## 92.8 Categories fibred in groupoids representable by algebraic spaces

A slightly weaker notion than being representable is the notion of being representable by algebraic spaces which we discuss in this section. This discussion might have been avoided had we worked with some category $\textit{Spaces}_{fppf}$ of algebraic spaces instead of the category $\mathit{Sch}_{fppf}$. However, it seems to us natural to consider the category of schemes as the natural collection of “test objects” over which the fibre categories of an algebraic stack are defined.

In analogy with Categories, Definitions 4.40.1 we make the following definition.

Definition 92.8.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. A category fibred in groupoids $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ is called representable by an algebraic space over $S$ if there exists an algebraic space $F$ over $S$ and an equivalence $j : \mathcal{X} \to \mathcal{S}_ F$ of categories over $(\mathit{Sch}/S)_{fppf}$.

We continue our abuse of notation in suppressing the equivalence $j$ whenever we encounter such a situation. It follows formally from the above that if $\mathcal{X}$ is representable (by a scheme), then it is representable by an algebraic space. Here is the analogue of Categories, Lemma 4.40.2.

Lemma 92.8.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ be a category fibred in groupoids. Then $\mathcal{X}$ is representable by an algebraic space over $S$ if and only if the following conditions are satisfied:

1. $\mathcal{X}$ is fibred in setoids1, and

2. the presheaf $U \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)/\! \! \cong$ is an algebraic space.

Proof. Omitted, but see Categories, Lemma 4.40.2. $\square$

If $\mathcal{X}, \mathcal{Y}$ are fibred in groupoids and representable by algebraic spaces $F, G$ over $S$, then we have

92.8.2.1
$$\label{algebraic-equation-morphisms-spaces} \mathop{Mor}\nolimits _{\textit{Cat}/(\mathit{Sch}/S)_{fppf}}(\mathcal{X}, \mathcal{Y}) \Big/ 2\text{-isomorphism} = \mathop{Mor}\nolimits _{\mathit{Sch}/S}(F, G)$$

see Categories, Lemma 4.39.6. More precisely, any $1$-morphism $\mathcal{X} \to \mathcal{Y}$ gives rise to a morphism $F \to G$. Conversely, given a morphism of sheaves $F \to G$ over $S$ there exists a $1$-morphism $\phi : \mathcal{X} \to \mathcal{Y}$ which gives rise to $F \to G$ and which is unique up to unique $2$-isomorphism.

[1] This means that it is fibred in groupoids and objects in the fibre categories have no nontrivial automorphisms, see Categories, Definition 4.38.2.

Comment #5938 by Dario Weißmann on

typo: give a morphism -> given a morphism

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