Lemma 81.7.1. In Situation 81.2.1 let $X,Y/B$ be good and let $f : X \to Y$ be a morphism over $B$. If $Z \subset X$ is an integral closed subspace, then there exists a unique integral closed subspace $Z' \subset Y$ such that there is a commutative diagram

$\xymatrix{ Z \ar[r] \ar[d] & X \ar[d]^ f \\ Z' \ar[r] & Y }$

with $Z \to Z'$ dominant. If $f$ is proper, then $Z \to Z'$ is proper and surjective.

Proof. Let $\xi \in |Z|$ be the generic point. Let $Z' \subset Y$ be the integral closed subspace whose generic point is $\xi ' = f(\xi )$, see Remark 81.2.3. Since $\xi \in |f^{-1}(Z')| = |f|^{-1}(|Z'|)$ by Properties of Spaces, Lemma 65.4.3 and since $Z$ is the reduced with $|Z| = \overline{\{ \xi \} }$ we see that $Z \subset f^{-1}(Z')$ as closed subspaces of $X$ (see Properties of Spaces, Lemma 65.12.4). Thus we obtain our morphism $Z \to Z'$. This morphism is dominant as the generic point of $Z$ maps to the generic point of $Z'$. Uniqueness of $Z'$ is clear. If $f$ is proper, then $Z \to Y$ is proper as a composition of proper morphisms (Morphisms of Spaces, Lemmas 66.40.3 and 66.40.5). Then we conclude that $Z \to Z'$ is proper by Morphisms of Spaces, Lemma 66.40.6. Surjectivity then follows as the image of a proper morphism is closed. $\square$

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Your email address will not be published.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