Lemma 42.63.2. Let $(S, \delta )$ be as above. Let $X$ be a smooth scheme over $S$, equidimensional of dimension $d$. The map

$A^ p(X) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{d - p}(X),\quad c \longmapsto c \cap [X]_ d$

is an isomorphism. Via this isomorphism composition of bivariant classes turns into the intersection product defined above.

Proof. Denote $g : X \to S$ the structure morphism. The map is the composition of the isomorphisms

$A^ p(X) \to A^{p - d + 1}(X \to S) \to \mathop{\mathrm{CH}}\nolimits _{d - p}(X)$

The first is the isomorphism $c \mapsto c \circ g^*$ of Proposition 42.59.2 and the second is the isomorphism $c \mapsto c \cap [S]_1$ of Lemma 42.62.2. From the proof of Lemma 42.62.2 we see that the inverse to the second arrow sends $\alpha \in \mathop{\mathrm{CH}}\nolimits _{d - p}(X)$ to the bivariant class $c_\alpha$ which sends $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Y)$ for $Y$ locally of finite type over $k$ to $\alpha \times \beta$ in $\mathop{\mathrm{CH}}\nolimits _*(X \times _ k Y)$. From the proof of Proposition 42.59.2 we see the inverse to the first arrow in turn sends $c_\alpha$ to the bivariant class which sends $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Y)$ for $Y \to X$ locally of finite type to $\Delta ^!(\alpha \times \beta ) = \alpha \cdot \beta$. From this the final result of the lemma follows. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).