I deleted by previous questions, seems they are too vague. Let me try to ask a more precise question.


Let $f:G\rightarrow K$ a morphism of simplicial groups such that $f$ is a weak homotopy equivalence of underlying simplicial sets. We will make to assumptions:
1) for each natural number $i$, $G\_{i}$ is a free group.
2) for each natural number $i$, $K\_{i}$ is a subgroup of a product of free groups, I mean that $K\_{i}\subset \prod\_{j\in J} F\_{j}$ where each $F\_{j}$ is free. 


Now lets take the corresponding abelianization, we get $f^{ab}:G^{ab}\rightarrow K^{ab}$. Is $f^{ab}$ a weak homotopy equivalence of underlying simplicial sets?



abelianization and homotopy
The map is not necessarily a weak equivalence, even if $K\_i$ is actually the whole product of free groups rather than just a subgroup.


Let $K$ be the constant simplicial set which is $\Bbb Z^2$ in each degree. The map $K \to K\_{ab}$ is an isomorphism, and $K = K\_{ab}$ has no higher homotopy groups.


Let $G$ be a "cofibrant replacement" of $K$. This is work to describe explicitly, but there exists a version of $G$ which, in degree $n$, is free on $(n+2)$ generators. Namely, we let $G\_n$ be the free group on generators $x, r\_1, \dots, r\_n, y$, subject to relations:


* all the face & degeneracy maps take $x$ to $x$ and $y$ to $y$,
* the generator $r\_1$ in $G\_1$ satisfies $d\_0(r\_1) = [x,y]$ and $d\_1(r\_1) = 1$, and
* the elements $r\_i$ in $G\_n$ are the images of $r\_1$ under the $i$ different degeneracy maps $G\_1 \to G\_n$.


This forces the rest of the face and degeneracy maps by multiplicativity.


There is a map of simplicial groups $G \to K$ sending $x$ to $(1,0)$, $y$ to $(0,1)$, and $r\_i$ to $0$. One can check that this is an equivalence. (It roughly corresponds to the presentation of a torus, which is a $K(\Bbb Z^2,1)$, using two 1-dimensional cells and a 2-dimensional cell.)


However, in $G\_{ab}$ the first homotopy group is nonzero: the element $r\_1$ becomes a cycle because its boundary $[x,y]$ became trivial, but it's not in the image of the boundary map.




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All of this is an instance of an insight of Quillen's. If you take a simplicial group $K$, replace it by a weak equivalence $G \to K$ from a simplicial group $G$ that is levelwise free, and form $G\_{ab}$, then the homotopy group $\pi\_k G\_{ab}$ is the homology group $H\_{k+1}(BK;\Bbb Z)$. In this respect, group homology is a kind of derived abelianization procedure. Because, in your question, the groups $K\_i$ are subgroups of products of free groups and may have nontrivial higher group homology, we can't use them as replacements for free groups when forming these resolutions.


