## 77.27 Gerbes and quotient stacks

In this section we relate quotient stacks to the discussion Stacks, Section 8.11 and especially gerbes as defined in Stacks, Definition 8.11.4. The stacks in groupoids occurring in this section are generally speaking not algebraic stacks!

Lemma 77.27.1. Notation and assumption as in Lemma 77.21.1. 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 77.24.1 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 77.22.1 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. $\square$

Lemma 77.27.2. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $G$ be a group algebraic space over $B$. Endow $B$ with the trivial action of $G$. The morphism

$[B/G] \longrightarrow \mathcal{S}_ B$

(Lemma 77.20.2) turns $[B/G]$ into a gerbe over $B$.

Proof. Immediate from Lemma 77.27.1 as the morphisms $B \to B$ and $B \times _ B G \to B$ are surjective as morphisms of sheaves. $\square$

Comment #2049 by Sasha on

In Lemma 66.26.2, is $[B/G]$ a gerbe over $B$ or the other way round? I am confused because there is a canonical map $B \to [B/G]$, but seemingly no natural map in the other direction.

Comment #2050 by on

There is a map of $[B/G]$ to $B$ because the action is trivial. It is the displayed arrow in the statement of the lemma and there is even a reference to another lemma which carefully defines this morphism. Maybe you got confused because this discussion happens before the introduction of the abuse of language in Section 99.2, hence instead of mapping to $B$ as you might have expected, we map to the algebraic stack $\mathcal{S}_B$ associated to $B$ (in the literature these two things are always identified).

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