Proposition 100.28.11. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

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

2. $f : \mathcal{X} \to \mathcal{Y}$ and $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ are surjective, flat, and locally of finite presentation.

Proof. The implication (1) $\Rightarrow$ (2) follows from Lemmas 100.28.8 and 100.28.10.

Assume (2). It suffices to prove (1) for the base change of $f$ by a surjective, flat, and locally finitely presented morphism $\mathcal{Y}' \to \mathcal{Y}$, see Lemma 100.28.5 (note that the base change of the diagonal of $f$ is the diagonal of the base change). Thus we may assume $\mathcal{Y}$ is a scheme $Y$. In this case $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is a base change of $\Delta$ and we conclude that $\mathcal{X}$ is a gerbe by Proposition 100.28.9. We still have to show that $\mathcal{X}$ is a gerbe over $Y$. Let $\mathcal{X} \to X$ be the morphism of Lemma 100.28.2 turning $\mathcal{X}$ into a gerbe over the algebraic space $X$ classifying isomorphism classes of objects of $\mathcal{X}$. It is clear that $f : \mathcal{X} \to Y$ factors as $\mathcal{X} \to X \to Y$. Since $f$ is surjective, flat, and locally of finite presentation, we conclude that $X \to Y$ is surjective as a map of fppf sheaves (for example use Lemma 100.27.13). On the other hand, $X \to Y$ is injective too: for any scheme $T$ and any two $T$-valued points $x_1, x_2$ of $X$ which map to the same point of $Y$, we can first fppf locally on $T$ lift $x_1, x_2$ to objects $\xi _1, \xi _2$ of $\mathcal{X}$ over $T$ and second deduce that $\xi _1$ and $\xi _2$ are fppf locally isomorphic by our assumption that $\Delta : \mathcal{X} \to \mathcal{X} \times _ Y \mathcal{X}$ is surjective, flat, and locally of finite presentation. Whence $x_1 = x_2$ by construction of $X$. Thus $X = Y$ and the proof is complete. $\square$

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