## 66.8 Quasi-compact morphisms

By Section 66.3 we know what it means for a representable morphism of algebraic spaces to be quasi-compact. In order to formulate the definition for a general morphism of algebraic spaces we make the following observation.

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. $\square$

This motivates the following definition.

Definition 66.8.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We say $f$ is quasi-compact if for every quasi-compact algebraic space $Z$ and morphism $Z \to Y$ the fibre product $Z \times _ Y X$ is quasi-compact.

By Lemma 66.8.1 above this agrees with the already existing notion for representable morphisms of algebraic spaces.

Lemma 66.8.3. Let $S$ be a scheme. If $f : X \to Y$ is a quasi-compact morphism of algebraic spaces over $S$, then the underlying map $|f| : |X| \to |Y|$ of topological space is quasi-compact.

Proof. Let $V \subset |Y|$ be quasi-compact open. By Properties of Spaces, Lemma 65.4.8 there is an open subspace $Y' \subset Y$ with $V = |Y'|$. Then $Y'$ is a quasi-compact algebraic space by Properties of Spaces, Lemma 65.5.2 and hence $X' = Y' \times _ Y X$ is a quasi-compact algebraic space by Definition 66.8.2. On the other hand, $X' \subset X$ is an open subspace (Spaces, Lemma 64.12.3) and $|X'| = |f|^{-1}(|X'|) = |f|^{-1}(V)$ by Properties of Spaces, Lemma 65.4.3. We conclude using Properties of Spaces, Lemma 65.5.2 again that $|X'|$ is a quasi-compact open of $|X|$ as desired. $\square$

Lemma 66.8.4. The base change of a quasi-compact morphism of algebraic spaces by any morphism of algebraic spaces is quasi-compact.

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

Lemma 66.8.5. The composition of a pair of quasi-compact morphisms of algebraic spaces is quasi-compact.

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

Lemma 66.8.6. Let $S$ be a scheme.

1. If $X \to Y$ is a surjective morphism of algebraic spaces over $S$, and $X$ is quasi-compact then $Y$ is quasi-compact.

2. If

$\xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[dl]^ q \\ & Z }$

is a commutative diagram of morphisms of algebraic spaces over $S$ and $f$ is surjective and $p$ is quasi-compact, then $q$ is quasi-compact.

Proof. Assume $X$ is quasi-compact and $X \to Y$ is surjective. By Definition 66.5.2 the map $|X| \to |Y|$ is surjective, hence we see $Y$ is quasi-compact by Properties of Spaces, Lemma 65.5.2 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 66.5.5 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. $\square$

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. $\square$

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). $\square$

The following (and the next) lemma guarantees in particular that a morphism $X \to \mathop{\mathrm{Spec}}(A)$ is quasi-compact as soon as $X$ is a quasi-compact algebraic space

Lemma 66.8.9. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. If $g \circ f$ is quasi-compact and $g$ is quasi-separated then $f$ is quasi-compact.

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 66.4.7 because it is a section of the quasi-separated morphism $X \times _ Z Y \to X$ (a base change of $g$, see Lemma 66.4.4). The second map is quasi-compact as it is the base change of $f$, see Lemma 66.8.4. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 66.8.5. $\square$

Lemma 66.8.10. Let $f : X \to Y$ be a morphism of algebraic spaces over a scheme $S$.

1. If $X$ is quasi-compact and $Y$ is quasi-separated, then $f$ is quasi-compact.

2. If $X$ is quasi-compact and quasi-separated and $Y$ is quasi-separated, then $f$ is quasi-compact and quasi-separated.

3. A fibre product of quasi-compact and quasi-separated algebraic spaces is quasi-compact and quasi-separated.

Proof. Part (1) follows from Lemma 66.8.9 with $Z = S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Part (2) follows from (1) and Lemma 66.4.10. 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 66.8.4 and 66.4.4. Hence $X \times _ Y Z$ is quasi-compact and quasi-separated as an algebraic space quasi-compact and quasi-separated over $Z$, see Lemmas 66.4.9 and 66.8.5. $\square$

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