Lemma 100.45.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be an unramified morphism of algebraic stacks. The following are equivalent

1. $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ is an isomorphism, and

2. $f$ induces an isomorphism between automorphism groups at $x$ and $f(x)$ (Remark 100.19.5) for all $x \in |\mathcal{X}|$.

Proof. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Denote $G \to H$ the pullback of the morphism $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ to $U$. By Remark 100.45.2 and Lemma 100.36.9 the morphism $G \to H$ is étale. Condition (1) is equivalent to the condition that $G \to H$ is an isomorphism (this follows for example by applying Properties of Stacks, Lemma 99.3.3). Condition (2) is equivalent to the condition that for every $u \in U$ the morphism $G_ u \to H_ u$ of fibres is an isomorphism. Thus (1) $\Rightarrow$ (2) is trivial. If (2) holds, then $G \to H$ is a surjective, universally injective, étale morphism of algebraic spaces. Such a morphism is an isomorphism by Morphisms of Spaces, Lemma 66.51.2. $\square$

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