Lemma 82.8.2. In Situation 82.2.1 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 82.7.1 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 82.7.4. 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 72.5.3 to conclude. \square
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