Lemma 101.15.5. 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 a universal homeomorphism, then $f$ is a universal homeomorphism.

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

$\xymatrix{ W \times _\mathcal {Y} \mathcal{X} \ar[d]^ g & U \times _\mathcal {Y} \mathcal{X} \ar[d] \ar[l] \ar[r] & \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \ar[d] \\ W & U \ar[l] \ar[r] & \mathcal{Z} }$

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. $\square$

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).