Proposition 90.13.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$ be an algebraic stack over $S$. The following are equivalent

1. $\mathcal{X}$ is a stack in setoids,

2. the canonical $1$-morphism $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is an equivalence, and

3. $\mathcal{X}$ is representable by an algebraic space.

Proof. The equivalence of (1) and (2) follows from Stacks, Lemma 8.7.2. The implication (3) $\Rightarrow$ (1) follows from Lemma 90.13.2. Finally, assume (1). By Stacks, Lemma 8.6.3 there exists an equivalence $j : \mathcal{X} \to \mathcal{S}_ F$ where $F$ is a sheaf on $(\mathit{Sch}/S)_{fppf}$. By Lemma 90.9.5 the fact that $\Delta _\mathcal {X}$ is representable by algebraic spaces, means that $\Delta _ F : F \to F \times F$ is representable by algebraic spaces. Let $U$ be a scheme and let $x : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ be a surjective smooth morphism. The composition $j \circ x : (\mathit{Sch}/U)_{fppf} \to \mathcal{S}_ F$ corresponds to a morphism $h_ U \to F$ of sheaves. By Bootstrap, Lemma 76.5.1 this morphism is representable by algebraic spaces. Hence by Lemma 90.10.4 we conclude that $h_ U \to F$ is surjective and smooth. In particular it is surjective, flat and locally of finite presentation (by Lemma 90.10.9 and the fact that a smooth morphism of algebraic spaces is flat and locally of finite presentation, see Morphisms of Spaces, Lemmas 63.37.5 and 63.37.7). Finally, we apply Bootstrap, Theorem 76.10.1 to see that $F$ is an algebraic space. $\square$

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