Lemma 100.28.7. Let $\pi : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

1. $\mathcal{X}$ is a gerbe over $\mathcal{Y}$, and

2. there exists an algebraic space $U$, a group algebraic space $G$ flat and locally of finite presentation over $U$, and a surjective, flat, and locally finitely presented morphism $U \to \mathcal{Y}$ such that $\mathcal{X} \times _\mathcal {Y} U \cong [U/G]$ over $U$.

Proof. Assume (2). By Lemma 100.28.5 to prove (1) it suffices to show that $[U/G]$ is a gerbe over $U$. This is immediate from Groupoids in Spaces, Lemma 77.27.2.

Assume (1). Any base change of $\pi$ is a gerbe, see Lemma 100.28.3. As a first step we choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Thus we may assume that $\pi : \mathcal{X} \to V$ is a gerbe over a scheme. This means that there exists an fppf covering $\{ V_ i \to V\}$ such that the fibre category $\mathcal{X}_{V_ i}$ is nonempty, see Stacks, Lemma 8.11.3 (2)(a). Note that $U = \coprod V_ i \to V$ is surjective, flat, and locally of finite presentation. Hence we may replace $V$ by $U$ and assume that $\pi : \mathcal{X} \to U$ is a gerbe over a scheme $U$ and that there exists an object $x$ of $\mathcal{X}$ over $U$. By Lemma 100.28.6 we see that $\mathcal{X} = [U/G]$ over $U$ for some flat and locally finitely presented group algebraic space $G$ over $U$. $\square$

Comment #4889 by SDIGR on

Typo: "Note that $U=\coprod V_i\to U$ is..." should be "Note that $U=\coprod V_i\to V$ is..."

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