The Stacks project

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.9.5. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is universally closed. We have to show that $f$ is universally closed. To see this, using Morphisms of Spaces, Lemma 67.9.5 again, it is enough to show that for every affine scheme $Y$ and morphism $Y \to Z$ the map $|Y \times _ Z X| \to |Y|$ is closed. Consider the cube (74.11.1.1). The assumption that $f'$ is universally closed implies that $|Y \times _ Z Z' \times _ Z X| \to |Y \times _ Z Z'|$ is closed. As $Y \times _ Z Z' \to Y$ is quasi-compact, surjective, and flat as a base change of $Z' \to Z$ we see the map $|Y \times _ Z Z'| \to |Y|$ is submersive, see Morphisms, Lemma 29.25.12. Moreover the map

is surjective, see Properties of Spaces, Lemma 66.4.3. It follows by elementary topology that $|Y \times _ Z X| \to |Y|$ is closed. $\square$