The Stacks project

Lemma 70.6.5. Notation and assumptions as in Situation 70.6.1. If

  1. $f$ is universally injective,

  2. $f_0$ is locally of finite type,

then $f_ i$ is universally injective for some $i \geq 0$.

Proof. Recall that a morphism $X \to Y$ is universally injective if and only if the diagonal $X \to X \times _ Y X$ is surjective (Morphisms of Spaces, Definition 67.19.3 and Lemma 67.19.2). Observe that $X_0 \to X_0 \times _{Y_0} X_0$ is of locally of finite presentation (Morphisms of Spaces, Lemma 67.28.10). Hence the lemma follows from Lemma 70.6.4 by considering the morphism $X_0 \to X_0 \times _{Y_0} X_0$. $\square$

Comments (2)

Comment #791 by Kestutis Cesnavicius on

One can add a similar lemma for 'open immersion' by using that etale universally injective morphisms are open immersions, i.e., by combining this lemma, 67.51.2, and 70.6.2.

Comment #802 by on

OK, you could, but actually this follows from the slightly stronger Lemma 70.5.2. Right?

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Your email address will not be published.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 084X. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 70.6.5 as pdf