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I am working on the following problem.

$X$ is any space. Let $f:S^1\rightarrow X$ be a loop. Show that $[f]=0$ in $H_1(X)$ if and only if $f$ extends to a map $F:\Sigma \rightarrow X$ where $\Sigma$ is some compact orientable genus $g$ surface with one boundary component. ($Hint.$ $H_1$ is the abelianization of $\pi_1$.)

I could not solve only if direction ($\implies$). I have no idea how to approach. I hope to have any help on this problem.

I prove one half of this $(\leftarrow)$ as follows.

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Then note that $C=[a_1,b_1]\cdots [a_g,b_g]$ in $\pi_1(\Sigma).$

Suppose that $f$ extends to a map $F:\Sigma \rightarrow X$. Then observe that $$[f]=[F(C)]=F_*[C]=F_*\left[[a_1,b_1]\cdots [a_g,b_g] \right]$$

where $F_*:H_1(\Sigma)\rightarrow H_1(X)$ is the induced homomorphism. Now, recall that $H_1(X)$ and $H_1(\Sigma)$ are abelianizations of $\pi_1(X)$ and $\pi_1(\Sigma)$ respectively. Thus, $[[a_i,b_i]]=0$ $\forall i$ in $H_1(\Sigma).$ Therefore, $$[f]=F_*[[a_1,b_1]\cdots [a_g,b_g]]=F_*(0)=0\hspace{1cm } \text{ in }H_1(X).$$

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    $\begingroup$ Hint for the other direction: let $<f>$ be the class of $f$ in $\pi_1(X)$ (angle brackets bad notation here but square brackets used). By the hypothesis on the class of $f$ in $H_1(X)$, you can write it as a product of commutators. Now build a CW structure for the appropriate $\Sigma$ by attaching a cell along that word. $\endgroup$ – hunter Mar 8 at 14:36
  • $\begingroup$ @hunter Thank you! I have not thought about such direction. Could you give me more hint? I was trying to follow your hint but could not build the extended map. $\endgroup$ – Lev Ban Mar 8 at 15:01

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