Lemma 94.17.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $(U, R, s, t, c)$ be a smooth groupoid in algebraic spaces over $S$. Then the morphism $\mathcal{S}_ U \to [U/R]$ is smooth 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 and smooth. We already know that the left hand side is representable by an algebraic space $F$, see Lemmas 94.17.1 and 94.10.11. Hence we have to show the corresponding morphism $F \to T$ of algebraic spaces is surjective and smooth. 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 smooth as the base change of the smooth 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. $\square$
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