This question is regarding the proof strategy presented in the paper, "On The Grayson Spectral Sequence", which its overview is explained in page 1 and 2. It seems a very general approach is presented for proving when two bigraded cohomology theories are the same, here one of the cohomology theories is the motivic cohomology and the other one is the Grayson motivic cohomology. I will give a brief overview of the method and then draw contradiction out of it, which has made me confused and I cannot figure out what am I missing.
Here is the short version of my issue:
The proof outline given in the fist two pages, seems to implies that $\mathbb{Z}(n)$ is quasi-isomorphic to $\mathbb{Z}(n)\oplus \mathbb{Z}(n-1)$, by simply replacing $\mathbb{Z}^{Gr}(n)$ by $\mathbb{Z}(n)\oplus \mathbb{Z}(n-1)$.
longer version:
We are going to work with the motivic cohomology $\mathbb{Z}(n)$ and another weighted cohomology theory $\mathbb{Z}'(n)$ with a natural map $\mathbb{Z}'(n)\rightarrow \mathbb{Z}(n)$, that we intend to show that it is a quasi-isomorphism. If the cohomology theory $\mathbb{Z}'(n)$ satisfies certain properties it can be shown that the quasi-isomorphism can be reduced to the case of fields. We won't care about these properties and only focus on the case of the fields. The idea of proving the quasi-isomorphism for the fields is induction on $n$ (weight). One should check manually that that quasi-isomorphism holds for $n\leq 0$. Let $\Delta^m$ be the $m$-dimensional algebraic simplex over the field $F$. We denote by $\hat{\Delta}^m$ be the semi-localization of the simplex on its vertices. Let $\mathcal{Z}$ be the family of supports consisting of all closed sub-schemes in the simplex that does not intersect the vertices. Because of purity (excision) property of the motivic cohomology we have the long exact sequence of the following form(check the paper for more details of the definition of these notations): $$H^{m+p-1}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z}(n))\rightarrow H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))\rightarrow H^{m+p}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n)) \rightarrow H^{m+p}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z}(n))$$
It been proved in the paper that $H^{m+p}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z}(n))=0$ for $m+p>n$.
The same sequence can written for $\mathbb{Z}'(n)$. Let's assume that we know that for example $H^{m+p}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z'}(n))=0$ for $m+p>n+1$. Then for a fixed $p$ and $n$ if $m$ is large enough we have the following isomorphisms:
$$H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))\cong H^{m+p}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))\cong H^p(F, \mathbb{Z}(n))$$ The last isomorphism is proved in the paper and follows from the homotopy invariance of motivic cohomology. We will also assume that $\mathbb{Z}'(n)$ is also homotopy invariant. Note that we just needed to assume that $m$ is large enough to get the above isomorphism. Because of the similar assumption on $\mathbb{Z}'(n)$ we get the isomorphism $H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z'}(n))\cong H^p(F, \mathbb{Z'}(n))$. Because of our inductive hypothesis that $\mathbb{Z}$ and $\mathbb{Z}'$ are quasi-isomorphic for weights $\leq n$ and the cohomological purity we have $H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z'}(n)) \cong H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))$, which implies that $H^p(F,\mathbb{Z}(n))\cong H^p(F, \mathbb{Z}'(n))$, which is the quasi-isomorphism for the weight $n$ and finishes the proof.
As you can see the proof seems to be very general, now I can give a counter-example for this. Assume that $\mathbb{Z}'(n)=\mathbb{Z}(n)\oplus \mathbb{Z}(n-1)$. Then it is easy to check that $\mathbb{Z}'(n)$ is a homotopy invariant sheaf which satisfies cohomological purity and at weights $\leq 0$ it coincides with the motivic cohomology. So the above argument shows that $\mathbb{Z}'(n)$ should be quasi-isomorphic to the motivic cohomology $\mathbb{Z}(n)$ on fields (and consequently on smooth schemes), which is obviously not correct.
So what am I missing about the proof in the paper? I'd really appreciate if anyone point me to the right direction and save me from the confusion.
