Is the 2-category of groupoids a topos?

Question

I have no justification for this, but I am wondering if the 2-category of groupoids is a topos.

Answer 1

If the category you're considering has small groupoids as objects and functors between them as morphisms, with equality of morphisms being exact equality of functors: then this category has no subobject classifier.

To see why, note that the objects functor $\operatorname{Ob} : \mathbf{Groupoids} \to \mathbf{Sets}$ is representable by the groupoid $1$ with one object and one morphism. Similarly, the arrows functor $\operatorname{Arr} : \mathbf{Groupoids} \to \mathbf{Sets}$ is representable by the groupoid $2$ with two objects and four morphisms (one between each pair of objects). In particular, given any monomorphism $F : G_1 \to G_2$, we then have that $\operatorname{Ob}(F)$ and $\operatorname{Arr}(F)$ are injective functions, which implies that $F$ is the composition of an isomorphism from $G_1$ to a subgroupoid of $G_2$ and the inclusion functor for that subgroupoid. It follows that $\mathbf{Groupoids}$ is well-powered, with $\operatorname{Sub}(G)$ being the set of subgroupoids of $G$.

Now, suppose we had a subclassifier object $\Omega$ of $\mathbf{Groupoids}$. Then we would have to have: $$\operatorname{Ob}(\Omega) \simeq \operatorname{Hom}(1, \Omega) \simeq \operatorname{Sub}(1) = \{ \emptyset, 1 \} $$ and $$\operatorname{Arr}(\Omega) \simeq \operatorname{Hom}(2, \Omega) \simeq \operatorname{Sub}(2).$$ Now, the subgroupoids of 2 are: 2 itself; the disconnected groupoid $2_d$ with the two objects of 2 but no morphisms between them; the groupoid $\{ 0 \}$ with the first object from 2; the groupoid $\{ 1 \}$ with the second object from 2; and the empty groupoid. Furthermore, the source and destination morphisms $\operatorname{Arr} \to \operatorname{Ob}$ are induced by the morphisms $1 \to 2$ mapping the object of 1 to $0 \in \operatorname{Ob}(2)$, respectively $1 \in \operatorname{Ob}(2)$. Therefore, we must have: $$ 2 \in \operatorname{Hom}_\Omega(1, 1); \\ 2_d \in \operatorname{Hom}_\Omega(1, 1); \\ \{ 0 \} \in \operatorname{Hom}_\Omega(1, \emptyset); \\ \{ 1 \} \in \operatorname{Hom}_\Omega(\emptyset, 1); \\ \emptyset \in \operatorname{Hom}_\Omega(\emptyset, \emptyset).$$ This means that $\Omega$ has a morphism from $\emptyset$ to 1, but $\operatorname{End}_\Omega(\emptyset)$ and $\operatorname{End}_\Omega(1)$ are not isomorphic groups, giving a contradiction.

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If, on the other hand, the category you're considering has small groupoids as objects, with morphisms being isomorphism classes of functors - then I'm not sure what the answer would be in that case. (In this case, the global sections functor $\operatorname{Hom}(1, -)$ would be the connected components functor. The functor $\operatorname{Hom}(\mathbb{Z}, -)$, where $\mathbb{Z}$ is the single-object groupoid with morphisms $\mathbb{Z}$, would then give some information about the $\pi_1$ groups.)