Lemma 29.3.5. Let $f : Y \to X$ be a morphism of schemes. If for all $y \in Y$ there is an open subscheme $f(y) \in U \subset X$ such that $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is an immersion, then $f$ is an immersion.

Proof. This statement follows readily from the discussion of closed subschemes at the end of Schemes, Section 26.10 but we will also give a detailed proof. Let $Z \subset X$ be the closure of $f(X)$. Since taking closures commutes with restricting to opens, we see from the assumption that $f(Y) \subset Z$ is open. Hence $Z' = Z \setminus f(Y)$ is closed. Hence $X' = X \setminus Z'$ is an open subscheme of $X$ and $f$ factors as $f : Y \to X'$ followed by the inclusion. If $y \in Y$ and $U \subset X$ is as in the statement of the lemma, then $U' = X' \cap U$ is an open neighbourhood of $f'(y)$ such that $(f')^{-1}(U') \to U'$ is an immersion (Lemma 29.3.1) with closed image. Hence it is a closed immersion, see Schemes, Lemma 26.10.4. Since being a closed immersion is local on the target (for example by Lemma 29.2.1) we conclude that $f'$ is a closed immersion as desired. $\square$

## Comments (0)

There are also:

• 2 comment(s) on Section 29.3: Immersions

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Your email address will not be published.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