Lemma 100.15.5. 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 100.14.8. 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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)