Independence of the path implies that the fundamental group is abelian

Question

a) Prove that two paths $f,g$ from $x$ to $y$ give rise to the same isomorphism from $\pi(X,x)$ to $\pi(X,y)$ (i.e. $u_f=u_g)$ if and only if $[g*\bar{f}] \in Z(\pi(X,x))$.

b) Let $u_f: \pi(X,x) \rightarrow \pi(X,y)$ be the isomorphism determined by a path $f$ from $x$ to $y$. Prove that $u_f$ is independent of $f$ if and only if $\pi(X,x)$ is abelian.

I manage to prove part a) but not part b). I prove the backward direction of part b) by using part a). Can anyone give some hints on how to prove the forward direction?

Answer 1

If $u_f$ is independent of $f$ then for any path $g$ from $x$ to $y$ we have $u_f=u_g$ and so by the first part $[g\ast \overline{f}]\in Z(\pi_1(X,x))$. Now let $[\gamma],[\gamma']\in\pi_1(X,x)$ for loops $\gamma,\gamma'$ based at $x$. Remember that $[f\ast\overline{f}]=[c]$ where $c$ is the constant loops at $x$ and so $[\gamma]=[\gamma\ast f\ast\overline{f}]$ and $[\gamma']=[\gamma'\ast f\ast\overline{f}]$ but now note that $\gamma\ast f$ and $\gamma'\ast f$ are paths from $x$ to $y$ and so by using the first part we get

$$\begin{array}{rcl}[\gamma][\gamma'] & = &[\gamma\ast f\ast\overline{f}][\gamma'\ast f\ast\overline{f}]\\ &=&[(\gamma\ast f)\ast\overline{f}][(\gamma'\ast f)\ast\overline{f}]\\ &=&[(\gamma'\ast f)\ast\overline{f}][(\gamma\ast f)\ast\overline{f}]\\ &=&[\gamma'][\gamma]\end{array}$$

and so $\pi_1(X,x)$ is abelian.

Answer 2

Let $\gamma$ be a loop based at $x$, and suppose that you modify the path $f$, by concatenating it with the loop $\gamma$ to get a new path $g$. How do the isomorphisms $u_f$ and $u_g$ differ?