Lemma 42.48.4. In Lemma 42.48.1 let $Y \to X$ be a morphism locally of finite type and let $c \in A^*(Y \to X)$ be a bivariant class. Then

$P'_ p(Q) \circ c = c \circ P'_ p(Q) \quad \text{resp.}\quad c'_ p(Q) \circ c = c \circ c'_ p(Q)$

in $A^*(Y \times _ X Z \to X)$.

Proof. Let $E \subset W_\infty$ be the inverse image of $Z$. Recall that $P'_ p(Q) = (E \to Z)_* \circ P'_ p(Q|_ E) \circ C$, resp. $c'_ p(Q) = (E \to Z)_* \circ c'_ p(Q|_ E) \circ C$ where $C$ is as in Lemma 42.47.1 and $P'_ p(Q|_ E)$, resp. $c'_ p(Q|_ E)$ are as in Lemma 42.46.1. By Lemma 42.47.4 we see that $C$ commutes with $c$ and by Lemma 42.46.5 we see that $P'_ p(Q|_ E)$, resp. $c'_ p(Q|_ E)$ commutes with $c$. Since $c$ is a bivariant class it commutes with proper pushforward by $E \to Z$ by definition. This finishes the proof. $\square$

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