I want to show that if $X$ is a non-empty path connected space, then the fundamental group is abelian if and only if given any points $y, z\in X$ and paths $\alpha, \beta$ from $y$ to $z$, $\hat{\alpha} = \hat{\beta}$, where $(\hat{\alpha}([f]) = [\bar{\alpha}]*[f]*[\alpha]$ and similarly for $\beta$. I have one part of the iff:
For some $x\in X$, suppose $\pi_1(X, x)$ is abelian. Take $y, z\in X$ and paths $\alpha, \beta$ from $y$ to $z$. Then by isomorphism between the fundamental group at different base points of the same space, $\pi_1(X, z)$ is abelian. Let $f$ be a loop at $y$. Then $[\bar{\alpha}] * [f] * [\beta]$ is a loop at $z$, as is $[\bar{\beta}]*[\alpha]$, so
\begin{equation}
\begin{split}
\hat{\alpha}([f]) &= [\bar{\alpha}]*[f]*[\alpha]\\
&= ([\bar{\alpha}] * [f] * [\beta])*([\bar{\beta}]*[\alpha])\\
&= ([\bar{\beta}]*[\alpha])*([\bar{\alpha}] * [f] * [\beta])\\
&= [\bar{\beta}]*[f]*[\beta]\\
&= \hat{\beta}([f])
\end{split}
\end{equation}
However, I'm having trouble with the other direction, which is to show that if $f, g$ are loops, then $[f]*[g] = [g]*[f]$. Any ideas?
Best Answer
Hint: note that $f \ast \alpha := \gamma$ is a path from $x$ to $y$, and calculate $\bar{\gamma}$ by its definition. Then you are given:
$\hat{\gamma} ([g])= \hat{\alpha} ([g])$,
which should cancel to what you want.