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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?

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2 Answers 2

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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. 

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Here's another way, quite similar to gnometorule's answer, but a bit different. Consider any $[f],[g] \in \pi_1(X,x)$. Your sufficient condition in particular implies $\hat{f} = \hat{g}$, so \begin{align} [f] = \hat{f}([f]) = \hat{g}([f]) = [\bar{g}] * [f] *[g] \ , \end{align} i.e. $[f]$ and $[g]$ commute.

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