## 101.13 Universally closed morphisms

Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 100.3 we have defined what it means for $f$ to be universally closed. Here is another characterization.

Lemma 101.13.1. 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 universally closed (as in Properties of Stacks, Section 100.3), and

for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces $|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$ is closed.

**Proof.**
Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times _\mathcal {Y} \mathcal{X} \to V$ of algebraic spaces is universally closed, in particular the map $|V \times _\mathcal {Y} \mathcal{X}| \to |V|$ is closed. By Properties of Stacks, Section 100.4 in the commutative diagram

\[ \xymatrix{ |V \times _\mathcal {Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| } \]

the horizontal arrows are open and surjective, and moreover

\[ |V \times _\mathcal {Y} \mathcal{X}| \longrightarrow |V| \times _{|\mathcal{Z}|} |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \]

is surjective. Hence as the left vertical arrow is closed it follows that the right vertical arrow is closed. This proves (2). The implication (2) $\Rightarrow $ (1) follows from the definitions.
$\square$

Thus we may use the following natural definition.

Definition 101.13.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

We say $f$ is *closed* if the map of topological spaces $|\mathcal{X}| \to |\mathcal{Y}|$ is closed.

We say $f$ is *universally closed* if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces

\[ |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}| \]

is closed, i.e., the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is closed.

Lemma 101.13.3. The base change of a universally closed morphism of algebraic stacks by any morphism of algebraic stacks is universally closed.

**Proof.**
This is immediate from the definition.
$\square$

Lemma 101.13.4. The composition of a pair of (universally) closed morphisms of algebraic stacks is (universally) closed.

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

Lemma 101.13.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

$f$ is universally closed,

for every scheme $Z$ and every morphism $Z \to \mathcal{Y}$ the projection $|Z \times _\mathcal {Y} \mathcal{X}| \to |Z|$ is closed,

for every affine scheme $Z$ and every morphism $Z \to \mathcal{Y}$ the projection $|Z \times _\mathcal {Y} \mathcal{X}| \to |Z|$ is closed, and

there exists an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$ such that $V \times _\mathcal {Y} \mathcal{X} \to V$ is a universally closed morphism of algebraic stacks.

**Proof.**
We omit the proof that (1) implies (2), and that (2) implies (3).

Assume (3). Choose a surjective smooth morphism $V \to \mathcal{Y}$. We are going to show that $V \times _\mathcal {Y} \mathcal{X} \to V$ is a universally closed morphism of algebraic stacks. Let $\mathcal{Z} \to V$ be a morphism from an algebraic stack to $V$. Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W = \coprod W_ i$ is a disjoint union of affine schemes. Then we have the following commutative diagram

\[ \xymatrix{ \coprod _ i |W_ i \times _\mathcal {Y} \mathcal{X}| \ar@{=}[r] \ar[d] & |W \times _\mathcal {Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \ar[d] \ar@{=}[r] & |\mathcal{Z} \times _ V (V \times _\mathcal {Y} \mathcal{X})| \ar[ld] \\ \coprod |W_ i| \ar@{=}[r] & |W| \ar[r] & |\mathcal{Z}| } \]

We have to show the south-east arrow is closed. The middle horizontal arrows are surjective and open (Properties of Stacks, Lemma 100.4.7). By assumption (3), and the fact that $W_ i$ is affine we see that the left vertical arrows are closed. Hence it follows that the right vertical arrow is closed.

Assume (4). We will show that $f$ is universally closed. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider the diagram

\[ \xymatrix{ |(V \times _\mathcal {Y} \mathcal{Z}) \times _ V (V \times _\mathcal {Y} \mathcal{X})| \ar@{=}[r] \ar[rd] & |V \times _\mathcal {Y} \mathcal{X}| \ar[r] \ar[d] & |Z \times _\mathcal {Y} \mathcal{X}| \ar[d] \\ & |V \times _\mathcal {Y} \mathcal{Z}| \ar[r] & |\mathcal{Z}| } \]

The south-west arrow is closed by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic stacks are surjective and smooth (see reference above). It follows that the right vertical arrow is closed.
$\square$

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