The Stacks project

Proof. Consider the element $[\mathop{\mathrm{Spec}}(k)] \in \mathop{\mathrm{CH}}\nolimits _0(\mathop{\mathrm{Spec}}(k))$. We get a map $A^ p(X \to \mathop{\mathrm{Spec}}(k)) \to \mathop{\mathrm{CH}}\nolimits _{-p}(X)$ by sending $c$ to $c \cap [\mathop{\mathrm{Spec}}(k)]$.

Conversely, suppose we have $\alpha \in \mathop{\mathrm{CH}}\nolimits _{-p}(X)$. 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 $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ n(X')$ we let

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

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 42.33.2. 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

by Lemma 42.33.4. 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

when $X'$ is an integral scheme locally of finite type over $\mathop{\mathrm{Spec}}(k)$, see Lemma 42.35.3. 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. $\square$