The Stacks project

Lemma 35.22.2. A universally injective étale morphism is an open immersion.

First proof. Let $f : X \to Y$ be an étale morphism which is universally injective. Then $f$ is open (Morphisms, Lemma 29.36.13) hence we can replace $Y$ by $f(X)$ and we may assume that $f$ is surjective. Then $f$ is bijective and open hence a homeomorphism. Hence $f$ is quasi-compact. Thus by Lemma 35.22.1 it suffices to show that $f$ is a monomorphism. As $X \to Y$ is étale the morphism $\Delta _{X/Y} : X \to X \times _ Y X$ is an open immersion by Morphisms, Lemma 29.35.13 (and Morphisms, Lemma 29.36.16). As $f$ is universally injective $\Delta _{X/Y}$ is also surjective, see Morphisms, Lemma 29.10.2. Hence $\Delta _{X/Y}$ is an isomorphism, i.e., $X \to Y$ is a monomorphism. $\square$

Second proof. Let $f : X \to Y$ be an étale morphism which is universally injective. Then $f$ is open (Morphisms, Lemma 29.36.13) hence we can replace $Y$ by $f(X)$ and we may assume that $f$ is surjective. Since the hypotheses remain satisfied after any base change, we conclude that $f$ is a universal homeomorphism. Therefore $f$ is integral, see Morphisms, Lemma 29.45.5. It follows that $f$ is finite by Morphisms, Lemma 29.44.4. It follows that $f$ is finite locally free by Morphisms, Lemma 29.48.2. To finish the proof, it suffices that $f$ is finite locally free of degree $1$ (a finite locally free morphism of degree $1$ is an isomorphism). There is decomposition of $Y$ into open and closed subschemes $V_ d$ such that $f^{-1}(V_ d) \to V_ d$ is finite locally free of degree $d$, see Morphisms, Lemma 29.48.5. If $V_ d$ is not empty, we can pick a morphism $\mathop{\mathrm{Spec}}(k) \to V_ d \subset Y$ where $k$ is an algebraically closed field (just take the algebraic closure of the residue field of some point of $V_ d$). Then $\mathop{\mathrm{Spec}}(k) \times _ Y X \to \mathop{\mathrm{Spec}}(k)$ is a disjoint union of copies of $\mathop{\mathrm{Spec}}(k)$, by Morphisms, Lemma 29.36.7 and the fact that $k$ is algebraically closed. However, since $f$ is universally injective, there can only be one copy and hence $d = 1$ as desired. $\square$


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