Remark 99.3.7. Let $\mathcal{Y}$ be an algebraic stack. Consider the following $2$-category:

An object is a morphism $f : \mathcal{X} \to \mathcal{Y}$ which is representable by algebraic spaces,

a $1$-morphism $(g, \beta ) : (f_1 : \mathcal{X}_1 \to \mathcal{Y}) \to (f_2 : \mathcal{X}_2 \to \mathcal{Y})$ consists of a morphism $g : \mathcal{X}_1 \to \mathcal{X}_2$ and a $2$-morphism $\beta : f_1 \to f_2 \circ g$, and

a $2$-morphism between $(g, \beta ), (g', \beta ') : (f_1 : \mathcal{X}_1 \to \mathcal{Y}) \to (f_2 : \mathcal{X}_2 \to \mathcal{Y})$ is a $2$-morphism $\alpha : g \to g'$ such that $(\text{id}_{f_2} \star \alpha ) \circ \beta = \beta '$.

Let us denote this $2$-category $\textit{Spaces}/\mathcal{Y}$ by analogy with the notation of Topologies on Spaces, Section 72.2. Now we claim that in this $2$-category the morphism categories

are all setoids. Namely, a $2$-morphism $\alpha $ is a rule which to each object $x_1$ of $\mathcal{X}_1$ assigns an isomorphism $\alpha _{x_1} : g(x_1) \longrightarrow g'(x_1)$ in the relevant fibre category of $\mathcal{X}_2$ such that the diagram

commutes. But since $f_2$ is faithful (see Algebraic Stacks, Lemma 93.15.2) this means that if $\alpha _{x_1}$ exists, then it is unique! In other words the $2$-category $\textit{Spaces}/\mathcal{Y}$ is very close to being a category. Namely, if we replace $1$-morphisms by isomorphism classes of $1$-morphisms we obtain a category. We will often perform this replacement without further mention.

## Comments (0)