Prove that $ \pi_{1}(X, x) $ is abelian Let $ X $ be a path-connected space with $ x, y \in X $. Denote $ [f] $ the equivalence class of a path $ f $ and $ \pi_{1}(X, x) $ the collection of homotopy classes of closed paths at $ x $. Prove that $ [f].[g] = [g].[f] $ for every $ [f], [g] \in \pi_{1}(X, x) $ iff for every pairs of paths $ \alpha, \beta $ from $ y $ to $ x $, we have $ a_{\alpha} = a_{\beta} $ as homomorphisms from $ \pi_{1}(X, x) $ to $ \pi_{1}(X, y) $. We define $ a_{\alpha}: \pi_{1}(X, x) \to \pi_{1}(X, y) $ as $ a_{\alpha}([f]) = [f_{\alpha}] $. 
Can anyone drop a hint/suggestion on how to proceed for this problem? 
 A: Consider that if $\alpha: [0,1]\longrightarrow X,$ is any path from $x$ to $y$ then $\alpha\circ(1-t):[0,1]\longrightarrow X$ is a path  from $y$ to $x.$ Since $x$ is path connected, then $X$ is contained in a single path component, so $a_\alpha$ is an isomorphism of the fundamental groups, and this will hopefully help you.
A: $(\implies)$ Hint: If $[f]\cdot [g] = [g]\cdot [f]$ for all $[f],[g]\in\pi_{1}(X,x)$, then 
\begin{align*}
[\alpha]^{-1}\cdot \Big([\beta]\cdot [f]\cdot [\beta]^{-1}\Big)\cdot [\alpha] &= \Big([\alpha]^{-1}\cdot [\beta]\Big)\cdot [f]\cdot \Big([\beta]^{-1}\cdot [\alpha]\Big) \\
&= \Big([\alpha]^{-1}\cdot [\beta]\Big)\cdot\Big([\beta]^{-1}\cdot [\alpha]\Big)\cdot [f] = [f].
\end{align*}
$(\impliedby)$ Hint: Suppose for every $\alpha,\beta$ from $y$ to $x$ we had $a_{\alpha} = a_{\beta}$. Since $X$ is path-connected, we know some $\alpha$ exists, and we'll fix this $\alpha$. Then let $[f],[g]$ be any two loops about $x$. Then $a_{\alpha}\circ a_{f} = a_{\alpha}$ and so 
$$a_{\alpha}\circ a_{f} ([g]) = a_{\alpha}([g]).$$
