The Stacks project

Lemma 106.13.7. Let $h : \mathcal{X}' \to \mathcal{X}$ be a morphism of algebraic stacks. Assume $\mathcal{X}$ is well-nigh affine, $h$ is étale, $h$ is separated, and $h$ induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 101.19.5). Then there exists a cartesian diagram

\[ \xymatrix{ \mathcal{X}' \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ M' \ar[r] & M } \]

where $M' \to M$ is a separated étale morphism of schemes and $\mathcal{X} \to M$ is the moduli space constructed in Lemma 106.13.4.

Proof. Choose an affine scheme $U$ and a surjective, flat, finite, and locally finitely presented morphism $U \to \mathcal{X}$. Since $h$ is representable by algebraic spaces (Morphisms of Stacks, Lemmas 101.45.3 and 101.45.1) we see that $U' = \mathcal{X}' \times _\mathcal {X} U$ is an algebraic space. Since $U' \to U$ is separated and étale, we see that $U'$ is a scheme and that every finite set of points of $U'$ is contained in an affine open, see Morphisms of Spaces, Lemma 67.51.1 and More on Morphisms, Lemma 37.45.1. Setting $R' = U' \times _{\mathcal{X}'} U'$ we see that $s', t' : R' \to U'$ are finite locally free. By Groupoids, Lemma 39.24.1 there exists an open covering $U' = \bigcup U'_ i$ by $R'$-invariant affine open subschemes $U'_ i \subset U'$. Let $\mathcal{X}'_ i \subset \mathcal{X}'$ be the corresponding open substacks. These are well-nigh affine as $U'_ i \to \mathcal{X}'_ i$ is surjective, flat, finite and of finite presentation. By Lemma 106.13.5 we obtain cartesian diagrams

\[ \xymatrix{ \mathcal{X}'_ i \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ M'_ i \ar[r] & M } \]

with $M'_ i \to M$ an étale morphism of affine schemes and vertical arrows as in Lemma 106.13.4. Observe that $\mathcal{X}'_{ij} = \mathcal{X}'_ i \cap \mathcal{X}'_ j$ is an open subspace of $\mathcal{X}'_ i$ and $\mathcal{X}'_ j$. Hence we get corresponding open subschemes $V_{ij} \subset M'_ i$ and $V_{ji} \subset M'_ j$. By the result of Lemma 106.13.6 we see that both $\mathcal{X}'_{ij} \to V_{ij}$ and $\mathcal{X}'_{ji} \to V_{ji}$ are categorical moduli spaces! Thus we get a unique isomorphism $\varphi _{ij} : V_{ij} \to V_{ji}$ such that

\[ \xymatrix{ \mathcal{X}'_ i \ar[d] & & \mathcal{X}'_ i \cap \mathcal{X}'_ j \ar[rr] \ar[ll] \ar[ld] \ar[rd] & & \mathcal{X}'_ j \ar[d] \\ M'_ i & V_{ij} \ar[l] \ar[rr]^{\varphi _{ij}} & & V_{ji} \ar[r] & M'_ j } \]

is commutative. These isomorphisms satisfy the cocyclce condition of Schemes, Section 26.14 by a computation (and another application of the previous lemma) which we omit. Thus we can glue the affine schemes in to scheme $M'$, see Schemes, Lemma 26.14.1. Let us identify the $M'_ i$ with their image in $M'$. We claim there is a morphism $\mathcal{X}' \to M'$ fitting into cartesian diagrams

\[ \xymatrix{ \mathcal{X}'_ i \ar[r] \ar[d] & \mathcal{X}' \ar[d] \\ M'_ i \ar[r] & M' } \]

This is clear from the description of the morphisms into the glued scheme $M'$ in Schemes, Lemma 26.14.1 and the fact that to give a morphism $\mathcal{X}' \to M'$ is the same thing as given a morphism $T \to M'$ for any morphism $T \to \mathcal{X}'$. Similarly, there is a morphism $M' \to M$ restricting to the given morphisms $M'_ i \to M$ on $M'_ i$. The morphism $M' \to M$ is étale (being étale on the members of an étale covering) and the fibre product property holds as it can be checked on members of the (affine) open covering $M' = \bigcup M'_ i$. Finally, $M' \to M$ is separated because the composition $U' \to \mathcal{X}' \to M'$ is surjective and universally closed and we can apply Morphisms, Lemma 29.41.11. $\square$


Comments (0)


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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DUS. Beware of the difference between the letter 'O' and the digit '0'.