Lemma 42.59.13. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$ such that both $X$ and $Y$ are quasi-compact, regular, have affine diagonal, and finite dimension. Then $f$ is a local complete intersection morphism. Assume moreover the gysin map exists for $f$ and that $f$ is proper. Then

$f_*(\alpha \cdot f^!\beta ) = f_*\alpha \cdot \beta$

in $\mathop{\mathrm{CH}}\nolimits ^*(Y) \otimes \mathbf{Q}$ where the intersection product is as in Section 42.58.

Proof. The first statement follows from More on Morphisms, Lemma 37.59.11. Observe that $f^![Y] = [X]$, see Lemma 42.59.8. Write $\alpha = ch(\alpha ') \cap [X]$ and $\beta = ch(\beta ') \cap [Y]$ $\alpha ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ and $\beta ' \in K_0(\textit{Vect}(Y)) \otimes \mathbf{Q}$ as in Section 42.58. Set $c = ch(\alpha ')$ and $c' = ch(\beta ')$. We have

\begin{align*} f_*(\alpha \cdot f^!\beta ) & = f_*(c \cap f^!(c' \cap [Y]_ e)) \\ & = f_*(c \cap c' \cap f^![Y]_ e) \\ & = f_*(c \cap c' \cap [X]_ d) \\ & = f_*(c' \cap c \cap [X]_ d) \\ & = c' \cap f_*(c \cap [X]_ d) \\ & = \beta \cdot f_*(\alpha ) \end{align*}

The first equality by the construction of the intersection product. By Lemma 42.59.7 we know that $f^!$ commutes with $c'$. The fact that Chern classes are in the center of the bivariant ring justifies switching the order of capping $[X]$ with $c$ and $c'$. Commuting $c'$ with $f_*$ is allowed as $c'$ is a bivariant class. The final equality is again the construction of the intersection product. $\square$

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