Lemma 42.48.5. In Lemma 42.48.1 let $f : Y \to X$ be a morphism locally of finite type and $c \in A^*(Y \to X)$. Then $C \circ c = c \circ C$ in $A^*(W_\infty \times _ X Y \to X)$.
Proof. Consider the commutative diagram
\[ \xymatrix{ W_\infty \times _ X Y \ar@{=}[r] & W_{Y, \infty } \ar[r]_{i_{Y, \infty }} \ar[d] & W_ Y \ar[r]_{b_ Y} \ar[d] & \mathbf{P}^1_ Y \ar[r]_{p_ Y} \ar[d] & Y \ar[d]^ f \\ & W_\infty \ar[r]^{i_\infty } & W \ar[r]^ b & \mathbf{P}^1_ X \ar[r]^ p & X } \]
with cartesian squares. For an elemnent $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$ choose $\beta \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W)$ whose restriction to $b^{-1}(\mathbf{A}^1_ X)$ is the flat pullback of $\alpha $. Then $c \cap \beta $ is a class in $\mathop{\mathrm{CH}}\nolimits _*(W_ Y)$ whose restriction to $b_ Y^{-1}(\mathbf{A}^1_ Y)$ is the flat pullback of $c \cap \alpha $. Next, we have
\[ i_{Y, \infty }^*(c \cap \beta ) = c \cap i_\infty ^*\beta \]
because $c$ is a bivariant class. This exactly says that $C \cap c \cap \alpha = c \cap C \cap \alpha $. The same argument works after any base change by $X' \to X$ locally of finite type. This proves the lemma. $\square$
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