Proposition

Let $X$ be a path-connected topological space, and let $p, q \in X$. Show that all paths from $p$ to $q$ give the same isomorphism of $\pi_1(X, p)$ with $\pi_1(X, q)$ if and only if $\pi_1(X, p)$ is abelian.

Solution

Suppose that all paths from $p$ to $q$ give the same isomorphism of $\pi_1(X, p)$ with $\pi_1(X, q)$.

Let $[f], [g] \in \pi_1(X, p)$. We will show that $[f] \cdot [g] = [g] \cdot [f]$. Let $h$ be a path from $p$ to $q$. Such a path must exist since $X$ is path-connected. Then $f \cdot h$ and $g \cdot h$ are both paths from $p$ to $q$. By our assumption, $\beta_{f \cdot h} = \beta_{g \cdot h}$.

\[\begin{align*} \beta_{f \cdot h}([g]) = \beta_{g \cdot h}([g]) &\implies [\overline{f \cdot h} \cdot g \cdot f \cdot h] = [\overline{g \cdot h} \cdot g \cdot g \cdot h] \\ &\implies [\overline{h} \cdot \overline{f} \cdot g \cdot f \cdot h] = [\overline{h} \cdot \overline{g} \cdot g \cdot g \cdot h] \\ &\implies [\overline{h} \cdot \overline{f} \cdot g \cdot f \cdot h] = [\overline{h} \cdot g \cdot h] \\ &\implies [\overline{f} \cdot g \cdot f] = [g] \\ &\implies [g \cdot f] = [f \cdot g] \\ &\implies [g] \cdot [f] = [f] \cdot [g]. \end{align*}\]

Therefore, $\pi_1(X, p)$ is abelian.

On the other hand, suppose that $\pi_1(X, p)$ is abelian. Let $h_1, h_2$ be paths from $p$ to $q$. We will show that $\beta_{h_1}$ and $\beta_{h_2}$ are the same isomorphism.

$h_1 \cdot \overline{h_2}$ is a loop at $p$. Thus $[h_1 \cdot \overline{h_2}] \in \pi_1(X, p)$. Since $\pi_1(X, p)$ is abelian, we know that $[f] \cdot [h_1 \cdot \overline{h_2}] = [h_1 \cdot \overline{h_2}] \cdot [f]$.

\[\begin{align*} [f] \cdot [h_1 \cdot \overline{h_2}] = [h_1 \cdot \overline{h_2}] \cdot [f] &\implies [f \cdot h_1 \cdot \overline{h_2}] = [h_1 \cdot \overline{h_2} \cdot f] \\ &\implies [f \cdot h_1] = [h_1 \cdot \overline{h_2} \cdot f \cdot h_2] \\ &\implies [\overline{h_1} \cdot f \cdot h_1] = [\overline{h_2} \cdot f \cdot h_2] \\ &\implies \beta_{h_1}([f]) = \beta_{h_2}([f]). \end{align*}\]

Therefore, all paths from $p$ to $q$ give the same isomorphism.