Proof.
The implication (1) \Rightarrow (2) follows from Lemmas 101.28.8 and 101.28.10.
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 101.28.5 (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 101.28.9. We still have to show that \mathcal{X} is a gerbe over Y. Let \mathcal{X} \to X be the morphism of Lemma 101.28.2 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 101.27.13). 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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