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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.

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up vote 7 down vote accepted

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 $H_1(X)$.

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Great and clear answer, thank you! – iwriteonbananas Feb 15 at 14:41

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