Quasi-compact morphisms of algebraic spaces are preserved under pullback and local on the target.

Lemma 66.8.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

1. $f$ is quasi-compact,

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

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

4. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is a quasi-compact morphism of algebraic spaces, and

5. there exists a surjective étale morphism $Y' \to Y$ of algebraic spaces such that $Y' \times _ Y X \to Y'$ is a quasi-compact morphism of algebraic spaces, and

6. there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is quasi-compact.

Proof. We will use Lemma 66.8.4 without further mention. It is clear that (1) implies (2) and that (2) implies (3). Assume (3). Let $Z$ be a quasi-compact algebraic space over $S$, and let $Z \to Y$ be a morphism. By Properties of Spaces, Lemma 65.6.3 there exists an affine scheme $U$ and a surjective étale morphism $U \to Z$. Then $U \times _ Y X \to Z \times _ Y X$ is a surjective morphism of algebraic spaces, see Lemma 66.5.5. By assumption $|U \times _ Y X|$ is quasi-compact. It surjects onto $|Z \times _ Y X|$, hence we conclude that $|Z \times _ Y X|$ is quasi-compact, see Topology, Lemma 5.12.7. This proves that (3) implies (1).

The implications (1) $\Rightarrow$ (4), (4) $\Rightarrow$ (5) are clear. The implication (5) $\Rightarrow$ (1) follows from Lemma 66.8.7 and the fact that an étale morphism of algebraic spaces is universally open (see discussion following Definition 66.6.2).

Of course (1) implies (6) by taking the covering $Y = Y$. Assume $Y = \bigcup Y_ i$ is as in (6). Let $Z$ be affine and let $Z \to Y$ be a morphism. Then there exists a finite standard affine covering $Z = Z_1 \cup \ldots \cup Z_ n$ such that each $Z_ j \to Y$ factors through $Y_{i_ j}$ for some $i_ j$. Hence the algebraic space

$Z_ j \times _ Y X = Z_ j \times _{Y_{i_ j}} f^{-1}(Y_{i_ j})$

is quasi-compact. Since $Z \times _ Y X = \bigcup _{j = 1, \ldots , n} Z_ j \times _ Y X$ is a Zariski covering we see that $|Z \times _ Y X| = \bigcup _{j = 1, \ldots , n} |Z_ j \times _ Y X|$ (see Properties of Spaces, Lemma 65.4.8) is a finite union of quasi-compact spaces, hence quasi-compact. Thus we see that (6) implies (3). $\square$

Comment #1109 by Evan Warner on

Suggested slogan: Quasi-compact morphisms of algebraic spaces are preserved under pullback and local on the target.

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