Lemma 101.19.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ be a point. The following are equivalent

1. for some morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the class of $x$ setting $y = f \circ x$ the map $G_ x \to G_ y$ of automorphism group algebraic spaces is an isomorphism, and

2. for any morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the class of $x$ setting $y = f \circ x$ the map $G_ x \to G_ y$ of automorphism group algebraic spaces is an isomorphism.

Proof. This comes down to the fact that being an isomorphism is fpqc local on the target, see Descent on Spaces, Lemma 74.11.15. Namely, suppose that $k'/k$ is an extension of fields and denote $x' : \mathop{\mathrm{Spec}}(k') \to \mathcal{X}$ the composition and set $y' = f \circ x'$. Then the morphism $G_{x'} \to G_{y'}$ is the base change of $G_ x \to G_ y$ by $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. Hence $G_ x \to G_ y$ is an isomorphism if and only if $G_{x'} \to G_{y'}$ is an isomorphism. Thus we see that the property propagates through the equivalence class if it holds for one. $\square$

There are also:

• 2 comment(s) on Section 101.19: Automorphism groups

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).