Problem Statement

For a path-connected space $X$, show that $\pi_1(X)$ is abelian iff all basepoint-change homomorphisms $\beta_h$ depend only on the endpoints of the path $h$.

Solution

First, suppose that $\pi_1(X)$ is abelian.

Let $x_0, x_1 \in X$ be given. Let $h_1, h_2$ be paths from $x_0$ to $x_1$. We claim that $\beta_{h_1} = \beta_{h_2}$.

Let $[f] \in \pi_1(X, x_1)$ be given. Then $[h_1 \cdot f \cdot \overline{h_2}], [h_2 \cdot \overline{h_1}] \in \pi_1(X, x_0)$.

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

Therefore, $\beta_{h_1} = \beta_{h_2}$.

On the other hand, suppose that all basepoint-change homomorphisms $\beta_h$ depend only on the endpoints of the path $h$. Let $x_0 \in X$ be given. Let $f, g \in \pi_1(X, x_0)$ be given. Let $x_1 = g(1 / 2)$. Define two paths $h_1, h_2$ from $x_1$ to $x_0$ such that

  • $h_1(t) = g((1 - t) / 2)$.
  • $h_2(t) = g((1 + t) / 2)$.

Then $[g] = [\overline{h_1} \cdot h_2]$.

By our assumption, $\beta_{h_1} = \beta_{h_2}$.

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

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