The Stacks project

Lemma 78.27.1. Notation and assumption as in Lemma 78.21.1. The morphism of quotient stacks

\[ [f] : [U/R] \longrightarrow [U'/R'] \]

turns $[U/R]$ into a gerbe over $[U'/R']$ if $f : U \to U'$ and $R \to R'|_ U$ are surjective maps of fppf sheaves. Here $R'|_ U$ is the restriction of $R'$ to $U$ via $f : U \to U'$.

Proof. We will verify that Stacks, Lemma 8.11.3 properties (2) (a) and (2) (b) hold. Property (2)(a) holds because $U \to U'$ is a surjective map of sheaves (use Lemma 78.24.1 to see that objects in $[U'/R']$ locally come from $U'$). To prove (2)(b) let $x, y$ be objects of $[U/R]$ over a scheme $T/S$. Let $x', y'$ be the images of $x, y$ in the category $[U'/'R]_ T$. Condition (2)(b) requires us to check the map of sheaves

\[ \mathit{Isom}(x, y) \longrightarrow \mathit{Isom}(x', y') \]

on $(\mathit{Sch}/T)_{fppf}$ is surjective. To see this we may work fppf locally on $T$ and assume that come from $a, b \in U(T)$. In that case we see that $x', y'$ correspond to $f \circ a, f \circ b$. By Lemma 78.22.1 the displayed map of sheaves in this case becomes

\[ T \times _{(a, b), U \times _ B U} R \longrightarrow T \times _{f \circ a, f \circ b, U' \times _ B U'} R' = T \times _{(a, b), U \times _ B U} R'|_ U. \]

Hence the assumption that $R \to R'|_ U$ is a surjective map of fppf sheaves on $(\mathit{Sch}/S)_{fppf}$ implies the desired surjectivity. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 78.27: Gerbes and quotient stacks

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Your email address will not be published.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 06PE. Beware of the difference between the letter 'O' and the digit '0'.