Lemma 66.8.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $g : Y' \to Y$ be a universally open and surjective morphism of algebraic spaces such that the base change $f' : X' \to Y'$ is quasi-compact. Then $f$ is quasi-compact.

**Proof.**
Let $Z \to Y$ be a morphism of algebraic spaces with $Z$ quasi-compact. As $g$ is universally open and surjective, we see that $Y' \times _ Y Z \to Z$ is open and surjective. As every point of $|Y' \times _ Y Z|$ has a fundamental system of quasi-compact open neighbourhoods (see Properties of Spaces, Lemma 65.5.5) we can find a quasi-compact open $W \subset |Y' \times _ Y Z|$ which surjects onto $Z$. Denote $f'' : W \times _ Y X \to W$ the base change of $f'$ by $W \to Y'$. By assumption $W \times _ Y X$ is quasi-compact. As $W \to Z$ is surjective we see that $W \times _ Y X \to Z \times _ Y X$ is surjective. Hence $Z \times _ Y X$ is quasi-compact by Lemma 66.8.6. Thus $f$ is quasi-compact.
$\square$

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