Lemma 92.9.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are necessary and sufficient conditions for $f$ to be representable by algebraic spaces:

1. for each scheme $U/S$ the functor $f_ U : \mathcal{X}_ U \longrightarrow \mathcal{Y}_ U$ between fibre categories is faithful, and

2. for each $U$ and each $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ the presheaf

$(h : V \to U) \longmapsto \{ (x, \phi ) \mid x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ V), \phi : h^*y \to f(x)\} /\cong$

is an algebraic space over $U$.

Here we have made a choice of pullbacks for $\mathcal{Y}$.

Proof. This follows from the description of fibre categories of the $2$-fibre products $(\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$ in Categories, Lemma 4.41.3 combined with Lemma 92.8.2. $\square$

Comment #69 by Quoc Ho on

Probably you might want to point back to tag 02ZY (categories-lemma-criterion-representable-map-stack-in-groupoids)since the two are basically the same result.

Comment #70 by on

@#69: I think you mean 4.41.7. And the sentence just preceding 92.9.2 does point back to that lemma. So no change needed I think. Right?

Just to make sure: in the stacks project the base category is always the category of schemes, so "being representable" always refers to being representable by schemes''. Whereas being representable by algebraic spaces is what 92.9.2 is about.

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• 2 comment(s) on Section 92.9: Morphisms representable by algebraic spaces

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