The Stacks project

[Proposition 3.5, rydh_quotients] and [Proposition 2.5, alper_quotient]

Lemma 100.45.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume

  1. $f$ is representable by algebraic spaces and unramified, and

  2. $\mathcal{I}_\mathcal {Y} \to \mathcal{Y}$ is proper.

Then the set of $x \in |\mathcal{X}|$ such that $f$ induces an isomorphism between automorphism groups at $x$ and $f(x)$ (Remark 100.19.5) is open. Letting $\mathcal{U} \subset \mathcal{X}$ be the corresponding open substack, the morphism $\mathcal{I}_\mathcal {U} \to \mathcal{U} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ is an isomorphism.

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 ├ętale. Since $f$ is representable by algebraic spaces, we see that $G \to H$ is a monomorphism. Hence $G \to H$ is an open immersion, see Morphisms of Spaces, Lemma 66.51.2. By assumption $H \to U$ is proper.

With these preparations out of the way, we can prove the lemma as follows. The inverse image of the subset of $|\mathcal{X}|$ of the lemma is clearly the set of $u \in U$ such that $G_ u \to H_ u$ is an isomorphism (since after all $G_ u$ is an open sub group algebraic space of $H_ u$). This is an open subset because the complement is the image of the closed subset $|H| \setminus |G|$ and $|H| \to |U|$ is closed. By Properties of Stacks, Lemma 99.9.12 we can consider the corresponding open substack $\mathcal{U}$ of $\mathcal{X}$. The final statement of the lemma follows from applying Lemma 100.45.3 to $\mathcal{U} \to \mathcal{Y}$. $\square$


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