Lemma 67.40.6. Let $S$ be a scheme. Consider a commutative diagram of algebraic spaces
over $S$.
If $X \to B$ is universally closed and $Y \to B$ is separated, then the morphism $X \to Y$ is universally closed. In particular, the image of $|X|$ in $|Y|$ is closed.
If $X \to B$ is proper and $Y \to B$ is separated, then the morphism $X \to Y$ is proper.
Proof. Assume $X \to B$ is universally closed and $Y \to B$ is separated. We factor the morphism as $X \to X \times _ B Y \to Y$. The first morphism is a closed immersion, see Lemma 67.4.6 hence universally closed. The projection $X \times _ B Y \to Y$ is the base change of a universally closed morphism and hence universally closed, see Lemma 67.9.3. Thus $X \to Y$ is universally closed as the composition of universally closed morphisms, see Lemma 67.9.4. This proves (1). To deduce (2) combine (1) with Lemmas 67.4.10, 67.8.9, and 67.23.6. $\square$
Your email address will not be published.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)