CatDat

Sifted colimits in groupoids

While the combination of this result and this result already implies that groupoids have sifted colimits, we can make these colimits more explicit and also prove their existence without using any non-trivial theorem. We also do this in a more general setting.

Let $D : \I \to \C$ be a sifted diagram in a category and $i_0 \in \I$. We call $D$ constant after $i_0$ when for all morphisms $i_0 \to i$ the morphism $D(i_0) \to D(i)$ is an isomorphism. Of course, in a groupoid, this is satisfied for every $i_0 \in \I$. If such an object $i_0$ exists, we call $D$ eventually constant.

In particular, every groupoid has sifted colimits.

Proof. For every $i \in \I$, define a morphism $u_i : D(i) \to D(i_0)$ as follows. Choose a cospan

$$i \xrightarrow{a} k \xleftarrow{b} i_0,$$

and set

$$u_i \coloneqq D(b)^{-1} D(a) : D(i) \to D(k) \to D(i_0).$$

This does not depend on the chosen cospan: since the category of cospans is connected, it suffices to check this for any morphism of cospans

$$\begin{CD} i @>a>> k @<b<< i_0 \\ @V{=}VV @VV{f}V @VV{=}V \\ i @>>{a'}> k' @<<{b'}< i_0 \end{CD}$$

Since $D(b)$ and $D(f) D(b) = D(b')$ are isomorphisms, also $D(f)$ is an isomorphism. We compute:

$$D(b)^{-1} D(a) = D(b)^{-1} D(f)^{-1} D(f) D(a) = D(fb)^{-1} D(fa) = D(b')^{-1} D(a').$$

Thus, $u_i : D(i) \to D(i_0)$ is well defined. Note that $u_{i_0}$ is the identity of $D(i_0)$.

To show that this defines a cocone, let $f : j \to i$ be a morphism in $\I$, and choose a cospan $i \xrightarrow{a} k \xleftarrow{b} i_0$ as above to compute $u_i$. We use the cospan $j \xrightarrow{af} k \xleftarrow{b} i_0$ to compute $u_j$. Hence,

$$u_j = D(b)^{-1} D(af) = D(b)^{-1} D(a) D(f) = u_i D(f),$$

which proves the claim.

Finally, we verify the universal property. Let $(v_i : D(i) \to T)$ be another cocone, and set $h \coloneqq v_{i_0} : D(i_0) \to T$. (Since $\C$ is not assumed to be a groupoid, the morphisms $v_i$ need not be isomorphisms; the argument does not use this.)

We check that $v_i = h u_i$ for every $i \in \I$. Choose a cospan $i \xrightarrow{a} k \xleftarrow{b} i_0$ as above. Then

$$h u_i = v_{i_0} D(b)^{-1} D(a) = v_k D(a) = v_i.$$

Moreover, $h$ is uniquely determined: if $v_i = h u_i$ for all $i \in \I$, then for $i = i_0$ we obtain $v_{i_0} = h$. $\square$

Proof. Let $D : \I \to \C$ be a diagram and choose $i_0 \in \I$ such that $D$ is constant after $i_0$. We obtain a commutative diagram:

$$\begin{CD} \Hom(X,D(i_0)) @>{=}>> \Hom(X,D(i_0)) \\ @VVV @VVV \\ \colim_{i \in \I} \Hom(X,D(i)) @>>> \Hom(X,\colim_{i \in \I} D(i)) \end{CD}$$

The top horizontal map is the identity. By Lemma 1 applied to $D$, the right vertical map is an isomorphism. By Lemma 2 applied to the diagram $\Hom(X,D(-))$ in $\Set$, the left vertical map is an isomorphism. Hence the bottom horizontal map is an isomorphism. $\square$

Proof. This follows directly from Lemma 1 and Corollary 2. $\square$

Proof. Since sifted colimits exist, filtered colimits also exist. Moreover, every object is strongly finitely presentable and therefore also finitely presentable. $\square$