Lemma 106.13.10. Let $h : \mathcal{X}' \to \mathcal{X}$ be a morphism of algebraic stacks. Assume

$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is finite,

$h$ is étale, separated, and 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 algebraic spaces and the vertical arrows are the moduli spaces constructed in Theorem 106.13.9.

**Proof.**
By Morphisms of Stacks, Lemma 101.45.3 we see that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is finite as a base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$. Let $f' : \mathcal{X}' \to M'$ and $f : \mathcal{X} \to M$ be as in Theorem 106.13.9. We obtain a commutative diagram as in the lemma because $f'$ is categorical moduli space. Choose $I$ and $g'_ i : \mathcal{X}'_ i \to \mathcal{X}'$ as in Lemma 106.13.8. Observe that $g_ i = h \circ g'_ i$ is étale, separated, and induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 101.19.5). Let $f'_ i : \mathcal{X}'_ i \to M'_ i$ be as in Lemma 106.13.4. In the proof of Theorem 106.13.9 we have seen that the diagrams

\[ \xymatrix{ \mathcal{X}'_ i \ar[d]_{f'_ i} \ar[r]_{g'_ i} & \mathcal{X}' \ar[d]^{f'} \\ M'_ i \ar[r] & M' } \quad \text{and}\quad \xymatrix{ \mathcal{X}'_ i \ar[d]_{f'_ i} \ar[r]_{g_ i} & \mathcal{X} \ar[d]^ f \\ M'_ i \ar[r] & M } \]

are cartesian and that $M'_ i \to M'$ and $M'_ i \to M$ are étale (this also follows directly from the properties of the morphisms $g'_ i, g_ i, f', f'_ i, f$ listed sofar by arguing in exactly the same way). This first implies that $M' \to M$ is étale and second that the diagram in the lemma is cartesian. We still need to show that $M' \to M$ is separated. To do this we contemplate the diagram

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

The top horizontal arrow is universally closed as $\mathcal{X}' \to \mathcal{X}$ is separated. The vertical arrows are as in Theorem 106.13.9 (as flat base changes of $\mathcal{X} \to M$) hence universal homeomorphisms. Thus the lower horizontal arrow is universally closed. This (combined with it being an étale monomorphism of algebraic spaces) proves it is a closed immersion as desired.
$\square$

## Comments (2)

Comment #2549 by Matthew Emerton on

Comment #2550 by Johan on