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. $\square$

Comment #4893 by Peng DU on

In line 3 and 7 of the proof, the term (f∘g)∗[W] should be (g∘f)∗[W].

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