Lemma 100.45.7. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume $f$ is separated, étale, $f$ induces an isomorphism between automorphism groups at points (Remark 100.19.5) and for every algebraically closed field $k$ the functor

$f : \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), \mathcal{X}) \longrightarrow \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), \mathcal{Y})$

is an equivalence. Then $f$ is an isomorphism.

Proof. By Lemma 100.14.5 we see that $f$ is universally injective. Combining Lemmas 100.45.1 and 100.45.3 we see that $f$ is representable by algebraic spaces. Hence $f$ is an open immersion by Morphisms of Spaces, Lemma 66.51.2. To finish we remark that the condition in the lemma also guarantees that $f$ is surjective. $\square$

Comment #7856 by Rachel Webb on

How is it used that $f$ is separated?

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