## 100.7 Quasi-compact morphisms

Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 99.3 we have defined what it means for $f$ to be quasi-compact. Here is another characterization.

Lemma 100.7.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent:

$f$ is quasi-compact (as in Properties of Stacks, Section 99.3), and

for every quasi-compact algebraic stack $\mathcal{Z}$ and any morphism $\mathcal{Z} \to \mathcal{Y}$ the algebraic stack $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is quasi-compact.

**Proof.**
Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks with $\mathcal{Z}$ quasi-compact. By Properties of Stacks, Lemma 99.6.2 there exists a quasi-compact scheme $U$ and a surjective smooth morphism $U \to \mathcal{Z}$. Since $f$ is representable by algebraic spaces and quasi-compact we see by definition that $U \times _\mathcal {Y} \mathcal{X}$ is an algebraic space, and that $U \times _\mathcal {Y} \mathcal{X} \to U$ is quasi-compact. Hence $U \times _ Y X$ is a quasi-compact algebraic space. The morphism $U \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is smooth and surjective (as the base change of the smooth and surjective morphism $U \to \mathcal{Z}$). Hence $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is quasi-compact by another application of Properties of Stacks, Lemma 99.6.2

Assume (2). Let $Z \to \mathcal{Y}$ be a morphism, where $Z$ is a scheme. We have to show that the morphism of algebraic spaces $p : Z \times _\mathcal {Y} \mathcal{X} \to Z$ is quasi-compact. Let $U \subset Z$ be affine open. Then $p^{-1}(U) = U \times _\mathcal {Y} \mathcal{Z}$ and the algebraic space $U \times _\mathcal {Y} \mathcal{Z}$ is quasi-compact by assumption (2). Hence $p$ is quasi-compact, see Morphisms of Spaces, Lemma 66.8.8.
$\square$

This motivates the following definition.

Definition 100.7.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is *quasi-compact* if for every quasi-compact algebraic stack $\mathcal{Z}$ and morphism $\mathcal{Z} \to \mathcal{Y}$ the fibre product $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is quasi-compact.

By Lemma 100.7.1 above this agrees with the already existing notion for morphisms of algebraic stacks representable by algebraic spaces. In particular this notion agrees with the notions already defined for morphisms between algebraic stacks and schemes.

Lemma 100.7.3. The base change of a quasi-compact morphism of algebraic stacks by any morphism of algebraic stacks is quasi-compact.

**Proof.**
Omitted.
$\square$

Lemma 100.7.4. The composition of a pair of quasi-compact morphisms of algebraic stacks is quasi-compact.

**Proof.**
Omitted.
$\square$

Lemma 100.7.5. A closed immersion of algebraic stacks is quasi-compact.

**Proof.**
This follows from the fact that immersions are always representable and the corresponding fact for closed immersion of algebraic spaces.
$\square$

Lemma 100.7.6. Let

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

be a $2$-commutative diagram of morphisms of algebraic stacks. If $f$ is surjective and $p$ is quasi-compact, then $q$ is quasi-compact.

**Proof.**
Let $\mathcal{T}$ be a quasi-compact algebraic stack, and let $\mathcal{T} \to \mathcal{Z}$ be a morphism. By Properties of Stacks, Lemma 99.5.3 the morphism $\mathcal{T} \times _\mathcal {Z} \mathcal{X} \to \mathcal{T} \times _\mathcal {Z} \mathcal{Y}$ is surjective and by assumption $\mathcal{T} \times _\mathcal {Z} \mathcal{X}$ is quasi-compact. Hence $\mathcal{T} \times _\mathcal {Z} \mathcal{Y}$ is quasi-compact by Properties of Stacks, Lemma 99.6.2.
$\square$

Lemma 100.7.7. Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. 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 $(1, f) : \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{Y}$. The first map is quasi-compact by Lemma 100.4.9 because it is a section of the quasi-separated morphism $\mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{X}$ (a base change of $g$, see Lemma 100.4.4). The second map is quasi-compact as it is the base change of $f$, see Lemma 100.7.3. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 100.7.4.
$\square$

Lemma 100.7.8. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

If $\mathcal{X}$ is quasi-compact and $\mathcal{Y}$ is quasi-separated, then $f$ is quasi-compact.

If $\mathcal{X}$ is quasi-compact and quasi-separated and $\mathcal{Y}$ is quasi-separated, then $f$ is quasi-compact and quasi-separated.

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

**Proof.**
Part (1) follows from Lemma 100.7.7. Part (2) follows from (1) and Lemma 100.4.12. For (3) let $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Z} \to \mathcal{Y}$ be morphisms of quasi-compact and quasi-separated algebraic stacks. Then $\mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{Z}$ is quasi-compact and quasi-separated as a base change of $\mathcal{X} \to \mathcal{Y}$ using (2) and Lemmas 100.7.3 and 100.4.4. Hence $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ is quasi-compact and quasi-separated as an algebraic stack quasi-compact and quasi-separated over $\mathcal{Z}$, see Lemmas 100.4.11 and 100.7.4.
$\square$

Lemma 100.7.9. Let $f : \mathcal{X} \to \mathcal{Y}$ be a quasi-compact morphism of algebraic stacks. Let $y \in |\mathcal{Y}|$ be a point in the closure of the image of $|f|$. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathcal{Y} } \]

such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $y$.

**Proof.**
Choose an affine scheme $V$ and a point $v \in V$ and a smooth morphism $V \to \mathcal{Y}$ sending $v$ to $y$. Consider the base change diagram

\[ \xymatrix{ V \times _\mathcal {Y} \mathcal{X} \ar[r] \ar[d]_ g & \mathcal{X} \ar[d]^ f \\ V \ar[r] & \mathcal{Y} } \]

Recall that $|V \times _\mathcal {Y} \mathcal{X}| \to |V| \times _{|\mathcal{Y}|} |\mathcal{X}|$ is surjective (Properties of Stacks, Lemma 99.4.3). Because $|V| \to |\mathcal{Y}|$ is open (Properties of Stacks, Lemma 99.4.7) we conclude that $v$ is in the closure of the image of $|g|$. Thus it suffices to prove the lemma for the quasi-compact morphism $g$ (Lemma 100.7.3) which we do in the next paragraph.

Assume $\mathcal{Y} = Y$ is an affine scheme. Then $\mathcal{X}$ is quasi-compact as $f$ is quasi-compact (Definition 100.7.2). Choose an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. Then the image of $|f|$ is the image of $W \to Y$. By Morphisms, Lemma 29.6.5 we can choose a diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & W \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y \ar[r] & Y } \]

such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $y$. Composing with $W \to \mathcal{X}$ we obtain a solution.
$\square$

Lemma 100.7.10. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be surjective, flat, and locally of finite presentation where $W$ is an algebraic space. If the base change $W \times _\mathcal {Y} \mathcal{X} \to W$ is quasi-compact, then $f$ is quasi-compact.

**Proof.**
Assume $W \times _\mathcal {Y} \mathcal{X} \to W$ is quasi-compact. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism with $\mathcal{Z}$ a quasi-compact algebraic stack. Choose a scheme $U$ and a surjective smooth morphism $U \to W \times _\mathcal {Y} \mathcal{Z}$. Since $U \to \mathcal{Z}$ is flat, surjective, and locally of finite presentation and $\mathcal{Z}$ is quasi-compact, we can find a quasi-compact open subscheme $U' \subset U$ such that $U' \to \mathcal{Z}$ is surjective. Then $U' \times _\mathcal {Y} \mathcal{X} = U' \times _ W (W \times _\mathcal {Y} \mathcal{X})$ is quasi-compact by assumption and surjects onto $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$. Hence $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is quasi-compact as desired.
$\square$

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