Lemma 41.50.2. Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$. Then we have a canonical identification

$A^ p(X \to \mathop{\mathrm{Spec}}(k)) = A_{-p}(X)$

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

Proof. Consider the element $[\mathop{\mathrm{Spec}}(k)] \in A_0(\mathop{\mathrm{Spec}}(k))$. We get a map $A^ p(X \to \mathop{\mathrm{Spec}}(k)) \to A_{-p}(X)$ by sending $c$ to $c \cap [\mathop{\mathrm{Spec}}(k)]$.

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

$c_\alpha \cap \alpha ' = \alpha \times \alpha '$

in $A_{n - p}(X \times _ k X')$. To show that this is a bivariant class we write $\alpha = \sum n_ i[X_ i]$ as in Definition 41.8.1. Consider the composition

$\coprod X_ i \xrightarrow {g} X \to \mathop{\mathrm{Spec}}(k)$

and denote $f : \coprod X_ i \to \mathop{\mathrm{Spec}}(k)$ the composition. Then $g$ is proper and $f$ is flat of relative dimension $-p$. Pullback along $f$ is a bivariant class $f^* \in A^ p(\coprod X_ i \to \mathop{\mathrm{Spec}}(k))$ by Lemma 41.31.6. Denote $\nu \in A^0(\coprod X_ i)$ the bivariant class which multiplies a cycle by $n_ i$ on the $i$th component. Thus $\nu \circ f^* \in A^ p(\coprod X_ i \to X)$. Finally, we have a bivariant class

$g_* \circ \nu \circ f^*$

by Lemma 41.31.8. The reader easily verifies that $c_\alpha$ is equal to this class and hence is itself a bivariant class.

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

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

when $X'$ is an integral scheme locally of finite type over $\mathop{\mathrm{Spec}}(k)$, see Lemma 41.31.12. However, then $p' : X' \to \mathop{\mathrm{Spec}}(k)$ is flat of relative dimension $\dim (X')$ and hence $[X'] = (p')^*[\mathop{\mathrm{Spec}}(k)]$. Thus the fact that the bivariant classes $c$ and $c_\alpha$ agree on $[\mathop{\mathrm{Spec}}(k)]$ implies they agree when capped against $[X']$ and the proof is complete. $\square$

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