The Stacks project

Proof. This lemma is the analogue of Morphisms, Lemma 29.35.15 and in fact we will deduce the lemma from it. By Definition 67.38.1 the set $\{ x \in |X| : f \text{ is unramified at }x\} $ is open in $X$. Hence we only need to prove the final statement. By Lemma 67.23.6 the morphism $X \to Y$ is locally of finite type. By Lemma 67.23.3 the morphism $X' \to Y'$ is locally of finite type.

Choose a scheme $W$ and a surjective étale morphism $W \to Z$. Choose a scheme $V$ and a surjective étale morphism $V \to W \times _ Z Y$. Choose a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Finally, choose a scheme $W'$ and a surjective étale morphism $W' \to W \times _ Z Z'$. Set $V' = W' \times _ W V$ and $U' = W' \times _ W U$, so that we obtain surjective étale morphisms $V' \to Y'$ and $U' \to X'$. We will use without further mention an étale morphism of algebraic spaces induces an open map of associated topological spaces (see Properties of Spaces, Lemma 66.16.7). This combined with Lemma 67.38.5 implies that $U(f)$ is the image in $|X|$ of the set $T$ of points in $U$ where the morphism $U \to V$ is unramified. Similarly, $U(f')$ is the image in $|X'|$ of the set $T'$ of points in $U'$ where the morphism $U' \to V'$ is unramified. Now, by construction the diagram

is cartesian (in the category of schemes). Hence the aforementioned Morphisms, Lemma 29.35.15 applies to show that $T'$ is the inverse image of $T$. Since $|U'| \to |X'|$ is surjective this implies the lemma. $\square$