:
Assume $\mathbf{Top}$ is coaccessible. Let $p\colon S \to I$ be the identity map from the Sierpinski space to the two-element indiscrete space. Then, a topological space is discrete if and only if it is projective to the morphism $p$. This implies that the full subcategory spanned by all discrete spaces, which is equivalent to $\mathbf{Set}$, is coaccessible by Prop. 4.7 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. However, since $\mathbf{Set}$ is not coaccessible, this is a contradiction.


category_id: Top
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\mathbf{Set} \to \mathbf{Top}$ which equips a set with the indiscrete topology.


category_id: Top*
property_id: locally small
reason:
There is a forgetful functor $\mathbf{Top}_* \to \mathbf{Set}_*$ and $\mathbf{Set}_*$ is locally small.


category_id: Top*
property_id: complete
reason:
This follows from $\mathbf{Top}_* \cong 1 / \mathbf{Top}$ and the fact that $\mathbf{Top}$ is complete.


category_id: Top*
property_id: coproducts
reason:
This follows from $\mathbf{Top}_* \cong 1 / \mathbf{Top}$ and the fact that $\mathbf{Top}$ has wide pushouts.


category_id: Top*
property_id: generator
reason:
The discrete space $\{0,1\}$ with base point $0$ is a generator since it represents the forgetful functor $\mathbf{Top}_* \to \mathbf{Set}$.


category_id: Top*
property_id: cogenerator
reason:
It is easily checked that the indiscrete two-point space $\{0,1\}$ with base point $1$ is a cogenerator.


category_id: Top*
property_id: regular subobject classifier
reason:
The indiscrete two-point space $\{0,1\}$ with base point $1$ is a regular subobject classifier since pointed continuous maps $X \to \{0,1\}$ correspond to pointed subsets of $X$ (by taking the fiber of $1$ as usual).


category_id: Top*
property_id: counital
reason:
Since embeddings are regular monomorphisms in this category (see below) and hence strong monomorphisms, it suffices to prove that the canonical morphism $X \vee Y \hookrightarrow X \times Y$ is an embedding. For a proof, see <a href="https://math.stackexchange.com/questions/4055988" target="_blank">MSE/4055988</a>.


category_id: Top*
property_id: CIP
reason:
This follows since $\mathbf{Set}_*$ has this property and the forgetful functor preserves products and coproducts.


category_id: Top*
property_id: cocartesian cofiltered limits
reason:
We continue the proof for <a href="/category/Set*">$\mathbf{Set}_*$</a> by showing that the natural bijective map
	$$\textstyle \alpha : X \vee \lim_i Y_i \to \lim_i (X \vee Y_i)$$
	is open. It suffices to consider open sets of two types: (1) If $U \subseteq X$ is open, the $\alpha$-image of $U \vee \lim_i Y_i$ is $p_{i_0}^{-1}(U \vee Y_{i_0})$ for any chosen index $i_0$, hence open. (2) If $i$ is an index and $V_i \subseteq Y_i$ is open, then the $\alpha$-image of $X \vee (p_i^{-1}(V_i) \cap \lim_i Y_i)$ is $p_i^{-1}(X \vee V_i)$, hence open.


category_id: Top*
property_id: coregular
reason:
Regular monomorphisms coincide with the embeddings (see below). Since $\mathbf{Top}$ is coregular, they are stable under pushouts, and pushouts in $\mathbf{Top}_*$ are the same.


category_id: Top*
property_id: regular
reason:
See Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a>. The proof also works for pointed spaces (resp. posets) by using the base points $a$ and $0$.


category_id: Top*
property_id: locally presentable
reason:
In fact, it does not have any small dense subcategory by <a href="https://math.stackexchange.com/questions/4097315/" target="_blank">MSE/4097315</a>. The proof easily adapts to pointed spaces.


category_id: Top*
property_id: balanced
reason:
If $X$ is a set with a base point $x_0$, consider the discrete space $X_d$ on $X$ and the indiscrete space $X_i$ on $X$. The identity map $X \to X$ lifts to a continuous map $X_d \to X_i$ preserving $x_0$, which is bijective and therefore both a mono- and an epimorphism, but it is not an isomorphism unless $X = \{x_0\}$.


category_id: Top*
property_id: cartesian filtered colimits
reason:
The functor $\mathbb{Q} \times - : \mathbf{Top}_* \to \mathbf{Top}_*$ does not preserve colimits, see <a href="https://math.stackexchange.com/questions/2969372" target="_blank">MSE/2969372</a>. The counterexample also works for pointed spaces.


category_id: Top*
property_id: co-Malcev
reason:
We can adjust the proof for $\mathbf{Top}$ as follows: Consider the forgetful functor $U : \mathbf{Top}_* \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(X) := \{(x,y) \in U(X)^2 : x \in \overline{\{y\}} \}$. Both are representable: $U$ by the discrete space $\{0,1\}$ with base point $0$ and $R$ by the Sierpinski space with an isolated base point added. It is clear that $R$ is reflexive, but not symmetric.


category_id: Top*
property_id: regular quotient object classifier
reason:
We can recycle the proof for the <a href="/category/Set*">category of pointed sets</a> using discrete topological spaces.


category_id: Top*
property_id: coaccessible
reason:
We can adjust the proof for $\mathbf{Top}$ as follows: Assume $\mathbf{Top}_*$ is coaccessible. Let $S_0=\{x,*\}$ be the pointed topological space such that $\{*\}$ is the only non-trivial open set, and let $S_1=\{x,*\}$ be the pointed space such that $\{x\}$ is the only non-trivial open set. Let $p_i\colon S_i \to \{x,*\}$ be the identity function to the two-element indiscrete pointed space. Then, a pointed topological space is discrete if and only if it is projective to the morphisms $p_0$ and $p_1$. This implies that the full subcategory spanned by all discrete pointed spaces, which is equivalent to $\mathbf{Set}_*$, is coaccessible by Prop. 4.7 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. However, since $\mathbf{Set}_*$ is not coaccessible, this is a contradiction.


category_id: Top*
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}_*$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\mathbf{Set}_* \to \mathbf{Top}_*$ that equips a pointed set with the indiscrete topology.


category_id: Vect
property_id: CSP
reason:
The canonical homomorphism $\bigoplus_{n \geq 0} K \to \prod_{n \geq 0} K$ is not surjective, hence no epimorphism.


category_id: Z
property_id: Malcev
reason:
Any counterexample for $\mathbf{Set}$ (i.e., any non-symmetric reflexive relation) yields one for this category by taking constant functors.


category_id: Z
property_id: semi-strongly connected
reason:
This is because already the full subcategory of representable functors is not semi-strongly connected, see <a href="/category/CRing">the entry for $\mathbf{CRing}$</a>. Specifically, there is no morphism between $\mathrm{Hom}(\mathbb{F}_2,-)$ and $\mathrm{Hom}(\mathbb{F}_3,-)$.


category_id: Z
property_id: cartesian closed
reason:
There are functors $F,G : \mathbf{CRing} \to \mathbf{Set}$ such that $\mathrm{Hom}(F,G)$ is not essentially small, see <a href="https://mathoverflow.net/questions/390611" target="_blank">MO/390611</a> for example. Now if the exponential $[F,G] : \mathbf{CRing} \to \mathbf{Set}$ exists, we get $[F,G](\mathbb{Z}) \cong \mathrm{Hom}(\mathrm{Hom}(\mathbb{Z},-),[F,G])$ by Yoneda, which simplifies to $\mathrm{Hom}(1,[F,G]) \cong \mathrm{Hom}(1 \times F,G) \cong \mathrm{Hom}(F,G)$, a contradiction.


category_id: Z
property_id: well-powered
reason:
Consider the functor $F$ from <a href="https://mathoverflow.net/questions/390611" target="_blank">MO/390611</a> for example. The collection of subobjects of $F$ is not isomorphic to a set: for each infinite cardinal $\kappa$, simply cut off the construction of $F$ at $\kappa$. This yields a different subobject for each $\kappa$.


category_id: Z
property_id: cofiltered-limit-stable epimorphisms
reason:
We already know that $\mathbf{Set}$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\mathbf{Set} \to [\mathbf{CRing}, \mathbf{Set}]$ that maps a set to its constant functor.


category_id: Z_div
property_id: products
reason:
Take the $\gcd$ of a subset. This also works for infinite sets, for example the $\gcd$ of all primes is $0$.


category_id: Z_div
property_id: distributive
reason:
We need to prove $\mathrm{lcm}_i \gcd(a, b_i) \cong \gcd(a, \mathrm{lcm}_i b_i)$ for finite families. If $x$ denotes the LHS and $y$ denotes the RHS, the relation $x \mid y$ is formal. If $v_p(-) : \mathbb{Z} \to \mathbb{N}_{\infty}$  denotes the multiplicity of a prime $p$, then $v_p(x)$ equals $\max_i \min(v_p(a),v_p(b_i))$, and $v_p(y)$ equals $\min(v_p(a), \max_i v_p(b_i))$. Since our family is finite, there is some $i_0$ with $\max_i v_p(b_i) = v_p(b_{i_0})$. Then $v_p(x) \geq \min(v_p(a),v_p(b_{i_0})) = v_p(y)$. This proves $y \mid x$.


category_id: Z_div
property_id: codistributive
reason:
We need to prove $\mathrm{lcm}(a, \gcd_i b_i) \cong \gcd_i \mathrm{lcm}(a,b_i)$ for finite families. This is clear for the empty family (both sides are $0$ then), so assume it is non-empty. If $x$ denotes the LHS and $y$ denotes the RHS, the relation $x \mid y$ is formal. If $v_p(-) : \mathbb{Z} \to \mathbb{N}_{\infty}$ denotes the multiplicity of a prime $p$, then $v_p(x)$ equals $\max(v_p(a), \min_i v_p(b_i))$, and $v_p(y)$ equals $\min_i \max(v_p(a), v_p(b_i))$. Choose some $i_0$ with $\min_i v_p(b_i) = v_p(b_{i_0})$. Then $v_p(x) = \max(v_p(a), v_p(b_{i_0})) \geq v_p(y)$.


category_id: Z_div
property_id: self-dual
reason:
The only integer with infinitely many divisors (up to isomorphism) is $0$. But many integers have infinitely many multiples (up to isomorphism).


category_id: Z_div
property_id: countably distributive
reason:
We have $2 \times \coprod_n 3^n = \gcd(2,\mathrm{lcm}_n(3^n)) = \gcd(2,0) = 2$, but $\coprod_n (2 \times 3^n) = \mathrm{lcm}_n \gcd(2,3^n) = \mathrm{lcm}_n 1  = 1$.


category_id: Z_div
property_id: countably codistributive
reason:
If $p$ runs through all odd primes, we have $2 \sqcup \prod_p p = \mathrm{lcm}(2,\mathrm{gcd}_p p) = \mathrm{lcm}(2,0) = 0$, but $\prod_p (2 \sqcup p) = \gcd_p (\mathrm{lcm}(2,p)) = \gcd_p (2 \cdot p) = 2$.


category_id: real_interval
property_id: locally ℵ₁-presentable
reason:
See <a href="https://math.stackexchange.com/questions/4481902/locally-presentable-vs-compactly-presentable-categories#comment9399784_4482185" target="_blank">MSE/4481902</a>.


category_id: real_interval
property_id: locally finitely presentable
reason:
It suffices to prove that $0$ (the initial object) is the only finitely presentable object. If $s > 0$, then $s = \sup_{n \in \mathbb{N}, \, s \geq 1/n } (s - 1/n)$, but there is no $n$ with $s \leq s - 1/n$.


category_id: sSet
property_id: locally small
reason:
This follows from the general fact that $[\mathcal{C},\mathcal{D}]$ is locally small when $\mathcal{C}$ is small and $\mathcal{D}$ is locally small, here applied to $\mathcal{C} = \Delta^{\mathrm{op}}$ and $\mathcal{D} = \mathbf{Set}$.


category_id: sSet
property_id: generator
reason:
Let $\Delta^n := \mathrm{Hom}([n],-)$ be the standard $n$-simplex for $n \geq 0$. The set $\{\Delta^n : n \geq 0\}$ is a <a href="/category-property/generating_set">generating set</a> by the Yoneda Lemma. For all $n,m$ there is a morphism $[n] \to [m]$ in $\Delta$ and hence a morphism $\Delta^m \to \Delta^n$ in $\mathbf{sSet}$. Then by <a href="/lemma/generator_construction">this lemma</a> the coproduct $\coprod_{n \geq 0} \Delta^n$ is a generator in $\mathbf{sSet}$.


category_id: sSet
property_id: locally strongly finitely presentable
reason:
This follows from the fact that every category of presheaves over a small category is locally strongly finitely presentable.


category_id: sSet
property_id: semi-strongly connected
reason:
Let $X,Y$ be two simplicial sets. Assume that $X_0$ is empty. Then $X_n$ is empty for all $n$ since there is a morphism $[0] \to [n]$, hence a map $X_n \to X_0$. So there is a morphism $X \to Y$ for trivial reasons. If $X_0$ is non-empty, pick an element. By the Yoneda Lemma it corresponds to a morphism $\Delta^0 \to X$. Since $\Delta^0 = 1$ is terminal, there is a morphism $Y \to \Delta^0$, and these compose to a morphism $Y \to X$.


category_id: sSet
property_id: Malcev
reason:
Any counterexample for $\mathbf{Set}$ (i.e., any non-symmetric reflexive relation) yields one for this category by taking constant simplicial sets.


category_id: sSet
property_id: finitary algebraic
reason:
A one-sorted finitary algebraic category has an object $F$ (the free algebra on one generator) such that $F$ is finitely presentable and every object $X$ admits an epimorphism $\coprod_{s \in S} F \to X$ for some index set $S$. Assume that such a simplicial set $F$ exists.  By using the sequence of $n$-skeletons of $F$, we see that there is some $n$ such that every $n$-simplex in $F$ is degenerate. Now take $X = \Delta^n$, which has a non-degenerate $n$-simplex. Then there cannot be an epimorphism $\coprod_{s \in S} F \to X$.


category_id: sSet
property_id: cofiltered-limit-stable epimorphisms
reason:
We show that epimorphisms are not stable under sequential limits, intersections to be precise. Let $X_n \in \mathbf{sSet}$ be the nerve of the poset $\mathbb{N}_{\geq n}$. This means that a $k$-simplex in $X_n$ is a chain of natural numbers $n \leq x_0 \leq \cdots \leq x_k$. Then $X_{n+1} \subseteq X_n$ and $\lim_n X_n = \bigcap_n X_n = 0$ (the initial simplicial set). Each $X_n \to 1$ is an epimorphism, but $\lim_n X_n \to 1$ is not.


category_id: walking_commutative_square
property_id: locally strongly finitely presentable
reason:
This is because the <a href="/category/walking_morphism">walking morphism</a> has this property. Alternatively, we may represent this category as the category of algebras for the finitary algebraic theory with two sorts $S_1,S_2$, the equation $x=y$ for $x,y \in S_1$, and the equation $x=y$ for $x,y \in S_2$.


category_id: walking_composable_pair
property_id: locally strongly finitely presentable
reason:
Take the two-sorted (finitary) algebraic theory with exactly one unary operation between them and the equation $x=y$ for each sort. There are exactly three algebras for this theory up to isomorphism: the identities on the empty set and the singleton, the morphism from the empty set to the singleton.
	Hence we get the equivalence to $\{0 \to  1 \to 2\}$.


category_id: walking_fork
property_id: locally cartesian closed
reason:
We need to check that every slice category is cartesian closed. The slice category over $0$ is the <a href="/category/1">trivial category</a>. The slice category over $1$ is the <a href="/category/walking_morphism">walking morphism</a>. Finally, the slice category over $2$ ist the <a href="/category/walking_commutative_square">walking commutative square</a>. All of these are cartesian closed, see their pages for details.


category_id: walking_fork
property_id: cofiltered-limit-stable epimorphisms
reason:
Let $(X_p)$ and $(Y_p)$ be diagrams in the walking fork indexed by a cofiltered poset $P$. Let $(X_p \to Y_p)_{p \in P}$ be compatible epimorphisms, we need to show that $\lim_p X_p \to \lim_p Y_p$ is an epimorphism as well. Assume it is not. The only non-epimorphism is $i : 0 \to 1$. Hence, $\lim_p X_p = 0$ and $\lim_p Y_p = 1$. So for every $p$ there is a morphism $1 \to Y_p$, meaning $Y_p \in \{1,2\}$. If $Y_p = 2$ for all $p$, the transition morphisms between them must be identities, so that $\lim_p Y_p = 2$, a contradiction. Choose $p$ with $Y_p = 1$. Then for all $q \leq p$ the morphism $Y_q \to Y_p$ shows that $Y_q = 1$ as well. Since $X_q \to Y_q = 1$ is an epimorphism by assumption, it cannot be $i : 0 \to 1$, and we see that $X_q = 1$. Then the transitions between the $X_q$ are identities, and we get the contradiction $\lim_p X_p = \lim_q X_q = 1$.


category_id: walking_fork
property_id: terminal object
reason:
$0$ and $1$ are not terminal since there is no morphism from $2$, and $2$ is not terminal since there are two different morphisms $1 \rightrightarrows 2$.


category_id: walking_fork
property_id: binary powers
reason:
Assume that $X := 2 \times 2$ exists. Since there is a diagonal morphism $2 \to X$, we must have $X = 2$, and the two projections $p_1,p_2 : X \rightrightarrows 2$ must be equal to the identity. But $f,g$ induce a morphism $(f,g) : 1 \to X$ with $p_1 (f,g) = f$ and $p_2 (f,g) = g$, so that $f=g$, a contradiction.


category_id: walking_idempotent
property_id: preadditive
reason:
The monoid $\{1,e\}$ with $e^2=e$ is the underlying multiplicative monoid of the ring $\mathbb{Z}/2$, where $e=0$. Thus, the (unique) preadditive structure is given by $1 + e = e + 1 = 1$, $e + e = e$ and $1 + 1 = e$.


category_id: walking_idempotent
property_id: reflexive coequalizers
reason:
The pair $e,\mathrm{id}_0$ is not reflexive, and the pairs $e,e$ and $\mathrm{id}_0,\mathrm{id}_0$ have a coequalizer for trivial reasons.


category_id: walking_idempotent
property_id: binary powers
reason:
If $0 \times 0 = 0$, then $\mathrm{Hom}(0,0) \cong \mathrm{Hom}(0,0)^2$. But the left hand side has $2$ elements, whereas the right hand has $4$ elements. This is a contradiction.


category_id: walking_morphism
property_id: finitary algebraic
reason:
Take the algebraic theory with no operations but with the equation $x=y$ that is supposed to hold for all elements $x,y$. The algebras for this theory are the empty set and the singleton set, hence we get the equivalence to $\{0 \to 1\}$.


category_id: walking_morphism
property_id: infinitary distributive
reason:
Clearly, this category is (co)complete. Also, the functors $X \mapsto X \times 1 = X$ and $X \mapsto X \times 0 = 0$ are clearly cocontinuous.


category_id: walking_pair
property_id: initial object
reason:
$0$ is not initial since it has two morphisms to $1$, and $1$ has not initial since it has no morphism to $0$.


category_id: walking_pair
property_id: binary products
reason:
We cannot have $0 \times 1 = 1$ since there is no morphism $1 \to 0$. So assume $0 \times 1 = 0$, say with projections $\mathrm{id}_0 : 0 \to 0$ and $a : 0 \to 1$. By applying the universal property to  $\mathrm{id}_0 : 0 \to 0$ and the other morphism $b : 0 \to 1$, it immediately follows $a=b$, which is a contradiction.


category_id: walking_pair
property_id: equalizers
reason:
The two morphisms $a,b : 0 \rightrightarrows 1$ have no equalizer, since it would have to be the identity of $0$, but $a \neq b$.


category_id: walking_pair
property_id: pullbacks
reason:
The two morphisms $a,b : 0 \rightrightarrows 1$ have no pullback, since it would have to consist of identities $0 \leftarrow 0 \rightarrow 0$, but $a \neq b$.


category_id: walking_pair
property_id: sifted colimits
reason:
A proof can be found <a href="/pdf/walking_parallel_pair_sifted_colimit.pdf" target="_blank">here</a>.


category_id: walking_span
property_id: locally cartesian closed
reason:
The slice category over $0$ is the <a href="/category/1">trivial category</a>, and the slice category over $1$ is the <a href="/category/walking_morphism">interval category</a>, which is cartesian closed (see there). The same holds for $2$ by symmetry.


category_id: walking_span
property_id: multi-algebraic
reason:
We first remark that for a set $X$, the identity span $(\mathrm{id},\mathrm{id})\colon X \leftarrow X \rightarrow X$ exhibits a product if and only if $X$ is either a singleton or the empty set. Therefore, there is a (finite product, coproduct)-sketch whose $\mathbf{Set}$-model is precisely a pair $(X,Y)$ of sets such that each of $X$ and $Y$ is either a singleton or the empty set and the product $X \times Y$ is the empty set. Any $\mathbf{Set}$-model of such a sketch is isomorphic to either $(\varnothing, \varnothing)$, $(\varnothing, 1)$, or $(1, \varnothing)$; hence the category of models is equivalent to the walking span.


category_id: walking_splitting
property_id: pointed
reason:
The object $0$ is initial and terminal. This also means that $\mathrm{id}_0, i, p, ip$ are zero morphisms. The only non-zero morphism is $\mathrm{id}_1$.


category_id: walking_splitting
property_id: equalizers
reason:
The only parallel pair of non-equal morphisms is $\mathrm{id}_1, ip : 1 \rightrightarrows 1$, and their equalizer is $i$.


category_id: walking_splitting
property_id: self-dual
reason:
There is an isomorphism $\mathrm{Split}^\mathrm{op} \to \mathrm{Split}$ defined by $0 \mapsto 0$, $1 \mapsto 1$, $i \mapsto p$, $p \mapsto i$.


category_id: walking_splitting
property_id: generator
reason:
The object $1$ a generator, since the only parallel pair of non-equal morphisms is $\mathrm{id}_1, ip : 1 \rightrightarrows 1$ with domain $1$.


category_id: walking_splitting
property_id: preadditive
reason:
We can define $\mathrm{id}_1 + \mathrm{id}_1 := ip$ (and it is clear how to add zero morphisms) and then verify that the axioms of a preadditive category hold. Alternatively, it suffices to find a preadditive category which is isomorphic to the walking splitting: Consider the full subcategory of $\mathbf{Vect}_{\mathbb{F}_2}$ that consists only of the trivial vector space $\{0\}$ and $\mathbb{F}_2$. Since $\mathbf{Vect}_{\mathbb{F}_2}$ is preadditive, it is preadditive as well. It has two objects, two identities, the morphisms $i : \{0\} \to \mathbb{F}_2$, $p : \mathbb{F}_2 \to \{0\}$, and the zero morphism $ip : \mathbb{F}_2 \to \mathbb{F}_2$. Clearly, $pi$ is the identity.


category_id: walking_splitting
property_id: sifted colimits
reason:
We work with the representation of the category as $\mathbf{Vect}^{\leq 1}_{\mathbb{F}_2}$, the category of vector spaces over $\mathbb{F}_2$ of dimension $\leq 1$. It suffices to show that it is closed under sifted colimits in $\mathbf{Vect}_{\mathbb{F}_2}$. More generally, we show this for $\mathbf{Vect}^{\leq d}_K \subseteq \mathbf{Vect}_K$, where $d \in \mathbb{N}$ and $K$ is a field. So let  $X : \mathcal{I} \to \mathbf{Vect}_K$ be a sifted diagram with colimit $(u_i : X_i \to X_\infty)_{i \in \mathcal{I}}$. Since $\mathcal{I}$ is sifted, for finitely many objects $i_1,\dotsc,i_n \in \mathcal{I}$ there is an object $k$ that admits morphisms $i_1 \to k, \dotsc, i_n \to k$; this is all we need to know about $\mathcal{I}$. Assume that each $X_i$ is of dimension $\leq d$, we need to show this for $X_\infty$ as well.<br>
	Every element in $X_\infty$ is a finite sum of elements of the form $u_i(x_i)$ with $x_i \in X_i$. Choose an object $k$ with morphisms $i \to k$ for every occurring $i$. If $y_i \in X_k$ denotes the image of $x_i$, we get $\sum_i u_i(x_i) = \sum_i u_k(y_i) = u_k(\sum_i y_i)$. Therefore, every element of $X_\infty$ has the form $u_i(x_i)$ for some $i \in \mathcal{I}$ and $x_i \in X_i$. Moreover, for finitely many elements in $X_\infty$ the index $i$ may be chosen uniformly.<br>
	Now, if $X_\infty$ has dimension $> d$, it would have linearly independent vectors $v_0,\dotsc,v_d$, all of which have a preimage in $X_i$ for some $i \in \mathcal{I}$. But then these preimages would be linearly independent as well, which contradicts $\dim(X_i) \leq d$.


category_id: walking_splitting
property_id: generalized variety
reason:
Again we work with $\mathbf{Vect}^{\leq 1}_{\mathbb{F}_2}$. We already know that it has sifted colimits and that the embedding to $\mathbf{Vect}_{\mathbb{F}_2}$ preserves them. The object $0$ is initial and hence strongly finitely presentable. The object $\mathbb{F}_2$ is strongly finitely presentable in $\mathbf{Vect}^{\leq 1}_{\mathbb{F}_2}$ since its hom-functor is the composition of the embedding and the forgetful functor $\mathbf{Vect}_{\mathbb{F}_2} \to \mathbf{Set}$, and the latter preserves sifted colimits by <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01, Lemma 3.3]</a> applied to $\mathbb{F}_2 \in \mathbf{Vect}_{\mathbb{F}_2}$.


category_id: walking_splitting
property_id: filtered-colimit-stable monomorphisms
reason:
We saw above that the embedding $\mathbf{Vect}^{\leq 1}_{\mathbb{F}_2} \hookrightarrow \mathbf{Vect}_{\mathbb{F}_2}$ is closed under sifted colimits, hence under filtered colimits. Hence, it preserves filtered colimits. Moreover, it preserves and reflects monomorphisms. Therefore, the claim follows from $\mathbf{Vect}_{\mathbb{F}_2}$.


category_id: Ab
property_id: locally essentially small
reason:
Since it is locally small, it is locally essentially small (by <a href="/category-implication/locally_small_consequence">this result</a>).


category_id: Ab
property_id: generator
reason:
Since it is finitary algebraic, it has a generator (by <a href="/category-implication/free-algebra-generates">this result</a>).


category_id: Ab
property_id: generating set
reason:
Since it has a generator, it has a generating set (by <a href="/category-implication/generator_consequence">this result</a>).


category_id: Ab
property_id: inhabited
reason:
Since it has a generator, it is inhabited (by <a href="/category-implication/generator_consequence">this result</a>).


category_id: Ab
property_id: locally strongly finitely presentable
reason:
Since it is finitary algebraic, it is locally strongly finitely presentable (by <a href="/category-implication/finitary_algebraic_consequence">this result</a>).


category_id: Ab
property_id: additive
reason:
Since it is abelian, it is additive (by <a href="/category-implication/abelian_definition">this result</a>).


category_id: Ab
property_id: finite products
reason:
Since it is additive, it has finite products (by <a href="/category-implication/additive_definition">this result</a>).


category_id: Ab
property_id: binary products
reason:
Since it has finite products, it has binary products (by <a href="/category-implication/finite_products_characterization">this result</a>).


category_id: Ab
property_id: terminal object
reason:
Since it has finite products, it has a terminal object (by <a href="/category-implication/finite_products_characterization">this result</a>).


category_id: Ab
property_id: finite powers
reason:
Since it has finite products, it has finite powers (by <a href="/category-implication/finite_products_include_finite_powers">this result</a>).


category_id: Ab
property_id: binary powers
reason:
Since it has binary products, it has binary powers (by <a href="/category-implication/binary_products_include_binary_powers">this result</a>).


category_id: Ab
property_id: connected
reason:
Since it has a terminal object, it is connected (by <a href="/category-implication/multi-terminal_with_connected">this result</a>).


category_id: Ab
property_id: multi-terminal object
reason:
Since it has a terminal object, it has a multi-terminal object (by <a href="/category-implication/multi-terminal_with_connected">this result</a>).


category_id: Ab
property_id: preadditive
reason:
Since it is additive, it is preadditive (by <a href="/category-implication/additive_definition">this result</a>).


category_id: Ab
property_id: zero morphisms
reason:
Since it is preadditive, it has zero morphisms (by <a href="/category-implication/preadditive_consequence">this result</a>).


category_id: Ab
property_id: strongly connected
reason:
Since it is inhabited and has zero morphisms, it is strongly connected (by <a href="/category-implication/zero_morphisms_mean_strongly_connected">this result</a>).


category_id: Ab
property_id: biproducts
reason:
Since it is additive, it has biproducts (by <a href="/category-implication/additive_biproducts">this result</a>).


category_id: Ab
property_id: finite coproducts
reason:
Since it has biproducts, it has finite coproducts (by <a href="/category-implication/biproducts_definition_consequences">this result</a>).


category_id: Ab
property_id: cokernels
reason:
Since it is abelian, it has cokernels (by <a href="/category-implication/abelian_definition">this result</a>).


category_id: Ab
property_id: conormal
reason:
Since it is abelian, it is conormal (by <a href="/category-implication/abelian_definition">this result</a>).


category_id: Ab
property_id: kernels
reason:
Since it is abelian, it has kernels (by <a href="/category-implication/abelian_definition">this result</a>).


category_id: Ab
property_id: quotients of congruences
reason:
Since it has cokernels and has kernels and is preadditive, it has quotients of congruences (by <a href="/category-implication/congruence_quotients_via_cokernels">this result</a>).


category_id: Ab
property_id: equalizers
reason:
Since it has kernels and is preadditive, it has equalizers (by <a href="/category-implication/equalizers_via_kernels">this result</a>).


category_id: Ab
property_id: coreflexive equalizers
reason:
Since it has equalizers, it has coreflexive equalizers (by <a href="/category-implication/coreflexive_equalizers_are_equalizers">this result</a>).


category_id: Ab
property_id: finitely complete
reason:
Since it has equalizers and has finite products, it is finitely complete (by <a href="/category-implication/finitely_complete_characterization">this result</a>).


category_id: Ab
property_id: pullbacks
reason:
Since it has binary products and has equalizers, it has pullbacks (by <a href="/category-implication/pullbacks_criterion">this result</a>).


category_id: Ab
property_id: Cauchy complete
reason:
Since it has equalizers, it is Cauchy complete (by <a href="/category-implication/equalizers_consequence">this result</a>).


category_id: Ab
property_id: unital
reason:
Since it has biproducts and is finitely complete, it is unital (by <a href="/category-implication/biproducts_unital">this result</a>).


category_id: Ab
property_id: normal
reason:
Since it is abelian, it is normal (by <a href="/category-implication/abelian_definition">this result</a>).


category_id: Ab
property_id: mono-regular
reason:
Since it is normal, it is mono-regular (by <a href="/category-implication/mono_regular_via_kernels">this result</a>).


category_id: Ab
property_id: balanced
reason:
Since it is mono-regular, it is balanced (by <a href="/category-implication/mono_regular_consequence">this result</a>).


category_id: Ab
property_id: regular
reason:
Since it is abelian, it is regular (by <a href="/category-implication/abelian_implies_regular">this result</a>).


category_id: Ab
property_id: well-copowered
reason:
Since it is finitary algebraic, it is well-copowered (by <a href="/category-implication/algebraic_well-copowered">this result</a>).


category_id: Ab
property_id: locally finitely presentable
reason:
Since it is locally strongly finitely presentable, it is locally finitely presentable (by <a href="/category-implication/locally_strongly_finitely_presentable_raise">this result</a>).


category_id: Ab
property_id: locally ℵ₁-presentable
reason:
Since it is locally finitely presentable, it is locally ℵ₁-presentable (by <a href="/category-implication/locally_finitely_presentable_raise">this result</a>).


category_id: Ab
property_id: locally presentable
reason:
Since it is locally ℵ₁-presentable, it is locally presentable (by <a href="/category-implication/locally_countably_presentable_raise">this result</a>).


category_id: Ab
property_id: exact filtered colimits
reason:
Since it is locally finitely presentable, it has exact filtered colimits (by <a href="/category-implication/locally_finitely_presentable_consequence">this result</a>).


category_id: Ab
property_id: filtered colimits
reason:
Since it has exact filtered colimits, it has filtered colimits (by <a href="/category-implication/exact_filtered_colimits_condition">this result</a>).


category_id: Ab
property_id: directed colimits
reason:
Since it has filtered colimits, it has directed colimits (by <a href="/category-implication/directed_colimits_suffice">this result</a>).


category_id: Ab
property_id: cartesian filtered colimits
reason:
Since it has biproducts and has filtered colimits, it has cartesian filtered colimits (by <a href="/category-implication/biproducts_cartesian_filtered_colimits">this result</a>).


category_id: Ab
property_id: filtered-colimit-stable monomorphisms
reason:
Since it has exact filtered colimits, it has filtered-colimit-stable monomorphisms (by <a href="/category-implication/exact_filtered_colimits_monos">this result</a>).


category_id: Ab
property_id: cocomplete
reason:
Since it is locally finitely presentable, it is cocomplete (by <a href="/category-implication/locally_presentable_definition_finite">this result</a>).


category_id: Ab
property_id: finitely accessible
reason:
Since it is locally finitely presentable, it is finitely accessible (by <a href="/category-implication/locally_presentable_definition_finite">this result</a>).


category_id: Ab
property_id: ℵ₁-accessible
reason:
Since it is locally ℵ₁-presentable, it is ℵ₁-accessible (by <a href="/category-implication/locally_presentable_definition_countable">this result</a>).


category_id: Ab
property_id: accessible
reason:
Since it is locally presentable, it is accessible (by <a href="/category-implication/locally_presentable_definition">this result</a>).


category_id: Ab
property_id: well-powered
reason:
Since it is accessible, it is well-powered (by <a href="/category-implication/accessible_well-powered">this result</a>).


category_id: Ab
property_id: generalized variety
reason:
Since it is locally strongly finitely presentable, it is a generalized variety (by <a href="/category-implication/locally_strongly_finitely_presentable_definition">this result</a>).


category_id: Ab
property_id: sifted colimits
reason:
Since it is a generalized variety, it has sifted colimits (by <a href="/category-implication/generalized_variety_require_sifted_colimit">this result</a>).


category_id: Ab
property_id: reflexive coequalizers
reason:
Since it has sifted colimits, it has reflexive coequalizers (by <a href="/category-implication/sifted_colimits_consequence">this result</a>).


category_id: Ab
property_id: multi-algebraic
reason:
Since it is locally strongly finitely presentable, it is multi-algebraic (by <a href="/category-implication/varieties_are_multi-algebraic">this result</a>).


category_id: Ab
property_id: locally finitely multi-presentable
reason:
Since it is multi-algebraic, it is locally finitely multi-presentable (by <a href="/category-implication/multi-algebraic_implies_locally_finitely_multi-presentable">this result</a>).


category_id: Ab
property_id: connected limits
reason:
Since it is locally finitely multi-presentable, it has connected limits (by <a href="/category-implication/locally_finitely_multi-presentable_definition">this result</a>).


category_id: Ab
property_id: wide pullbacks
reason:
Since it has connected limits, it has wide pullbacks (by <a href="/category-implication/connected_limits_characterization">this result</a>).


category_id: Ab
property_id: complete
reason:
Since it has a terminal object and has wide pullbacks, it is complete (by <a href="/category-implication/complete_criterion">this result</a>).


category_id: Ab
property_id: products
reason:
Since it is complete, it has products (by <a href="/category-implication/complete_characterization">this result</a>).


category_id: Ab
property_id: countable products
reason:
Since it has products, it has countable products (by <a href="/category-implication/products_consequences">this result</a>).


category_id: Ab
property_id: powers
reason:
Since it has products, it has powers (by <a href="/category-implication/products_consequences">this result</a>).


category_id: Ab
property_id: countable powers
reason:
Since it has countable products, it has countable powers (by <a href="/category-implication/countable_products_consequence">this result</a>).


category_id: Ab
property_id: cofiltered limits
reason:
Since it has wide pullbacks, it has cofiltered limits (by <a href="/category-implication/wide_pullbacks_criterion">this result</a>).


category_id: Ab
property_id: sequential limits
reason:
Since it has countable products and has equalizers, it has sequential limits (by <a href="/category-implication/sequential_limits_criterion">this result</a>).


category_id: Ab
property_id: multi-complete
reason:
Since it is complete, it is multi-complete (by <a href="/category-implication/multi-complete_generalize_limits">this result</a>).


category_id: Ab
property_id: CIP
reason:
Since it has biproducts and has filtered colimits and has filtered-colimit-stable monomorphisms and has products, it satisfies CIP (by <a href="/category-implication/CIP_criterion">this result</a>).


category_id: Ab
property_id: coproducts
reason:
Since it satisfies CIP, it has coproducts (by <a href="/category-implication/CIP_assumption">this result</a>).


category_id: Ab
property_id: Grothendieck abelian
reason:
Since it is abelian and has coproducts and has exact filtered colimits and has a generator, it is Grothendieck abelian (by <a href="/category-implication/grothendieck_abelian_definition">this result</a>).


category_id: Ab
property_id: cogenerator
reason:
Since it is Grothendieck abelian, it has a cogenerator (by <a href="/category-implication/grothendieck_abelian_cogenerator">this result</a>).


category_id: Ab
property_id: locally multi-presentable
reason:
Since it is accessible and has connected limits, it is locally multi-presentable (by <a href="/category-implication/locally_multi-presentable_definition">this result</a>).


category_id: Ab
property_id: multi-cocomplete
reason:
Since it is locally multi-presentable, it is multi-cocomplete (by <a href="/category-implication/locally_multi-presentable_another_definition">this result</a>).


category_id: Ab
property_id: locally poly-presentable
reason:
Since it is accessible and has wide pullbacks, it is locally poly-presentable (by <a href="/category-implication/locally_poly-presentable_definition">this result</a>).


category_id: Ab
property_id: semi-strongly connected
reason:
Since it is strongly connected, it is semi-strongly connected (by <a href="/category-implication/strongly_connected_consequence">this result</a>).


category_id: Ab
property_id: disjoint finite products
reason:
Since it has finite products and is strongly connected, it has disjoint finite products (by <a href="/category-implication/strongly_connected_disjoint_products">this result</a>).


category_id: Ab
property_id: Malcev
reason:
Since it is additive and is finitely complete, it is Malcev (by <a href="/category-implication/malcev_additive_criterion">this result</a>).


category_id: Ab
property_id: pointed
reason:
Since it is unital, it is pointed (by <a href="/category-implication/unital_assumptions">this result</a>).


category_id: Ab
property_id: initial object
reason:
Since it is pointed, it has an initial object (by <a href="/category-implication/pointed_characterization">this result</a>).


category_id: Ab
property_id: disjoint products
reason:
Since it is finitary algebraic and is pointed, it has disjoint products (by <a href="/category-implication/algebras_with_0_disjoint_products">this result</a>).


category_id: Ab
property_id: filtered
reason:
Since it has a terminal object, it is filtered (by <a href="/category-implication/terminal_object_yields_filtered">this result</a>).


category_id: Ab
property_id: sifted
reason:
Since it is filtered, it is sifted (by <a href="/category-implication/filtered_is_sifted">this result</a>).


category_id: Ab
property_id: connected colimits
reason:
Since it is cocomplete, it has connected colimits (by <a href="/category-implication/dual_complete_consequence">this result</a>).


category_id: Ab
property_id: finitely cocomplete
reason:
Since it is cocomplete, it is finitely cocomplete (by <a href="/category-implication/dual_complete_consequence">this result</a>).


category_id: Ab
property_id: cosifted limits
reason:
Since it has connected limits, it has cosifted limits (by <a href="/category-implication/dual_sifted_categories_are_connected">this result</a>).


category_id: Ab
property_id: coquotients of cocongruences
reason:
Since it has coreflexive equalizers, it has coquotients of cocongruences (by <a href="/category-implication/dual_congruence_quotients_are_reflexive_coequalizers">this result</a>).


category_id: Ab
property_id: copowers
reason:
Since it has coproducts, it has copowers (by <a href="/category-implication/dual_products_consequences">this result</a>).


category_id: Ab
property_id: countable coproducts
reason:
Since it has coproducts, it has countable coproducts (by <a href="/category-implication/dual_products_consequences">this result</a>).


category_id: Ab
property_id: binary coproducts
reason:
Since it has finite coproducts, it has binary coproducts (by <a href="/category-implication/dual_finite_products_characterization">this result</a>).


category_id: Ab
property_id: coequalizers
reason:
Since it has binary coproducts and has reflexive coequalizers, it has coequalizers (by <a href="/category-implication/dual_equalizers_via_coreflexive_equalizers">this result</a>).


category_id: Ab
property_id: countable copowers
reason:
Since it has countable coproducts, it has countable copowers (by <a href="/category-implication/dual_countable_products_consequence">this result</a>).


category_id: Ab
property_id: sequential colimits
reason:
Since it has directed colimits, it has sequential colimits (by <a href="/category-implication/dual_directed_limits_consequence">this result</a>).


category_id: Ab
property_id: pushouts
reason:
Since it has binary coproducts and has coequalizers, it has pushouts (by <a href="/category-implication/dual_pullbacks_criterion">this result</a>).


category_id: Ab
property_id: directed limits
reason:
Since it has cofiltered limits, it has directed limits (by <a href="/category-implication/dual_directed_colimits_suffice">this result</a>).


category_id: Ab
property_id: wide pushouts
reason:
Since it has connected colimits, it has wide pushouts (by <a href="/category-implication/dual_connected_limits_characterization">this result</a>).


category_id: Ab
property_id: finite copowers
reason:
Since it has finite coproducts, it has finite copowers (by <a href="/category-implication/dual_finite_products_include_finite_powers">this result</a>).


category_id: Ab
property_id: binary copowers
reason:
Since it has binary coproducts, it has binary copowers (by <a href="/category-implication/dual_binary_products_include_binary_powers">this result</a>).


category_id: Ab
property_id: multi-initial object
reason:
Since it is multi-cocomplete, it has a multi-initial object (by <a href="/category-implication/dual_multi-terminal_special_case">this result</a>).


category_id: Ab
property_id: counital
reason:
Since it has biproducts and is finitely cocomplete, it is counital (by <a href="/category-implication/dual_biproducts_unital">this result</a>).


category_id: Ab
property_id: cocartesian cofiltered limits
reason:
Since it has biproducts and has cofiltered limits, it has cocartesian cofiltered limits (by <a href="/category-implication/dual_biproducts_cartesian_filtered_colimits">this result</a>).


category_id: Ab
property_id: cogenerating set
reason:
Since it has a cogenerator, it has a cogenerating set (by <a href="/category-implication/dual_generator_consequence">this result</a>).


category_id: Ab
property_id: epi-regular
reason:
Since it is conormal, it is epi-regular (by <a href="/category-implication/dual_mono_regular_via_kernels">this result</a>).


category_id: Ab
property_id: coregular
reason:
Since it is abelian, it is coregular (by <a href="/category-implication/dual_abelian_implies_regular">this result</a>).


category_id: Ab
property_id: disjoint finite coproducts
reason:
Since it has finite coproducts and is strongly connected, it has disjoint finite coproducts (by <a href="/category-implication/dual_strongly_connected_disjoint_products">this result</a>).


category_id: Ab
property_id: disjoint coproducts
reason:
Since it has coproducts and has disjoint finite coproducts, it has disjoint coproducts (by <a href="/category-implication/disjoint_coproducts_equivalence">this result</a>).


category_id: Ab
property_id: co-Malcev
reason:
Since it is additive and is finitely cocomplete, it is co-Malcev (by <a href="/category-implication/dual_malcev_additive_criterion">this result</a>).


category_id: Ab
property_id: cosifted
reason:
Since it has binary products and is inhabited, it is cosifted (by <a href="/category-implication/dual_coproducts_implies_sifted">this result</a>).


category_id: Ab
property_id: cofiltered
reason:
Since it has an initial object, it is cofiltered (by <a href="/category-implication/dual_terminal_object_yields_filtered">this result</a>).


category_id: Ab
property_id: direct
reason:
Assume for contradiction that it is direct. Then it would be skeletal (by <a href="/category-implication/direct_implies_one-way">this result</a>) – contradiction.


category_id: Ab
property_id: gaunt
reason:
Assume for contradiction that it is gaunt. Then it would be skeletal (by <a href="/category-implication/gaunt_characterization">this result</a>) – contradiction.


category_id: Ab
property_id: discrete
reason:
Assume for contradiction that it is discrete. Then it would be direct (by <a href="/category-implication/discrete_consequences">this result</a>) – contradiction.


category_id: Ab
property_id: trivial
reason:
Assume for contradiction that it is trivial. Then it would be split abelian (by <a href="/category-implication/trivial_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: thin
reason:
Assume for contradiction that it is thin. Then it would have disjoint finite coproducts and would be thin, so it would be trivial (by <a href="/category-implication/disjoint_coproducts_thin_trivial">this result</a>) – contradiction.


category_id: Ab
property_id: strict terminal object
reason:
Assume for contradiction that it has a strict terminal object. Then it would have disjoint finite coproducts and would have a strict terminal object, so it would be thin (by <a href="/category-implication/disjoint_coproducts_strict">this result</a>) – contradiction.


category_id: Ab
property_id: strict initial object
reason:
Assume for contradiction that it has a strict initial object. Then it would be pointed and would have a strict initial object, so it would be trivial (by <a href="/category-implication/strict_initial_trivial">this result</a>) – contradiction.


category_id: Ab
property_id: left cancellative
reason:
Assume for contradiction that it is left cancellative. Then it would have an initial object and would be left cancellative, so it would have a strict initial object (by <a href="/category-implication/strict_initial_left_criterion">this result</a>) – contradiction.


category_id: Ab
property_id: right cancellative
reason:
Assume for contradiction that it is right cancellative. Then it would have an initial object and would be right cancellative, so it would have a strict initial object (by <a href="/category-implication/strict_initial_right_criterion">this result</a>) – contradiction.


category_id: Ab
property_id: distributive
reason:
Assume for contradiction that it is distributive. Then it would have a strict initial object (by <a href="/category-implication/distributive_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: countably distributive
reason:
Assume for contradiction that it is countably distributive. Then it would be distributive (by <a href="/category-implication/countably_distributive_trivial">this result</a>) – contradiction.


category_id: Ab
property_id: infinitary distributive
reason:
Assume for contradiction that it is infinitary distributive. Then it would be countably distributive (by <a href="/category-implication/infinitary_distributive_trivial">this result</a>) – contradiction.


category_id: Ab
property_id: cartesian closed
reason:
Assume for contradiction that it is cartesian closed. Then it would be cartesian closed and would have finite coproducts, so it would be distributive (by <a href="/category-implication/distributive_criterion">this result</a>) – contradiction.


category_id: Ab
property_id: extensive
reason:
Assume for contradiction that it is extensive. Then it would have a strict initial object (by <a href="/category-implication/extensive_consequences">this result</a>) – contradiction.


category_id: Ab
property_id: infinitary extensive
reason:
Assume for contradiction that it is infinitary extensive. Then it would be extensive (by <a href="/category-implication/infinitary_extensive_finitary">this result</a>) – contradiction.


category_id: Ab
property_id: groupoid
reason:
Assume for contradiction that it is a groupoid. Then it would be left cancellative (by <a href="/category-implication/groupoid_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: one-way
reason:
Assume for contradiction that it is one-way. Then it would be one-way and would have zero morphisms, so it would be thin (by <a href="/category-implication/one-way_zero">this result</a>) – contradiction.


category_id: Ab
property_id: core-thin
reason:
Assume for contradiction that it is core-thin. Then it would have binary powers and would be core-thin, so it would be thin (by <a href="/category-implication/core-thin_products_thin">this result</a>) – contradiction.


category_id: Ab
property_id: self-dual
reason:
Assume for contradiction that it is self-dual. Then it would be Grothendieck abelian and would be self-dual, so it would be trivial (by <a href="/category-implication/grothendieck_abelian_self-dual">this result</a>) – contradiction.


category_id: Ab
property_id: essentially discrete
reason:
Assume for contradiction that it is essentially discrete. Then it would be a groupoid (by <a href="/category-implication/essentially_discrete_characterization">this result</a>) – contradiction.


category_id: Ab
property_id: essentially small
reason:
Assume for contradiction that it is essentially small. Then it would be essentially small and would have powers, so it would be thin (by <a href="/category-implication/freyd_small">this result</a>) – contradiction.


category_id: Ab
property_id: small
reason:
Assume for contradiction that it is small. Then it would be essentially small (by <a href="/category-implication/small_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: essentially countable
reason:
Assume for contradiction that it is essentially countable. Then it would be essentially small (by <a href="/category-implication/essentially_countable_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: essentially finite
reason:
Assume for contradiction that it is essentially finite. Then it would be essentially countable (by <a href="/category-implication/essentially_finite_raise">this result</a>) – contradiction.


category_id: Ab
property_id: finite
reason:
Assume for contradiction that it is finite. Then it would be essentially finite (by <a href="/category-implication/finite_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: countable
reason:
Assume for contradiction that it is countable. Then it would be essentially countable (by <a href="/category-implication/countable_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: locally copresentable
reason:
Assume for contradiction that it is locally copresentable. Then it would be locally copresentable and would be locally presentable, so it would be essentially small (by <a href="/category-implication/locally_presentable_essentially_small">this result</a>) – contradiction.


category_id: Ab
property_id: elementary topos
reason:
Assume for contradiction that it is an elementary topos. Then it would be cartesian closed (by <a href="/category-implication/topos_definition">this result</a>) – contradiction.


category_id: Ab
property_id: regular subobject classifier
reason:
Assume for contradiction that it has a regular subobject classifier. Then it would be additive and would have a regular subobject classifier, so it would be trivial (by <a href="/category-implication/additive_trivial_condition">this result</a>) – contradiction.


category_id: Ab
property_id: subobject classifier
reason:
Assume for contradiction that it has a subobject classifier. Then it would have a regular subobject classifier (by <a href="/category-implication/regular_subobject_weaker">this result</a>) – contradiction.


category_id: Ab
property_id: locally cartesian closed
reason:
Assume for contradiction that it is locally cartesian closed. Then it would be locally cartesian closed and would have a terminal object, so it would be cartesian closed (by <a href="/category-implication/locally_cartesian_closed_with_terminal_is_closed">this result</a>) – contradiction.


category_id: Ab
property_id: Grothendieck topos
reason:
Assume for contradiction that it is a Grothendieck topos. Then it would be an elementary topos (by <a href="/category-implication/grothendieck_topos_definition">this result</a>) – contradiction.


category_id: Ab
property_id: natural numbers object
reason:
Assume for contradiction that it has a natural numbers object. Then it would have a natural numbers object and would be pointed, so it would be trivial (by <a href="/category-implication/nno_pointed_case">this result</a>) – contradiction.


category_id: Ab
property_id: codistributive
reason:
Assume for contradiction that it is codistributive. Then it would have a strict terminal object (by <a href="/category-implication/dual_distributive_consequence">this result</a>) – contradiction.


category_id: Ab
property_id: countably codistributive
reason:
Assume for contradiction that it is countably codistributive. Then it would be codistributive (by <a href="/category-implication/dual_countably_distributive_trivial">this result</a>) – contradiction.


category_id: Ab
property_id: infinitary codistributive
reason:
Assume for contradiction that it is infinitary codistributive. Then it would be countably codistributive (by <a href="/category-implication/dual_infinitary_distributive_trivial">this result</a>) – contradiction.


category_id: Ab
property_id: cocartesian coclosed
reason:
Assume for contradiction that it is cocartesian coclosed. Then it would be cocartesian coclosed and would have finite products, so it would be codistributive (by <a href="/category-implication/dual_distributive_criterion">this result</a>) – contradiction.


category_id: Ab
property_id: coextensive
reason:
Assume for contradiction that it is coextensive. Then it would have a strict terminal object (by <a href="/category-implication/dual_extensive_consequences">this result</a>) – contradiction.


category_id: Ab
property_id: infinitary coextensive
reason:
Assume for contradiction that it is infinitary coextensive. Then it would be coextensive (by <a href="/category-implication/dual_infinitary_extensive_finitary">this result</a>) – contradiction.


category_id: Ab
property_id: cofiltered-limit-stable epimorphisms
reason:
Assume for contradiction that it has cofiltered-limit-stable epimorphisms. Then it would have biproducts and would have cofiltered limits and would have cofiltered-limit-stable epimorphisms and would have coproducts, so it would satisfy CSP (by <a href="/category-implication/dual_CIP_criterion">this result</a>) – contradiction.


category_id: Ab
property_id: exact cofiltered limits
reason:
Assume for contradiction that it has exact cofiltered limits. Then it would have cofiltered-limit-stable epimorphisms (by <a href="/category-implication/dual_exact_filtered_colimits_monos">this result</a>) – contradiction.


category_id: Ab
property_id: inverse
reason:
Assume for contradiction that it is inverse. Then it would be one-way (by <a href="/category-implication/dual_direct_implies_one-way">this result</a>) – contradiction.


category_id: Ab
property_id: coaccessible
reason:
Assume for contradiction that it is coaccessible. Then it would be coaccessible and would be complete, so it would be locally copresentable (by <a href="/category-implication/dual_locally_presentable_definition">this result</a>) – contradiction.


category_id: Ab
property_id: regular quotient object classifier
reason:
Assume for contradiction that it has a regular quotient object classifier. Then it would be additive and would have a regular quotient object classifier, so it would be trivial (by <a href="/category-implication/dual_additive_trivial_condition">this result</a>) – contradiction.


category_id: Ab
property_id: quotient object classifier
reason:
Assume for contradiction that it has a quotient object classifier. Then it would have a regular quotient object classifier (by <a href="/category-implication/dual_regular_subobject_weaker">this result</a>) – contradiction.


category_id: Ab
property_id: locally cocartesian coclosed
reason:
Assume for contradiction that it is locally cocartesian coclosed. Then it would have an initial object and would be locally cocartesian coclosed, so it would be cocartesian coclosed (by <a href="/category-implication/dual_locally_cartesian_closed_with_terminal_is_closed">this result</a>) – contradiction.


category_id: Sh(X,Ab)
property_id: abelian
reason:
Since it is Grothendieck abelian, it is abelian (by <a href="/category-implication/grothendieck_abelian_definition">this result</a>).


category_id: Sh(X,Ab)
property_id: coproducts
reason:
Since it is Grothendieck abelian, it has coproducts (by <a href="/category-implication/grothendieck_abelian_definition">this result</a>).


category_id: Sh(X,Ab)
property_id: exact filtered colimits
reason:
Since it is Grothendieck abelian, it has exact filtered colimits (by <a href="/category-implication/grothendieck_abelian_definition">this result</a>).


category_id: Sh(X,Ab)
property_id: generator
reason:
Since it is Grothendieck abelian, it has a generator (by <a href="/category-implication/grothendieck_abelian_definition">this result</a>).


category_id: Sh(X,Ab)
property_id: locally presentable
reason:
Since it is Grothendieck abelian, it is locally presentable (by <a href="/category-implication/grothendieck_abelian_presentable">this result</a>).


category_id: Sh(X,Ab)
property_id: cocomplete
reason:
Since it is locally presentable, it is cocomplete (by <a href="/category-implication/locally_presentable_definition">this result</a>).


category_id: Sh(X,Ab)
property_id: sifted colimits
reason:
Since it has connected colimits, it has sifted colimits (by <a href="/category-implication/sifted_categories_are_connected">this result</a>).


category_id: Sh(X,Ab)
property_id: well-copowered
reason:
Since it is accessible and has pushouts, it is well-copowered (by <a href="/category-implication/accessible_well-copowered">this result</a>).


category_id: Sh(X,Ab)
property_id: multi-cocomplete
reason:
Since it is cocomplete, it is multi-cocomplete (by <a href="/category-implication/dual_multi-complete_generalize_limits">this result</a>).


category_id: Sh(X,Ab)
property_id: locally multi-presentable
reason:
Since it is accessible and is multi-cocomplete, it is locally multi-presentable (by <a href="/category-implication/locally_multi-presentable_another_definition">this result</a>).


category_id: Sh(X,Ab)
property_id: connected limits
reason:
Since it is locally multi-presentable, it has connected limits (by <a href="/category-implication/locally_multi-presentable_definition">this result</a>).


category_id: Sh(X,Ab)
property_id: disjoint products
reason:
Since it has disjoint finite products and has products, it has disjoint products (by <a href="/category-implication/dual_disjoint_coproducts_equivalence">this result</a>).


category_id: Alg(R)
property_id: regular
reason:
Since it is locally strongly finitely presentable, it is regular (by <a href="/category-implication/many_sorted_algebraic_implies_regular">this result</a>).


category_id: Alg(R)
property_id: finitely complete
reason:
Since it is regular, it is finitely complete (by <a href="/category-implication/regular_def">this result</a>).


category_id: Alg(R)
property_id: equalizers
reason:
Since it is finitely complete, it has equalizers (by <a href="/category-implication/finitely_complete_characterization">this result</a>).


category_id: Alg(R)
property_id: finite products
reason:
Since it is finitely complete, it has finite products (by <a href="/category-implication/finitely_complete_characterization">this result</a>).


category_id: Alg(R)
property_id: cartesian filtered colimits
reason:
Since it has exact filtered colimits, it has cartesian filtered colimits (by <a href="/category-implication/exact_includes_cartesian_filtered_colimits">this result</a>).


category_id: Alg(R)
property_id: quotients of congruences
reason:
Since it has reflexive coequalizers, it has quotients of congruences (by <a href="/category-implication/congruence_quotients_are_reflexive_coequalizers">this result</a>).


category_id: Alg(R)
property_id: coequalizers
reason:
Since it is cocomplete, it has coequalizers (by <a href="/category-implication/dual_complete_characterization">this result</a>).


category_id: Alg(R)
property_id: coproducts
reason:
Since it is cocomplete, it has coproducts (by <a href="/category-implication/dual_complete_characterization">this result</a>).


category_id: Alg(R)
property_id: finite coproducts
reason:
Since it has coproducts, it has finite coproducts (by <a href="/category-implication/dual_products_consequences">this result</a>).


category_id: Alg(R)
property_id: initial object
reason:
Since it has finite coproducts, it has an initial object (by <a href="/category-implication/dual_finite_products_characterization">this result</a>).


category_id: Alg(R)
property_id: extensive
reason:
Assume for contradiction that it is extensive. Then it would have cofiltered limits and would be extensive and would have a terminal object, so it would have cocartesian cofiltered limits (by <a href="/category-implication/extensive_cocartesian_cofiltered_limits">this result</a>) – contradiction.


category_id: Alg(R)
property_id: mono-regular
reason:
Assume for contradiction that it is mono-regular. Then it would be balanced (by <a href="/category-implication/mono_regular_consequence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: groupoid
reason:
Assume for contradiction that it is a groupoid. Then it would be mono-regular (by <a href="/category-implication/groupoid_consequence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: normal
reason:
Assume for contradiction that it is normal. Then it would be mono-regular (by <a href="/category-implication/mono_regular_via_kernels">this result</a>) – contradiction.


category_id: Alg(R)
property_id: abelian
reason:
Assume for contradiction that it is abelian. Then it would be normal (by <a href="/category-implication/abelian_definition">this result</a>) – contradiction.


category_id: Alg(R)
property_id: Grothendieck abelian
reason:
Assume for contradiction that it is Grothendieck abelian. Then it would be abelian (by <a href="/category-implication/grothendieck_abelian_definition">this result</a>) – contradiction.


category_id: Alg(R)
property_id: split abelian
reason:
Assume for contradiction that it is split abelian. Then it would be abelian (by <a href="/category-implication/split_abelian_condition">this result</a>) – contradiction.


category_id: Alg(R)
property_id: trivial
reason:
Assume for contradiction that it is trivial. Then it would be essentially discrete (by <a href="/category-implication/trivial_consequence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: subobject classifier
reason:
Assume for contradiction that it has a subobject classifier. Then it would be mono-regular (by <a href="/category-implication/subobject_classifier_consequence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: elementary topos
reason:
Assume for contradiction that it is an elementary topos. Then it would have a subobject classifier (by <a href="/category-implication/topos_definition">this result</a>) – contradiction.


category_id: Alg(R)
property_id: strongly connected
reason:
Assume for contradiction that it is strongly connected. Then it would be semi-strongly connected (by <a href="/category-implication/strongly_connected_consequence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: zero morphisms
reason:
Assume for contradiction that it has zero morphisms. Then it would be inhabited and would have zero morphisms, so it would be strongly connected (by <a href="/category-implication/zero_morphisms_mean_strongly_connected">this result</a>) – contradiction.


category_id: Alg(R)
property_id: kernels
reason:
Assume for contradiction that it has kernels. Then it would have zero morphisms (by <a href="/category-implication/kernels_condition">this result</a>) – contradiction.


category_id: Alg(R)
property_id: biproducts
reason:
Assume for contradiction that it has biproducts. Then it would have zero morphisms (by <a href="/category-implication/biproducts_definition_consequences">this result</a>) – contradiction.


category_id: Alg(R)
property_id: pointed
reason:
Assume for contradiction that it is pointed. Then it would have zero morphisms (by <a href="/category-implication/pointed_characterization">this result</a>) – contradiction.


category_id: Alg(R)
property_id: CIP
reason:
Assume for contradiction that it satisfies CIP. Then it would have zero morphisms (by <a href="/category-implication/CIP_assumption">this result</a>) – contradiction.


category_id: Alg(R)
property_id: preadditive
reason:
Assume for contradiction that it is preadditive. Then it would have zero morphisms (by <a href="/category-implication/preadditive_consequence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: additive
reason:
Assume for contradiction that it is additive. Then it would be preadditive (by <a href="/category-implication/additive_definition">this result</a>) – contradiction.


category_id: Alg(R)
property_id: unital
reason:
Assume for contradiction that it is unital. Then it would be pointed (by <a href="/category-implication/unital_assumptions">this result</a>) – contradiction.


category_id: Alg(R)
property_id: cokernels
reason:
Assume for contradiction that it has cokernels. Then it would have zero morphisms (by <a href="/category-implication/dual_kernels_condition">this result</a>) – contradiction.


category_id: Alg(R)
property_id: thin
reason:
Assume for contradiction that it is thin. Then it would have disjoint finite products and would be thin, so it would be trivial (by <a href="/category-implication/dual_disjoint_coproducts_thin_trivial">this result</a>) – contradiction.


category_id: Alg(R)
property_id: disjoint finite coproducts
reason:
Assume for contradiction that it has disjoint finite coproducts. Then it would have disjoint finite coproducts and would have a strict terminal object, so it would be thin (by <a href="/category-implication/disjoint_coproducts_strict">this result</a>) – contradiction.


category_id: Alg(R)
property_id: disjoint coproducts
reason:
Assume for contradiction that it has disjoint coproducts. Then it would have disjoint finite coproducts (by <a href="/category-implication/disjoint_coproducts_equivalence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: right cancellative
reason:
Assume for contradiction that it is right cancellative. Then it would have equalizers and would be right cancellative, so it would be thin (by <a href="/category-implication/cancellative_consequence">this result</a>) – contradiction.


category_id: Alg(R)
property_id: left cancellative
reason:
Assume for contradiction that it is left cancellative. Then it would have binary copowers and would be left cancellative, so it would be thin (by <a href="/category-implication/codiagonal-no-mono">this result</a>) – contradiction.


category_id: Alg(R)
property_id: one-way
reason:
Assume for contradiction that it is one-way. Then it would be core-thin (by <a href="/category-implication/one-way_implies_core-thin">this result</a>) – contradiction.


category_id: Alg(R)
property_id: cartesian closed
reason:
Assume for contradiction that it is cartesian closed. Then it would be cartesian closed and would have a strict terminal object, so it would be thin (by <a href="/category-implication/ccc_no_strict_terminal">this result</a>) – contradiction.


category_id: Alg(R)
property_id: regular subobject classifier
reason:
Assume for contradiction that it has a regular subobject classifier. Then it would have a regular subobject classifier and would have a strict terminal object, so it would be thin (by <a href="/category-implication/regular_subobject_classifier_strict">this result</a>) – contradiction.


category_id: Alg(R)
property_id: natural numbers object
reason:
Assume for contradiction that it has a natural numbers object. Then it would have a natural numbers object and would have a strict terminal object, so it would be one-way (by <a href="/category-implication/nno_terminal">this result</a>) – contradiction.


category_id: Alg(R)
property_id: countably distributive
reason:
Assume for contradiction that it is countably distributive. Then it would have a natural numbers object (by <a href="/category-implication/nno_criterion">this result</a>) – contradiction.


category_id: Alg(R)
property_id: distributive
reason:
Assume for contradiction that it is distributive. Then it would have cartesian filtered colimits and would have coproducts and would be distributive, so it would be infinitary distributive (by <a href="/category-implication/infinite_distributive_filtered_criterion">this result</a>) – contradiction.


