I was hoping someone could review my proof for correctness. Thanks in advance!
Problem: Let x$_0$ and x$_1$ be points of the path-connected space X. Show that $\pi_1$(X,x$_0$) is abelian if and only if for every pair $\alpha$ and $\beta$ of paths from x$_0$ and x$_1$, we have $\hat{\alpha}$ = $\hat{\beta}$
Note: $\hat{\alpha}$ is the isomorphism from $\pi_1$(X,x$_0$) to $\pi_1$(X,x$_1$) via the usual map using a path from x$_0$ to x$_1$.
Solution:
For the forward direction, suppose $\pi_1$(X,x$_0$) is abelian. Then since $\hat{\alpha}$ is an isomorphism we have that $\pi_1$(X,x$_1$) is also abelian. Then we have the following:
($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$) $\cdot$ ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$) = ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$) $\cdot$ ($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$)
Then using the fact that $\pi_1$(X,x$_0$) is also abelian we have:
($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $f$ $\cdot$ $\alpha$ $\cdot$ $\bar{\beta}$ $\cdot$ $\beta$) = ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $f$ $\cdot$ $\beta$ $\cdot$ $\bar{\alpha}$ $\cdot$ $\alpha$)
($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $f$ $\cdot$ $\alpha$ ) = ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $f$ $\cdot$ $\beta$)
Using the homomorphism property we have:
($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$) $\cdot$ ($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$) = ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$) $\cdot$ ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$)
Which yields:
($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$) = ($\bar{\alpha}$ $\cdot$ $f^{-1}$ $\cdot$ $\alpha$) $\cdot$ ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$) $\cdot$ ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$)
Which gives:
($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$) = ($\bar{\alpha}$ $\cdot$ $\alpha$) $\cdot$ ($\bar{\beta}$ $\cdot$ $f^{-1}$ $\cdot$ $f$ $\cdot$ $\beta$) $\cdot$ ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$)
=> ($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$ ) = ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$)
And so $\hat{\alpha}$ = $\hat{\beta}$
This part is wrong up to the horizontal line:
But we have that ($f$ $\cdot$ $f$) is homotopic to $f$ since:
Let i$_2$(s) = { (2s) for s $\in$ [0,$\frac{1}{2}$] and (2s -1) for s $\in$ [$\frac{1}{2}$,1] }, where i$_2$: [0,1] -> [0,1]. Also let i: [0,1] -> [0,1] be the identity map. Then since [0,1] is convex, we have i$_2$ is homotopic to $i$ via the straight-line homotopy. Then we have f $\circ$ i$_2$ is homotopic to f $\circ$ $i$ which implies $f$ $\cdot$ $f$ is homotopic to f. Then we have now established:
($\bar{\alpha}$ $\cdot$ $f$ $\cdot$ $\alpha$ ) = ($\bar{\beta}$ $\cdot$ $f$ $\cdot$ $\beta$)
Hence $\hat{\alpha}$ = $\hat{\beta}$.
For the reverse direction, suppose $\hat{\alpha}$ = $\hat{\beta}$ and that X is path connected. Now since X is path connected we have that all fundamental groups, $\pi_1$(X,x$_j$) are isomorphic, for any x$_j$ $\in$ X. Suppose $\pi_1$(X,x$_0$) is not abelian. Then there exists and $f$,$g$ $\in$ $\pi_1$(X,x$_0$) such that f and g do not commute. Then $g$ is a loop from x$_0$ to x$_0$ but can be written as follows:
Let $\phi$ be the path from x$_0$ to $w$ for some $w$ $\in$ image($g$), where the path $\phi$ follows the loop $g$ up to some point $w$ $\ne$ x$_0$. And let $\delta$ be the path from x$_0$ to $w$ using the remainder of the loop $g$. Note that such a $w$ $\ne$ x$_0$ exists since if $g$ is the constant loop onto x$_0$ then $g$ = e$_{x_0}$ and g must commute with f, a contradiction to the case we are in.
Then we have $g$ = $\phi$ $\cdot$ $\bar{\delta}$. Then using the equivalence of $\hat{\phi}$ and $\hat{\delta}$ we have: $\bar{\phi}$ $\cdot$ $f$ $\cdot$ $\phi$ = $\bar{\delta}$ $\cdot$ $f$ $\cdot$ $\delta$, which implies that we have $f$ $\cdot$ $\phi$ $\cdot$ $\bar{\delta}$ = $\phi$ $\cdot$ $\bar{\delta}$ $\cdot$ $f$, which in turn implies $f$ $\cdot$ $g$ = $g$ $\cdot$ $f$, a contradiction. Hence the fundamental group with base point x$_0$ must be abelian.
Note that we have the equivalence of $\hat{\phi}$ and $\hat{\delta}$ as both are derived from paths from x$_0$ to $w$, with both points in X, and hence by hypothesis the isomorphisms they induce between the groups $\pi_1$(X,x$_0$) and $\pi_1$(X,$w$) are equivalent, namely $\hat{\phi}$ and $\hat{\delta}$. (is this a correct interpreation of the hypothesis or are x$_0$ and x$_1$ fixed?)