**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 $U$.
$\square$

## Comments (2)

Comment #4889 by SDIGR on

Comment #5165 by Johan on

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