The quotient cycle map has exactly the smooth-point image

Apply the supplied comparisons under [F3]. Let $Z$ be the free abelian group on smooth closed points and $R$ the completed Cartier-curve relation subgroup. [F1] asserts a homomorphism on $Z\mathrm{/}R$ whose composite with the quotient sends $[x]$ to the classical residue class. By [F2] the affine perfect comparison transports that homomorphism to the current K-group and sends the same generator to $[i_{\ast }\kappa (x)]$. Therefore every member of $R$, including each divisor generator, maps to zero under this actual point-class map. This descent is the exact imported established input; it is not a new local reconstruction of the relation theorem.

The quotient $Z\to Z\mathrm{/}R$ is surjective, so composing any homomorphism from $Z\mathrm{/}R$ with it does not change its image. The image of the resulting free-point map consists exactly of all finite integer linear combinations of the residue classes. By [F2] this is $F_{\mathrm{s}\mathrm{m}}^{\dim T}{K}_0(T)$, and this subgroup lies in the same-weight motivic filtration by the completed definition. Restricting the map's codomain gives the stated surjection. No assertion about its kernel beyond $R$ being killed has been made.