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

1. $f$ is universally closed,

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

3. 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

4. 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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).