Lemma 97.8.3. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $\mathcal{X}$ is an algebraic stack and $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces, then $F$ is algebraic.
Proof. Choose a representable stack in groupoids $\mathcal{U}$ and a surjective smooth $1$-morphism $\mathcal{U} \to \mathcal{X}$. Let $T$ be a scheme and let $\xi $ be an object of $\mathcal{Y}$ over $T$. The morphism of $2$-fibre products
\[ (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{U} \longrightarrow (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{X} \]
is representable by algebraic spaces, surjective, and smooth as a base change of $\mathcal{U} \to \mathcal{X}$, see Algebraic Stacks, Lemmas 94.9.7 and 94.10.6. By our condition on the diagonal of $\mathcal{Y}$ we see that the source of this morphism is representable by an algebraic space, see Algebraic Stacks, Lemma 94.10.11. Hence the target is an algebraic stack by Algebraic Stacks, Lemma 94.15.3. $\square$
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Your email address will not be published.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)