Lemma 82.7.1. In Situation 82.2.1 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
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 82.2.3. Since $\xi \in |f^{-1}(Z')| = |f|^{-1}(|Z'|)$ by Properties of Spaces, Lemma 66.4.3 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 66.12.4). 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 67.40.3 and 67.40.5). Then we conclude that $Z \to Z'$ is proper by Morphisms of Spaces, Lemma 67.40.6. Surjectivity then follows as the image of a proper morphism is closed. $\square$
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