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
the horizontal arrows are open and surjective, and moreover
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. $\square$
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