The Stacks project

Lemma 78.12.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $T$ be an algebraic space over $S$, and let $(a, Z)$ be a pair as in If $f$ is separated, then $Z$ is closed in $T \times _ Y X$.

Proof. A finite morphism of algebraic spaces is universally closed by Morphisms of Spaces, Lemma 66.45.9. Since $f$ is separated so is the morphism $T \times _ Y X \to T$, see Morphisms of Spaces, Lemma 66.4.4. Thus the closedness of $Z$ follows from Morphisms of Spaces, Lemma 66.40.6. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 78.12: The finite part of a morphism

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Your email address will not be published.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 04PH. Beware of the difference between the letter 'O' and the digit '0'.