Lemma 76.48.11. 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 both $p$ and $q$ are flat and locally of finite presentation. Then there exists an open subspace $U(f) \subset X$ such that $|U(f)| \subset |X|$ is the set of points where $f$ is Koszul. 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 More on Morphisms, Lemma 37.62.21 and in fact we will deduce the lemma from it. By Definition 76.48.1 the set $\{ x \in |X| : f \text{ is Koszul at }x\}$ is open in $|X|$ hence by Properties of Spaces, Lemma 66.4.8 it corresponds to an open subspace $U(f)$ of $X$. Hence we only need to prove the final statement.

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). Note that by definition $U(f)$ is the image in $|X|$ of the set $T$ of points in $U$ where the morphism of schemes $U \to V$ is Koszul. Similarly, $U(f')$ is the image in $|X'|$ of the set $T'$ of points in $U'$ where the morphism of schemes $U' \to V'$ is Koszul. 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 More on Morphisms, Lemma 37.62.21 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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