Lemma 42.12.2. Let (S, \delta ) be as in Situation 42.7.1. Let X, Y, and Z be locally of finite type over S. Let f : X \to Y and g : Y \to Z be proper morphisms. 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 subscheme of dimension k. Consider W' = f(W) \subset Y and W'' = g(f(W)) \subset Z. Since f, g are proper we see that W' (resp. W'') is an integral closed subscheme of Y (resp. Z). We have to show that g_*(f_*[W]) = (g \circ f)_*[W]. If \dim _\delta (W'') < k, then both sides are zero. If \dim _\delta (W'') = k, then we see the induced morphisms
W \longrightarrow W' \longrightarrow W''
both satisfy the hypotheses of Lemma 42.11.1. Hence
g_*(f_*[W]) = \deg (W/W')\deg (W'/W'')[W''], \quad (g \circ f)_*[W] = \deg (W/W'')[W''].
Then we can apply Morphisms, Lemma 29.51.9 to conclude. \square
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