10
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $(G_i^{ab})$? $\endgroup$ – Praphulla Koushik Dec 20 '19 at 14:38
  • 1
    $\begingroup$ Abelianization is a functor, so yes it is obvious the face and degeneracy maps extend. $\endgroup$ – Chris Schommer-Pries Dec 20 '19 at 16:08
  • $\begingroup$ @ChrisSchommer-Pries I guessed that would be the case... I never came across abelianization of simplicial groups.. I knew something about abelianization of groups.. :) $\endgroup$ – Praphulla Koushik Dec 20 '19 at 16:45
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.