Lemma 42.63.2. Let $(S, \delta )$ be as above. Let $X$ be a scheme locally of finite type over $S$. Then we have a canonical identification

for all $p \in \mathbf{Z}$.

Lemma 42.63.2. Let $(S, \delta )$ be as above. Let $X$ be a scheme locally of finite type over $S$. Then we have a canonical identification

\[ A^ p(X \to S) = \mathop{\mathrm{CH}}\nolimits _{1 - p}(X) \]

for all $p \in \mathbf{Z}$.

**Proof.**
Consider the element $[S]_1 \in \mathop{\mathrm{CH}}\nolimits _1(S)$. We get a map $A^ p(X \to S) \to \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$ by sending $c$ to $c \cap [S]_1$.

Conversely, suppose we have $\alpha \in \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$. Then we can define $c_\alpha \in A^ p(X \to S)$ as follows: given $X' \to S$ and $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ n(X')$ we let

\[ c_\alpha \cap \alpha ' = \alpha \times \alpha ' \]

in $\mathop{\mathrm{CH}}\nolimits _{n - p}(X \times _ S X')$. To show that this is a bivariant class we write $\alpha = \sum _{i \in I} n_ i[X_ i]$ as in Definition 42.8.1. In particular the morphism

\[ g : \coprod \nolimits _{i \in I} X_ i \longrightarrow X \]

is proper. Pick $i \in I$. If $X_ i$ dominates an irreducible component of $S$, then the structure morphism $p_ i : X_ i \to S$ is flat and we have $\xi _ i = p_ i^* \in A^ p(X_ i \to S)$. On the other hand, if $p_ i$ factors as $p'_ i : X_ i \to s_ i$ followed by the inclusion $s_ i \to S$ of a closed point, then we have $\xi _ i = (p'_ i)^* \circ c_ i \in A^ p(X_ i \to S)$ where $c_ i \in A^1(s_ i \to S)$ is the gysin homomorphism and $(p'_ i)^*$ is flat pullback. Observe that

\[ A^ p(\coprod \nolimits _{i \in I} X_ i \to S) = \prod \nolimits _{i \in I} A^ p(X_ i \to S) \]

Thus we have

\[ \xi = \sum n_ i \xi _ i \in A^ p(\coprod \nolimits _{i \in I} X_ i \to S) \]

Finally, since $g$ is proper we have a bivariant class

\[ g_* \circ \xi \in A^ p(X \to S) \]

by Lemma 42.33.4. The reader easily verifies that $c_\alpha $ is equal to this class (please compare with the proof of Lemma 42.63.1) and hence is itself a bivariant class.

To finish the proof we have to show that the two constructions are mutually inverse. Since $c_\alpha \cap [S]_1 = \alpha $ this is clear for one of the two directions. For the other, let $c \in A^ p(X \to S)$ and set $\alpha = c \cap [S]_1$. It suffices to prove that

\[ c \cap [X'] = c_\alpha \cap [X'] \]

when $X'$ is an integral scheme locally of finite type over $S$, see Lemma 42.35.3. However, either $p' : X' \to S$ is flat of relative dimension $\dim _\delta (X') - 1$ and hence $[X'] = (p')^*[S]_1$ or $X' \to S$ factors as $X' \to s \to S$ and hence $[X'] = (p')^*(s \to S)^*[S]_1$. Thus the fact that the bivariant classes $c$ and $c_\alpha $ agree on $[S]_1$ implies they agree when capped against $[X']$ (since bivariant classes commute with flat pullback and gysin maps) and the proof is complete. $\square$

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