By definition, Voevodsky's motivic complex (an object of his $DM^{eff}_-$) is a complex of sheaves with transfers whose cohomology sheaves are homotopy invariant. Now, consider the complex (of functors) $C=X\mapsto (k(X)^*\to Div(X))$. As far as I understand, this is a complex of sheaves with transfers; its cohomology sheaves are $G_m$ and $Pic$. So, as an object of $DM^{eff}_-$ $C$ is isomorphic to a cone of a homomorphism from $\mathbb{Z}(1)[2]$ to $G_m[2]=\mathbb{Z}(1)[3]$. Since $Hom(\mathbb{Z}(1)[2], \mathbb{Z}(1)[3])\cong Hom(\mathbb{Z}, \mathbb{Z}[1])=0$, this complex should be quasi-isomorphic to the direct sum of its cohomology sheaves. Is there an explicit quasi-isomorphism here?

By definition, Voevodsky's motivic complex (an object of his $DM^{eff}_-$) is a complex of sheaves with transfers whose cohomology sheaves are homotopy invariant. Now, I consider the complex (of functors) $C=X\mapsto (k(X)^*\to Div(X))$. As far as I understand, this is a complex of sheaves with transfers; its cohomology presheaves are $G_m$ and $Pic$. So, it seems to be a motivic complex indeed.

In the previous version of this question I thought that $Pic$ gives a contribution of $\mathbb{Z}(1)[2]$ here. This seems to be very silly of me; since $Pic$ is given by the first (Zariski) cohomology of $G_m$, its Zariski (or Nisnevich) sheafification is zero, isn't it? So we just get $C\cong G_m$.

Besides, $X\mapsto H^iC(X)$ is a cohomology theory for varieties that satisfies the Mayer-Viertoris property (since both $X\mapsto k(X)^*$ and $X\mapsto Div(X)$ have this property). It seems to follow that the Zariski hypercohomology of $C$ coincides with its cohomology, though I don't know an easy proof.

By definition, Voevodsky's motivic complex (an object of his $DM^{eff}_-$) is a complex of sheaves with transfers whose cohomology sheaves are homotopy invariant. Now, consider the complex (of functors) $X\mapsto (k(X)^*\to Div(X))$. As far as I understand, this is a complex of sheaves with transfers; its cohomology sheaves are $G_m$ and $Pic$. So, as an object of $DM^{eff}_-$ it is isomorphic to a cone of a homomorphism from $\mathbb{Z}(1)[2]$ (or is it $\mathbb{Z}(1)[2]\oplus \mathbb{Z}$? sorry, I am somewhat confused here) to $G_m[2]=\mathbb{Z}(1)[3]$. It seems that there are no non-trivial homomorphisms here; so this complex should be quasi-isomorphic to the direct sum of its cohomology sheaves. Is there an explicit quasi-isomorphism here? Is my reasoning correct?

By definition, Voevodsky's motivic complex (an object of his $DM^{eff}_-$) is a complex of sheaves with transfers whose cohomology sheaves are homotopy invariant. Now, consider the complex (of functors) $C=X\mapsto (k(X)^*\to Div(X))$. As far as I understand, this is a complex of sheaves with transfers; its cohomology sheaves are $G_m$ and $Pic$. So, as an object of $DM^{eff}_-$ $C$ is isomorphic to a cone of a homomorphism from $\mathbb{Z}(1)[2]$ to $G_m[2]=\mathbb{Z}(1)[3]$. Since $Hom(\mathbb{Z}(1)[2], \mathbb{Z}(1)[3])\cong Hom(\mathbb{Z}, \mathbb{Z}[1])=0$, this complex should be quasi-isomorphic to the direct sum of its cohomology sheaves. Is there an explicit quasi-isomorphism here?