Lemma 100.17.8. Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $g \circ f : \mathcal{X} \to \mathcal{Z}$ is locally of finite type, then $f : \mathcal{X} \to \mathcal{Y}$ is locally of finite type.
Proof. We can find a diagram
\[ \xymatrix{ U \ar[r] \ar[d] & V \ar[r] \ar[d] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} } \]
where $U$, $V$, $W$ are schemes, the vertical arrow $W \to \mathcal{Z}$ is surjective and smooth, the arrow $V \to \mathcal{Y} \times _\mathcal {Z} W$ is surjective and smooth, and the arrow $U \to \mathcal{X} \times _\mathcal {Y} V$ is surjective and smooth. Then also $U \to \mathcal{X} \times _\mathcal {Z} V$ is surjective and smooth (as a composition of a surjective and smooth morphism with a base change of such). By definition we see that $U \to W$ is locally of finite type. Hence $U \to V$ is locally of finite type by Morphisms, Lemma 29.15.8 which in turn means (by definition) that $\mathcal{X} \to \mathcal{Y}$ is locally of finite type. $\square$
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Your email address will not be published.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)