The Stacks project

Proof. Assume (1), and let $Z \to Y$ be a morphism of algebraic spaces with $Z$ quasi-compact. By Properties of Spaces, Definition 66.5.1 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 67.3). 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. $\square$

Proof. Let $V \subset |Y|$ be quasi-compact open. By Properties of Spaces, Lemma 66.4.8 there is an open subspace $Y' \subset Y$ with $V = |Y'|$. Then $Y'$ is a quasi-compact algebraic space by Properties of Spaces, Lemma 66.5.2 and hence $X' = Y' \times _ Y X$ is a quasi-compact algebraic space by Definition 67.8.2. On the other hand, $X' \subset X$ is an open subspace (Spaces, Lemma 65.12.3) and $|X'| = |f|^{-1}(|X'|) = |f|^{-1}(V)$ by Properties of Spaces, Lemma 66.4.3. We conclude using Properties of Spaces, Lemma 66.5.2 again that $|X'|$ is a quasi-compact open of $|X|$ as desired. $\square$

Proof. Omitted. Hint: Transitivity of fibre products. $\square$

Proof. Omitted. Hint: Transitivity of fibre products. $\square$

Proof. Assume $X$ is quasi-compact and $X \to Y$ is surjective. By Definition 67.5.2 the map $|X| \to |Y|$ is surjective, hence we see $Y$ is quasi-compact by Properties of Spaces, Lemma 66.5.2 and the topological fact that the image of a quasi-compact space under a continuous map is quasi-compact, see Topology, Lemma 5.12.7. Let $f, p, q$ be as in (2). Let $T \to Z$ be a morphism whose source is a quasi-compact algebraic space. By assumption $T \times _ Z X$ is quasi-compact. By Lemma 67.5.5 the morphism $T \times _ Z X \to T \times _ Z Y$ is surjective. Hence by part (1) we see $T \times _ Z Y$ is quasi-compact too. Thus $q$ is quasi-compact. $\square$

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 66.5.5) 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 67.8.6. Thus $f$ is quasi-compact. $\square$

Proof. We will use Lemma 67.8.4 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 66.6.3 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 67.5.5. 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 67.8.7 and the fact that an étale morphism of algebraic spaces is universally open (see discussion following Definition 67.6.2).

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

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 66.4.8) is a finite union of quasi-compact spaces, hence quasi-compact. Thus we see that (6) implies (3). $\square$

Proof. This is true because $f$ equals the composition $(1, f) : X \to X \times _ Z Y \to Y$. The first map is quasi-compact by Lemma 67.4.7 because it is a section of the quasi-separated morphism $X \times _ Z Y \to X$ (a base change of $g$, see Lemma 67.4.4). The second map is quasi-compact as it is the base change of $f$, see Lemma 67.8.4. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 67.8.5. $\square$

Proof. Part (1) follows from Lemma 67.8.9 with $Z = S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Part (2) follows from (1) and Lemma 67.4.10. For (3) let $X \to Y$ and $Z \to Y$ be morphisms of quasi-compact and quasi-separated algebraic spaces. Then $X \times _ Y Z \to Z$ is quasi-compact and quasi-separated as a base change of $X \to Y$ using (2) and Lemmas 67.8.4 and 67.4.4. Hence $X \times _ Y Z$ is quasi-compact and quasi-separated as an algebraic space quasi-compact and quasi-separated over $Z$, see Lemmas 67.4.9 and 67.8.5. $\square$