Lemma 66.38.10. Let $S$ be a scheme. Consider a commutative diagram

$\xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z }$

of algebraic spaces over $S$. Assume that $X \to Z$ is locally of finite type. Then there exists an open subspace $U(f) \subset X$ such that $|U(f)| \subset |X|$ is the set of points where $f$ is unramified. Moreover, for any morphism of algebraic spaces $Z' \to Z$, if $f' : X' \to Y'$ is the base change of $f$ by $Z' \to Z$, then $U(f')$ is the inverse image of $U(f)$ under the projection $X' \to X$.

Proof. This lemma is the analogue of Morphisms, Lemma 29.35.15 and in fact we will deduce the lemma from it. By Definition 66.38.1 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 66.23.6 the morphism $X \to Y$ is locally of finite type. By Lemma 66.23.3 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 65.16.7). This combined with Lemma 66.38.5 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

$\xymatrix{ U' \ar[r] \ar[d] & U \ar[d] \\ V' \ar[r] & V }$

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. $\square$

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