Lemma 42.46.5. In Lemma 42.46.1 let $f : Y \to X$ be locally of finite type and say $c \in A^*(Y \to X)$. Then

$c \circ P'_ p(E_2) = P'_ p(Lf_2^*E_2) \circ c \quad \text{resp.}\quad c \circ c'_ p(E_2) = c'_ p(Lf_2^*E_2) \circ c$

in $A^*(Y_2 \to Y)$ where $f_2 : Y_2 \to X_2$ is the base change of $f$.

Proof. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$. We may write

$\alpha = \alpha _1 + \alpha _2$

with $\alpha _ i \in \mathop{\mathrm{CH}}\nolimits _ k(X_ i)$; we are omitting the pushforwards by the closed immersions $X_ i \to X$. The reader then checks that $c'_ p(E_2) \cap \alpha = c_ p(E_2) \cap \alpha _2$, $c \cap c'_ p(E_2) \cap \alpha = c \cap c_ p(E_2) \cap \alpha _2$, $c \cap \alpha = c \cap \alpha _1 + c \cap \alpha _2$, and $c'_ p(Lf_2^*E_2) \cap c \cap \alpha = c_ p(Lf_2^*E_2) \cap c \cap \alpha _2$. We conclude by Lemma 42.45.3. $\square$

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