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.
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, $$[f]=F_*[[a_1,b_1]\cdots [a_g,b_g]]=F_*(0)=0\hspace{1cm } \text{ in }H_1(X).$$