Universally injective étale maps are open immersions.

Lemma 66.51.2. Let $S$ be a scheme. Let $f : X \to Y$ be an étale and universally injective morphism of algebraic spaces over $S$. Then $f$ is an open immersion.

Proof. Let $T \to Y$ be a morphism from a scheme into $Y$. If we can show that $X \times _ Y T \to T$ is an open immersion, then we are done. Since being étale and being universally injective are properties of morphisms stable under base change (see Lemmas 66.39.4 and 66.19.5) we may assume that $Y$ is a scheme. Note that the diagonal $\Delta _{X/Y} : X \to X \times _ Y X$ is étale, a monomorphism, and surjective by Lemma 66.19.2. Hence we see that $\Delta _{X/Y}$ is an isomorphism (see Spaces, Lemma 64.5.9), in particular we see that $X$ is separated over $Y$. It follows that $X$ is a scheme too, by Proposition 66.50.2. Finally, $X \to Y$ is an open immersion by the fundamental theorem for étale morphisms of schemes, see Étale Morphisms, Theorem 41.14.1. $\square$

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Suggested slogan: Universally injective \'etale maps are open immersions.

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