The Stacks project

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 a universal homeomorphism, in particular the map $|V \times _\mathcal {Y} \mathcal{X}| \to |V|$ is a homeomorphism. 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 a homeomorphism it follows that the right vertical arrow is a homeomorphism. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. $\square$

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

Proof. Omitted. $\square$

Proof. Assume $g : W \times _\mathcal {Y} \mathcal{X} \to W$ is a universal homeomorphism. Then $g$ is universally injective, hence $f$ is universally injective by Lemma 101.14.8. On the other hand, let $\mathcal{Z} \to \mathcal{Y}$ be a morphism with $\mathcal{Z}$ an algebraic stack. Choose a scheme $U$ and a surjective smooth morphism $U \to W \times _\mathcal {Y} \mathcal{Z}$. Consider the diagram

The middle vertical arrow induces a homeomorphism on topological space by assumption on $g$. The morphism $U \to \mathcal{Z}$ and $U \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ are surjective, flat, and locally of finite presentation hence induce open maps on topological spaces. We conclude that $|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$ is open. Surjectivity is easy to prove; we omit the proof. $\square$