I am stuck at the following exercise from Hatcher section 3.3:

14. Let $X$ be the shrinking wedge of circles in Example 1.25, the subspace of $\mathbb{R}^2$ consisting of the circles of radius $1/n$ and center $(1/n, 0)$ for $n = 1, 2, \dots$

(a) If $f_n : I \to X$ is the loop based at the origin winding once around the nth circle, show that the infinite product of commutators $[f_1, f_2] [f_3, f_4]\dots$ defines a loop in $X$ that is nontrivial in $H_1(X)$. [Use Exercise 12.]

I am fine with the fact that $[f_1,f_2][f_3,f_4]\cdots$ actually defines a loop in $X$, but don't know how to prove that it's non-trivial in $H_1(X)$.

I tried following the hint. Exercise 12 is the following:

12. As an algebraic application of the preceding problem, show that in a free group $F$ with basis $x_1, \dots, x_{2k}$, the product of commutators $[x_1,x_2]\dots[x_{2k-1},x_{2k}]$ is not equal to a product of fewer than $k$ commutators $[v_i,w_i]$ of elements $v_i, w_i \in F$.

So from the fact that there exists a retraction $X\to\bigvee_{i=1}^n S^1$ for all $n\in \Bbb N$, we get that the inclusion $i:\bigvee_{i=1}^n S^1\to X$ is injective on $\pi_1$. Thus, from exercise 12, we can conclude that loops in $X$ defined by finite commutators $[f_1,f_2]\cdots[f_{2k-1},f_{2k}]$ are not homotopic to loops in $X$ expressed by fewer than $k$ commutators.

How can we use that to prove the original claim? How do we eventually pass to $H_1$? Of course all finite products of commutators are still trivial in $H_1(X)$, so I don't really know what to do with the conclusion above.


Since $H_1(X) = \pi_1(X)_{ab} = \pi_1(X) / [\pi_1(X), \pi_1(X)]$, if $f$ were trivial in $H_1(X)$, then it would be conjugate (in $\pi_1(X)$) to a finite product of commutators: $$f = g [u_1, v_1] \dots [u_k, v_k] g^{-1},$$ for some $g, u_i, v_i \in \pi_1(X)$. The retraction $X \to \bigvee_{i=1}^{k+1} S^1$ onto the $(k+1)$st circles induces on the fundamental group a map that sends $f$ to $[x_1, x_2] [x_3,x_4] \dots [x_{2k+1},x_{2k+2}]$, while it sends $g [u_1, v_1] \dots [u_k, v_k] g^{-1}$ to the conjugate of a product of $k$ commutators.

Using the same technique as in Exercise 12, you can use this relation to construct a map $M_k \to M_{k+1}$ of degree $1$, which is a contradiction by Exercise 11 (the conjugate does not really change much). So $f$ cannot be expressed as a conjugate of a finite number of commutators, and so is nontrivial in $H_1(X)$.

  • $\begingroup$ Great and clear answer, thank you! $\endgroup$ Feb 15 '16 at 14:41

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.