## 65.8 Quasi-compact morphisms

By Section 65.3 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 65.8.1. 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 65.3), 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 64.5.1 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 65.3). 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 65.8.2. 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 65.8.1 above this agrees with the already existing notion for representable morphisms of algebraic spaces.

Lemma 65.8.3. 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 64.4.8 there is an open subspace $Y' \subset Y$ with $V = |Y'|$. Then $Y'$ is a quasi-compact algebraic space by Properties of Spaces, Lemma 64.5.2 and hence $X' = Y' \times _ Y X$ is a quasi-compact algebraic space by Definition 65.8.2. On the other hand, $X' \subset X$ is an open subspace (Spaces, Lemma 63.12.3) and $|X'| = |f|^{-1}(|X'|) = |f|^{-1}(V)$ by Properties of Spaces, Lemma 64.4.3. We conclude using Properties of Spaces, Lemma 64.5.2 again that $|X'|$ is a quasi-compact open of $|X|$ as desired. $\square$

Lemma 65.8.4. 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 65.8.5. The composition of a pair of quasi-compact morphisms of algebraic spaces is quasi-compact.

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

Lemma 65.8.6. 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 65.5.2 the map $|X| \to |Y|$ is surjective, hence we see $Y$ is quasi-compact by Properties of Spaces, Lemma 64.5.2 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 65.5.5 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 65.8.7. 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 64.5.5) 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 65.8.6. Thus $f$ is quasi-compact. $\square$

Lemma 65.8.8. 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 65.8.4 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 64.6.3 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 65.5.5. 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 65.8.7 and the fact that an étale morphism of algebraic spaces is universally open (see discussion following Definition 65.6.2).

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 64.4.8) 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 65.8.9. 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 65.4.7 because it is a section of the quasi-separated morphism $X \times _ Z Y \to X$ (a base change of $g$, see Lemma 65.4.4). The second map is quasi-compact as it is the base change of $f$, see Lemma 65.8.4. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 65.8.5. $\square$

Lemma 65.8.10. 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 65.8.9 with $Z = S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Part (2) follows from (1) and Lemma 65.4.10. 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 65.8.4 and 65.4.4. Hence $X \times _ Y Z$ is quasi-compact and quasi-separated as an algebraic space quasi-compact and quasi-separated over $Z$, see Lemmas 65.4.9 and 65.8.5. $\square$

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