Lemma 66.8.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. The following are equivalent:

1. $f$ is quasi-compact (in the sense of Section 66.3), and

2. for every quasi-compact algebraic space $Z$ and any morphism $Z \to Y$ the algebraic space $Z \times _ Y X$ is quasi-compact.

Proof. Assume (1), and let $Z \to Y$ be a morphism of algebraic spaces with $Z$ quasi-compact. By Properties of Spaces, Definition 65.5.1 there exists a quasi-compact scheme $U$ and a surjective étale morphism $U \to Z$. Since $f$ is representable and quasi-compact we see by definition that $U \times _ Y X$ is a scheme, and that $U \times _ Y X \to U$ is quasi-compact. Hence $U \times _ Y X$ is a quasi-compact scheme. The morphism $U \times _ Y X \to Z \times _ Y X$ is étale and surjective (as the base change of the representable étale and surjective morphism $U \to Z$, see Section 66.3). Hence by definition $Z \times _ Y X$ is quasi-compact.

Assume (2). Let $Z \to Y$ be a morphism, where $Z$ is a scheme. We have to show that $p : Z \times _ Y X \to Z$ is quasi-compact. Let $U \subset Z$ be affine open. Then $p^{-1}(U) = U \times _ Y Z$ and the scheme $U \times _ Y Z$ is quasi-compact by assumption (2). Hence $p$ is quasi-compact, see Schemes, Section 26.19. $\square$

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