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
Comments (0)