Lemma 97.17.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $(U, R, s, t, c)$ be a groupoid in algebraic spaces over $S$. Assume $s, t$ are flat and locally of finite presentation. Then the morphism $\mathcal{S}_ U \to [U/R]$ is flat, locally of finite presentation, and surjective.

Proof. Let $T$ be a scheme and let $x : (\mathit{Sch}/T)_{fppf} \to [U/R]$ be a $1$-morphism. We have to show that the projection

$\mathcal{S}_ U \times _{[U/R]} (\mathit{Sch}/T)_{fppf} \longrightarrow (\mathit{Sch}/T)_{fppf}$

is surjective, flat, and locally of finite presentation. We already know that the left hand side is representable by an algebraic space $F$, see Algebraic Stacks, Lemmas 94.17.1 and 94.10.11. Hence we have to show the corresponding morphism $F \to T$ of algebraic spaces is surjective, locally of finite presentation, and flat. Since we are working with properties of morphisms of algebraic spaces which are local on the target in the fppf topology we may check this fppf locally on $T$. By construction, there exists an fppf covering $\{ T_ i \to T\}$ of $T$ such that $x|_{(\mathit{Sch}/T_ i)_{fppf}}$ comes from a morphism $x_ i : T_ i \to U$. (Note that $F \times _ T T_ i$ represents the $2$-fibre product $\mathcal{S}_ U \times _{[U/R]} (\mathit{Sch}/T_ i)_{fppf}$ so everything is compatible with the base change via $T_ i \to T$.) Hence we may assume that $x$ comes from $x : T \to U$. In this case we see that

$\mathcal{S}_ U \times _{[U/R]} (\mathit{Sch}/T)_{fppf} = (\mathcal{S}_ U \times _{[U/R]} \mathcal{S}_ U) \times _{\mathcal{S}_ U} (\mathit{Sch}/T)_{fppf} = \mathcal{S}_ R \times _{\mathcal{S}_ U} (\mathit{Sch}/T)_{fppf}$

The first equality by Categories, Lemma 4.31.10 and the second equality by Groupoids in Spaces, Lemma 78.22.2. Clearly the last $2$-fibre product is represented by the algebraic space $F = R \times _{s, U, x} T$ and the projection $R \times _{s, U, x} T \to T$ is flat and locally of finite presentation as the base change of the flat locally finitely presented morphism of algebraic spaces $s : R \to U$. It is also surjective as $s$ has a section (namely the identity $e : U \to R$ of the groupoid). This proves the lemma. $\square$

Comment #6817 by Anonymous on

In the second sentence of the proof, it says "We have to show that the projection (...) is surjective and smooth." Is this given projection really smooth? What if $U = T$ is a scheme and $R \rightarrow U$ is non-smooth? I think we're only trying to show that this map is surjective, flat, and locally of finite presentation.

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