The Stacks project

Lemma 67.12.6. Let $S$ be a scheme. Let $Z \to X$ be an immersion of algebraic spaces over $S$. Assume $Z \to X$ is quasi-compact. There exists a factorization $Z \to \overline{Z} \to X$ where $Z \to \overline{Z}$ is an open immersion and $\overline{Z} \to X$ is a closed immersion.

Proof. Let $U$ be a scheme and let $U \to X$ be surjective étale. As usual denote $R = U \times _ X U$ with projections $s, t : R \to U$. Set $T = Z \times _ U X$. Let $\overline{T} \subset U$ be the scheme theoretic image of $T \to U$. Note that $s^{-1}\overline{T} = t^{-1}\overline{T}$ as taking scheme theoretic images of quasi-compact morphisms commute with flat base change, see Morphisms, Lemma 29.25.16. Hence we obtain a closed subspace $\overline{Z} \subset X$ whose pullback to $U$ is $\overline{T}$, see Properties of Spaces, Lemma 66.12.2. By Morphisms, Lemma 29.7.7 the morphism $T \to \overline{T}$ is an open immersion. It follows that $Z \to \overline{Z}$ is an open immersion and we win. $\square$


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Your email address will not be published.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 081U. Beware of the difference between the letter 'O' and the digit '0'.