Lemma 81.8.2. In Situation 81.2.1 let $X, Y, Z/B$ be good. Let $f : X \to Y$ and $g : Y \to Z$ be proper morphisms over $B$. Then $g_* \circ f_* = (g \circ f)_*$ as maps $Z_ k(X) \to Z_ k(Z)$.

Proof. Let $W \subset X$ be an integral closed subspace of dimension $k$. Consider the integral closed subspaces $W' \subset Y$ and $W'' \subset Z$ we get by applying Lemma 81.7.1 to $f$ and $W$ and then to $g$ and $W'$. Then $W \to W'$ and $W' \to W''$ are surjective and proper. We have to show that $g_*(f_*[W]) = (f \circ g)_*[W]$. If $\dim _\delta (W'') < k$, then both sides are zero. If $\dim _\delta (W'') = k$, then we see $W \to W'$ and $W' \to W''$ both satisfy the hypotheses of Lemma 81.7.4. Hence

$g_*(f_*[W]) = \deg (W/W')\deg (W'/W'')[W''], \quad (f \circ g)_*[W] = \deg (W/W'')[W''].$

Then we can apply Spaces over Fields, Lemma 71.5.3 to conclude. $\square$

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