Results on subcategories

This page collects several useful results of the following form: if $U : \C \to \D$ is a faithful functor (perhaps even fully faithful, or satisfying additional assumptions) and $\D$ has a certain property, then $\C$ has this property as well.

Lemma 1.

Let $\D$ be a category with a (regular) subobject classifier $\Omega$ . Assume that $U : \C \to \D$ is a fully faithful functor such that (1) $U$ is coreflective, i.e. there is a functor $R : \D \to \C$ right adjoint to $U$ , and (2) every (regular) monomorphism $Y \to U(X)$ in $\D$ is the image of a (regular) monomorphism $X' \to X$ in $\C$ . Then $R(\Omega)$ is a (regular) subobject classifier in $\C$ .

Proof. If $X \in \C$ , then $\Hom(X,R(\Omega)) \cong \Hom(U(X),\Omega) \cong \Sub(U(X)) \cong \Sub(X).$ The same proof works for regular subobjects. $\square$

Lemma 2.

Let $\C$ be a category with filtered colimits. Assume that $U : \C \to \D$ is a faithful functor that preserves monomorphisms and filtered colimits. If monomorphisms in $\D$ are stable under filtered colimits, then the same is true in $\C$ .

For the record, here is the dual statement: let $\C$ be a category with cofiltered limits. Assume that $U : \C \to \D$ is a faithful functor that preserves epimorphisms and cofiltered limits. If epimorphisms in $\D$ are stable under cofiltered limits, then the same is true in $\C$ .

Proof. If $(f_i : X_i \to Y_i)$ is a filtered diagram of monomorphisms in $\C$ , it induces a filtered diagram $(U(f_i) : U(X_i) \to U(Y_i))$ of monomorphisms in $\D$ . Hence, its colimit $\colim_i U(f_i) : \colim_i U(X_i) \to \colim_i U(Y_i)$ is a monomorphism in $\D$ . This morphism is isomorphic to $U(\colim_i f_i) : U(\colim_i X_i) \to U(\colim_i Y_i)$ . Since $U(\colim_i f_i)$ is a monomorphism in $\D$ and $U$ is faithful, it follows that $\colim_i f_i$ is a monomorphism in $\C$ . $\square$

Lemma 3.

Let $U : \C \to \D$ be a fully faithful functor with a left adjoint $L : \D \to \C$ (i.e. $\C$ is equivalent to a reflective subcategory of $\D$ ). Assume that $\D$ has exact filtered colimits, that $\C$ has finite limits, and that $L$ preserves finite limits. Then $\C$ also has exact filtered colimits.

Proof. It is well known (and easy to prove) that the colimit of a diagram $(X_j)$ in $\C$ is given by $L(\colim_j U(X_j))$ , provided that the colimit in $\D$ exists. In particular, $\C$ has filtered colimits. By assumption, it also has finite limits, and $U$ preserves them since it is a right adjoint. Now let $X : \I \times \J \to \C$ be a diagram, where $\I$ is finite and $\J$ is filtered. We compute:

$\begin{align*} \colim_j {\lim}_i X(i,j) & \cong L(\colim_j U({\lim}_i X(i,j))) \\ & \cong L(\colim_j {\lim}_i U(X(i,j))) \\ & \cong L({\lim}_i \colim_j U(X(i,j))) \\ & \cong {\lim}_i L(\colim_j U(X(i,j))) \\ & \cong {\lim}_i \colim_j X(i,j) \end{align*}$

$\square$

Lemma 4.

Let $U : \C \to \D$ be a fully faithful functor. Assume that $\C$ has finite products and filtered colimits, and that $U$ preserves binary products and filtered colimits. If $\D$ has cartesian filtered colimits, then so does $\C$ .

Proof. Let $X$ be an object of $\C$ and $Y : \I \to \C$ a filtered diagram. Then we have the canonical comparison map $c : \colim_{i\in\I} (X \times Y_i) \to X \times \colim_{i\in\I} Y_i$ . By the assumptions, $Uc$ is equivalent to the comparison map $\colim_{i\in\I} (UX \times UY_i) \to UX \times \colim_{i\in\I} UY_i$ , which is an isomorphism. Since $U$ is fully faithful and therefore conservative, we conclude that $c$ is an isomorphism. $\square$

Lemma 5.

Let $U : \C \to \D$ be a fully faithful functor with a right adjoint $R : \D \to \C$ (i.e. $\C$ is equivalent to a coreflective subcategory of $\D$ ). Assume that $\C$ has binary products, and that $U$ preserves these binary products. If $\D$ is cartesian closed, then so is $\C$ , with exponentials in $\C$ given by $[X, Y]_{\C} \cong R([UX, UY]_{\D}).$

Proof. For any objects $X, Y, Z$ of $\C$ we have natural isomorphisms

$\begin{align*} \Hom_\C(Z\times X, Y) & \cong \Hom_\D(UZ \times UX, UY) \\ & \cong \Hom_\D(UZ, [UX,UY]) \\ & \cong \Hom_\C\bigl(Z, R([UX,UY])\bigr). \end{align*}$

$\square$

Corollary 6.

If $\C$ is a cartesian closed category and $P$ is a subterminal object of $\C$ , then the slice category $\C / P$ is also cartesian closed, with exponentials in $\C / P$ given by $[X, Y]_{\C / P} \cong [X, Y]_{\C} \times P.$

Proof. The forgetful functor $\C / P \to \C$ is fully faithful; it has right adjoint ${-} \times P$ ; and it preserves binary products (in fact all inhabited limits). Hence, Lemma 5 applies. $\square$

Lemma 7.

Let $U : \C \to \D$ be a fully faithful functor. Assume that $\C$ has finite limits and coequalizers, and that $U$ preserves pullbacks and coequalizers. If $\D$ is regular, then so is $\C$ .

Proof. Since $\C$ has finite limits and coequalizers, the only nontrivial part of proving $\C$ is regular is to check that regular epimorphisms are stable under pullback in $\C$ . Since $U$ preserves pullbacks and regular epimorphisms, it suffices to show that $U$ reflects regular epimorphisms. Thus, suppose $f : X \to Y$ is a morphism in $\C$ with $Uf$ a regular epimorphism. Then in $\C$ we have the diagram $X \times_Y X \rightrightarrows X \to \im(f) \xrightarrow{i} Y$ where $X \times_Y X$ is the kernel pair of $f$ , and $\im(f)$ is the coequalizer. By the assumptions, the image under $U$ is equivalent to the diagram in $\D$ : $UX \times_{UY} UX \rightrightarrows UX \to \im(Uf) \xrightarrow{Ui} UY$ where $UX \times_{UY} UX$ is the kernel pair of $Uf$ , and $\im(Uf)$ is the coequalizer. Since $Uf$ is a regular epimorphism, we must have $Ui$ is an isomorphism. Since $U$ is fully faithful and therefore conservative, we get $i$ is an isomorphism as well, so $f$ is a regular epimorphism. $\square$

Lemma 8.

Let $U : \C \to \D$ be a fully faithful functor. Assume that $\C$ has finite limits and coequalizers, and that $U$ preserves inhabited finite limits and coequalizers. If $\D$ has effective congruences, then so does $\C$ .

Proof. Suppose we have a congruence $E \hookrightarrow X\times X$ in $\C$ . We can then form the quotient $X \to X/E$ as a coequalizer, along with the kernel pair $X \times_{X/E} X$ and the comparison map $i$ in the diagram below: $E \xrightarrow{i} X \times_{X/E} X \rightrightarrows X \to X/E.$ By the assumptions, the image under $U$ is equivalent to the diagram in $\D$ : $UE \xrightarrow{Ui} UX \times_{U(X/E)} UX \rightrightarrows UX \to U(X/E).$ Here, $UE \rightrightarrows UX$ is a congruence: the map $UE \to UX \times UX$ is a monomorphism since $U$ preserves pullbacks and therefore preserves monomorphisms; the reflexivity and symmetry morphisms for $E$ are easily seen to transform under $U$ to reflexivity and symmetry morphisms for $UE$ ; and similarly, since $U$ preserves pullbacks, the transitivity morphism for $E$ transforms under $U$ to a transitivity morphism for $UE$ . This congruence $UE$ of $\D$ is effective, so we must have $Ui$ is an isomorphism. Since $U$ is fully faithful and therefore conservative, we get $i$ is an isomorphism as well, so $E$ is effective. $\square$

Authors: Martin Brandenburg, Daniel Schepler

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