## 81.7 Preparation for proper pushforward

This section is the analogue of Chow Homology, Section 42.11.

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.
$\square$

Lemma 81.7.4. In Situation 81.2.1 let $X, Y/B$ be good and let $f : X \to Y$ be a morphism over $B$. Assume $X$, $Y$ integral and $\dim _\delta (X) = \dim _\delta (Y)$. Then either $f$ factors through a proper closed subspace of $Y$, or $f$ is dominant and the extension of function fields $R(X) / R(Y)$ is finite.

**Proof.**
By Lemma 81.7.1 there is a unique integral closed subspace $Z \subset Y$ such that $f$ factors through a dominant morphism $X \to Z$. Then $Z = Y$ if and only if $\dim _\delta (Z) = \dim _\delta (Y)$. On the other hand, by our construction of dimension functions (see Situation 81.2.1) we have $\dim _\delta (X) = \dim _\delta (Z) + r$ where $r$ the transcendence degree of the extension $R(X)/R(Z)$. Combining this with Spaces over Fields, Lemma 71.5.1 the lemma follows.
$\square$

Lemma 81.7.5. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a morphism over $B$. Assume $f$ is quasi-compact, and $\{ T_ i\} _{i \in I}$ is a locally finite collection of closed subsets of $|X|$. Then $\{ \overline{|f|(T_ i)}\} _{i \in I}$ is a locally finite collection of closed subsets of $|Y|$.

**Proof.**
Let $V \subset |Y|$ be a quasi-compact open subset. Then $|f|^{-1}(V) \subset |X|$ is quasi-compact by Morphisms of Spaces, Lemma 66.8.3. Hence the set $\{ i \in I : T_ i \cap |f|^{-1}(V) \not= \emptyset \} $ is finite by a simple topological argument which we omit. Since this is the same as the set

\[ \{ i \in I : |f|(T_ i) \cap V \not= \emptyset \} = \{ i \in I : \overline{|f|(T_ i)} \cap V \not= \emptyset \} \]

the lemma is proved.
$\square$

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