Let $X$ be a path-connected space. For any $x_0, x_1 \in X$ and for any path $\gamma$ from $x_0$ to $x_1$, let $$\beta_\gamma : \pi_1(X,x_0) \to \pi_1(X,x_1)$$ be the change-of-basepoint homomorphism. Show that if for any paths $\gamma, \gamma' \in X$ with $\gamma(0)=\gamma'(0)$ and $\gamma(1)=\gamma'(1)$ we have equality $\beta_\gamma = \beta_{\gamma'}$, then $\pi_1(X,x)$ is abelian for any $x \in X$.
Pick $[f],[g] \in \pi_1(X, x)$, then both of these are equivalence classes of loops based at $x$. I thought if I consider now $\beta_f([f])=[f] = \beta_g([f]) = [g^{-1}][f][g]$ I get that $$[f] = [g^{-1}][f][g] \iff [g][f] = [f][g]$$
which would mean that $\pi_1(X,x)$ is abelian as $[f], [g]$ were arbitary.
Can I argue like this? The condition $\gamma(0)=\gamma'(0)$ and $\gamma(1)=\gamma'(1)$ essentially says that $\gamma^{-1}\ast \gamma'$ is a loop based at $\gamma(1)=\gamma'(1)$ and $\gamma \ast \gamma'^{-1}$ is a loop at $\gamma(0)=\gamma'(0)$.
Best Answer
Your argument is correct.
The change-of-basepoint isomorphism $\beta_\gamma$ depends only the path homotopy class $\Gamma = [\gamma]$ of $\gamma$ and thus has the form $\beta_\Gamma(a) = \Gamma^{-1} * a * \Gamma$, where $*$ denotes composition of path homotopy classes. In particular, for each $g \in \pi_1(X,x_0)$ we get the automorphism $$\beta_g : \pi_1(X,x_0) \to \pi_1(X,x_0), \beta_g (a) = g^{-1} * a * g .$$ This is nothing else than the conjugation automorphism by $g$.
Under the assumption of your exercise all $g \in \pi_1(X,x_0)$ induce the same automorphism, hence $\beta_g = \beta_e = id$, where $e$ is the neutral element.
Therefore $h = \beta_g(h) = g^{-1} * h * g$ for all $g,h \in \pi_1(X,x_0)$, i.e. $g * h = h * g$.