Gpd as a presheaf category

Question

I wonder if there exists a way to see the category of groupoids Gpd as (isomorphic to, or maybe just equivalent to) a presheaf category (valued in Set) ?

Thanks

Answer 1

For any category $\mathcal{C}$, the presheaf category $[\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ is a regular category. In particular, the class of regular epimorphisms is closed under pullback in $[\mathcal{C}^\mathrm{op}, \mathbf{Set}]$. However, this is not true in $\mathbf{Grpd}$.

Indeed, let $\mathbb{I} = \{ 0 \cong 1 \}$ be the groupoid with two objects and a unique isomorphism between them. Let $\mathbb{A} = \mathbb{I} \amalg \mathbb{I}$. There is an evident regular epimorphism $\mathbb{A} \to \mathbb{B}$ obtained by identifying one copy of $1$ with the disjoint copy of $0$; so $\mathbb{B} = \{ 0 \cong 1 \cong 2 \}$ is the groupoid with three objects and a unique isomorphism between any pair of objects. Let $\mathbb{B}'$ be the subgroupoid $\{ 0 \cong 2 \} \subset \mathbb{B}$, and consider the evident pullback diagram: $$\require{AMScd} \begin{CD} \mathbb{A}^\prime @>>> \mathbb{A} \\ @VVV @VVV \\ \mathbb{B}^\prime @>>> \mathbb{B} \end{CD}$$ Clearly, $\mathbb{A}'$ is the discrete groupoid with two objects, and therefore the morphism $\mathbb{A}' \to \mathbb{A}$ is not a regular epimorphism.

Exactly the same argument shows $\mathbf{Cat}$ is not a regular category.