9
$\begingroup$

I'm little bit puzzled with the following question, I was wondering if someone could help with a suggestion or a hint.

Let start with some notations. Suppose that $G$ is a Hausdorff topological group such that

  1. As a space $G$ is contractible.
  2. As a group $G$ is isomorphic to the free product $\mathrm{Q}\ast F(X) $ where $\mathrm{Q}$ is the additive abelian group of rational numbers and $F(X)$ is a free group generated by a set $X$.
  3. We assume that $\mathrm{Q}$ is a topological subgroup of $G$ in an obvious way and that $H=\{1\}\ast_{\mathrm{Q}} G$ is a retract of a free topological group (generated by a CW-complex).

Let $\mathbf{Ab}: \mathsf{Tgr}\rightarrow \mathsf{AbTgr}$ be the abelianization functor from the category of topological groups to abelian topological groups.

Since $G$ is a contractible group, for any natural number $n>0$ the power maps $p^{n}: G\rightarrow G$, $t\mapsto t^n$ are weak homotopy equivalences of underlying topological spaces.

My question is the following: is there a hope that the power maps: $p^{n}: \mathbf{Ab}(G)\rightarrow \mathbf{Ab}(G)$ induce weak homotopy equivalences of underlying topological spaces? The same question of $H$ and power maps $p^{n}: H\rightarrow H$, are they weak homotopy equivalences ?

Thank you in advance for any help!

Improve this question
$\endgroup$
16
  • $\begingroup$ I'm having a hard time picturing such a $G$. For one thing, a Hausdorff contractible space is uncountable, so I guess the set $X$ must be uncountable (and have an interesting topology). For another, you say that $G$ is an amalgamated product of $Q$ and $F(X)$ but do not mention what subgroup it's amalgamated over. Could you give an example of such a $G$? $\endgroup$
    – Kevin Casto
    May 1, 2019 at 23:07
  • $\begingroup$ This a list of questions rather than a helpful comment. (1) Are there example of such $G$ for some specific examples of $X$? So, what is $G$ if $X=\{a,b\}$ or $X=\mathbb{R}$. (2) As to the above comment, do the above desired properties of $G$ depends on whether or not $X$ is countable or not? (3) If you choose $X$ to be a finite set then is there any reason to believe that the truth of `hope' depends on the divisibility of $|X|$ by $p$ or $n$? $\endgroup$
    – user51223
    May 2, 2019 at 5:48
  • $\begingroup$ @KevinCasto roughly speaking, an ideal example is homotopy theoretical, take $\mathrm{Q}\rightarrow \ast$ the obvious homomorphism of groups. There is a factorization $Q\rightarrow G\rightarrow \ast$ where the first map is a cofibration (of topological groups) and the second map is a trivial fibration. $\endgroup$
    – Ilias A.
    May 2, 2019 at 17:35
  • $\begingroup$ @user51223 I just wrote a comment of an ideal example. in that case my question has a positive answer. the set X has not to be countable in general. $\endgroup$
    – Ilias A.
    May 2, 2019 at 17:38
  • 2
    $\begingroup$ @TimCampion I'm little bit confused, in my question the condition is that the topological group is contractibe as a space (not as a topological group). Your comment suggests that you are talking about a topological group contractible as a topological group, then I do agree that its abelianization is also contractible. $\endgroup$
    – Ilias A.
    May 10, 2019 at 19:24

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.