abelianization and homotopy

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?

• can you kindly give a reference for abelianization of simplicial group $G=(G_i)$.. Is it just $(G_i^{ab})$ component wise? Does degeneracy map and face maps on $(G_i)$ extend obviously to that of $(G_i^{ab})$? – Praphulla Koushik Dec 20 '19 at 14:38
• Abelianization is a functor, so yes it is obvious the face and degeneracy maps extend. – Chris Schommer-Pries Dec 20 '19 at 16:08
• @ChrisSchommer-Pries I guessed that would be the case... I never came across abelianization of simplicial groups.. I knew something about abelianization of groups.. :) – Praphulla Koushik Dec 20 '19 at 16:45

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.

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.