The Stacks project

Lemma 76.14.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

  1. $f$ is a closed immersion,

  2. $f$ is universally closed, unramified, and a monomorphism,

  3. $f$ is universally closed, unramified, and universally injective,

  4. $f$ is universally closed, locally of finite type, and a monomorphism,

  5. $f$ is universally closed, universally injective, locally of finite type, and formally unramified.

Proof. The equivalence of (2) – (5) follows immediately from Lemma 76.14.8. Moreover, if (2) – (5) are satisfied then $f$ is representable. Similarly, if (1) is satisfied then $f$ is representable. Hence the result follows from the case of schemes, see Étale Morphisms, Lemma 41.7.2. $\square$

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Your email address will not be published.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 05W8. Beware of the difference between the letter 'O' and the digit '0'.