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
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