Lemma 42.69.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. The map
from Lemma 42.23.4 induces a bijection from $\mathop{\mathrm{CH}}\nolimits _ k(X)$ onto the image $B_ k(X)$ of the map
Lemma 42.69.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. The map
from Lemma 42.23.4 induces a bijection from $\mathop{\mathrm{CH}}\nolimits _ k(X)$ onto the image $B_ k(X)$ of the map
Proof. By Lemma 42.23.2 we have $Z_ k(X) = K_0(\textit{Coh}_{\leq k}(X)/\textit{Coh}_{\leq k - 1}(X))$ compatible with the map of Lemma 42.23.4. Thus, suppose we have an element $[A] - [B]$ of $K_0(\textit{Coh}_{\leq k}(X)/\textit{Coh}_{\leq k - 1}(X))$ which maps to zero in $B_ k(X)$, i.e., maps to zero in $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$. We have to show that $[A] - [B]$ corresponds to a cycle rationally equivalent to zero on $X$. Suppose $[A] = [\mathcal{A}]$ and $[B] = [\mathcal{B}]$ for some coherent sheaves $\mathcal{A}, \mathcal{B}$ on $X$ supported in $\delta $-dimension $\leq k$. The assumption that $[A] - [B]$ maps to zero in the group $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$ means that there exists coherent sheaves $\mathcal{A}', \mathcal{B}'$ on $X$ supported in $\delta $-dimension $\leq k - 1$ such that $[\mathcal{A} \oplus \mathcal{A}'] - [\mathcal{B} \oplus \mathcal{B}']$ is zero in $K_0(\textit{Coh}_{k + 1}(X))$ (use part (1) of Homology, Lemma 12.11.3). By part (2) of Homology, Lemma 12.11.3 this means there exists a $(2, 1)$-periodic complex $(\mathcal{F}, \varphi , \psi )$ in the category $\textit{Coh}_{\leq k + 1}(X)$ such that $\mathcal{A} \oplus \mathcal{A}' = H^0(\mathcal{F}, \varphi , \psi )$ and $\mathcal{B} \oplus \mathcal{B}' = H^1(\mathcal{F}, \varphi , \psi )$. By Lemma 42.69.2 this implies that
This proves that $[A] - [B]$ maps to a cycle rationally equivalent to zero by the map
of Lemma 42.23.2. This is what we had to prove and the proof is complete. $\square$
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