Lemma 100.37.5. Consider a commutative diagram

$\xymatrix{ \mathcal{X} \ar[rr] \ar[rd] & & \mathcal{Y} \ar[ld] \\ & \mathcal{Z} & }$

of algebraic stacks.

1. If $\mathcal{X} \to \mathcal{Z}$ is universally closed and $\mathcal{Y} \to \mathcal{Z}$ is separated, then the morphism $\mathcal{X} \to \mathcal{Y}$ is universally closed. In particular, the image of $|\mathcal{X}|$ in $|\mathcal{Y}|$ is closed.

2. If $\mathcal{X} \to \mathcal{Z}$ is proper and $\mathcal{Y} \to \mathcal{Z}$ is separated, then the morphism $\mathcal{X} \to \mathcal{Y}$ is proper.

Proof. Assume $\mathcal{X} \to \mathcal{Z}$ is universally closed and $\mathcal{Y} \to \mathcal{Z}$ is separated. We factor the morphism as $\mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{Y}$. The first morphism is proper (Lemma 100.4.8) hence universally closed. The projection $\mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{Y}$ is the base change of a universally closed morphism and hence universally closed, see Lemma 100.13.3. Thus $\mathcal{X} \to \mathcal{Y}$ is universally closed as the composition of universally closed morphisms, see Lemma 100.13.4. This proves (1). To deduce (2) combine (1) with Lemmas 100.4.12, 100.7.7, and 100.17.8. $\square$

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