The Stacks project

Lemma 95.17.1. 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 and smooth. We already know that the left hand side is representable by an algebraic space $F$, see Algebraic Stacks, Lemmas 92.17.1 and 92.10.11. 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 76.21.2. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06FH. Beware of the difference between the letter 'O' and the digit '0'.