# Extending loop to compact orientable genus $g$ space $\Sigma_g$.

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

• 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. – hunter Mar 8 at 14:36
• @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. – Lev Ban Mar 8 at 15:01