An étale universally injective morphism is an open immersion

First suppose that $f$ is injective on points and every residue-field map is an isomorphism. After any base change $T\to Y$ and for any $t\in T$ with image $y$, a point of $(X{\times }_{Y}T{)}_t$ lies over some $x\in f^{-1}(y)$ and, by [F3], corresponds to a prime of $\kappa (x){\otimes }_{\kappa (y)}\kappa (t)$. There is at most one such $x$; if it exists, the tensor product is $\kappa (t)$ and has one prime. Hence every base change of $f$ is injective on points, which is universal injectivity.

Now assume universal injectivity. By [F1], $f$ is unramified, so its diagonal $\mathrm{\Delta }:X\to X{\times }_{Y}X$ is an open immersion. Base-change $f$ along itself: the projection $p_1:X{\times }_{Y}X\to X$ is injective on points by universal injectivity. The diagonal is a section of $p_1$, so it supplies a point in every nonempty fibre of $p_1$; injectivity makes that the only point in each fibre. Hence $\mathrm{\Delta }$ is surjective on points. A surjective open immersion is an isomorphism, so $\mathrm{\Delta }$ is an isomorphism and $f$ is a monomorphism.

By [F1], $f$ is open. Put $U=f(X)\subseteq Y$, an open subscheme; then $f:X\to U$ is surjective, flat, locally of finite presentation and a monomorphism. Fix $y\in U$ and its unique preimage $x$. Choose an affine open $V=\mathrm{Spec}A\subseteq U$ around $y$ and an affine open $W=\mathrm{Spec}B\subseteq f^{-1}(V)$ around $x$. Because $f$ is open, $f(W)$ is an open neighbourhood of $y$ in $V$. Choose a principal open $D(a)\subseteq f(W)$ containing $y$. Since $f$ is injective on points, every point of $f^{-1}(D(a))$ already lies in $W$, so $f^{-1}(D(a))=W\cap f^{-1}(D(a))=\mathrm{Spec}{B}_{f^{\mathrm{\#}}(a)}$. The induced affine map $A_a\to B_{f^{\mathrm{\#}}(a)}$ is flat and surjective on spectra, hence faithfully flat by [F2].

Abbreviate the rings of step 2.1 by $A^{\mathrm{\prime }}\to B^{\mathrm{\prime }}$. Because $f$ is a monomorphism, its diagonal over $D(a)$ is an isomorphism, so the multiplication map $\mu :B^{\mathrm{\prime }}{\otimes }_{A^{\mathrm{\prime }}}{B}^{\mathrm{\prime }}\to B^{\mathrm{\prime }}$ is an isomorphism. By [F2], $A^{\mathrm{\prime }}\to B^{\mathrm{\prime }}$ is injective; let $C=B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}$. Tensoring the exact sequence $0\to A^{\mathrm{\prime }}\to B^{\mathrm{\prime }}\to C\to 0$ with the flat $A^{\mathrm{\prime }}$-module $B^{\mathrm{\prime }}$ gives $0⟶B^{\mathrm{\prime }}\stackrel{b\mapsto 1\otimes b}{\to }{B}^{\mathrm{\prime }}{\otimes }_{A^{\mathrm{\prime }}}{B}^{\mathrm{\prime }}⟶C{\otimes }_{A^{\mathrm{\prime }}}{B}^{\mathrm{\prime }}⟶0.$ The first map is inverse to $\mu$, hence an isomorphism. Thus $C{\otimes }_{A^{\mathrm{\prime }}}{B}^{\mathrm{\prime }}=0$; faithful flatness detects zero modules by [F2], so $C=0$ and $A^{\mathrm{\prime }}\to B^{\mathrm{\prime }}$ is an isomorphism. The resulting principal opens $D(a)$ cover $U$ as $y$ varies, and $f$ is an isomorphism over each of them. Therefore $f:X\stackrel{\sim }{\to }U\hookrightarrow Y$ is an open immersion.

Step 1.1 proves the residue-field criterion for universal injectivity; steps 1.2--3.1 prove the principal assertion. The proof makes only finitely many local chart choices at each point. The declared Axiom of Choice enters through the published suppliers identified in [F4]. [F1, F2, F3, F4, step 1.1, step 1.2, step 3.1] $\mathrm{\square }$