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
Fix points \(x_0,x_1\in X\) and paths \(h,k\) from \(x_0\) to \(x_1\). Recall that \[ \beta_h([f])=[h\cdot f\cdot\bar h] \] for \([f]\in\pi_1(X,x_1)\).
1. Suppose \(\pi_1(X)\) is abelian.Then \(\beta_h=\beta_k\) whenever \(h\) and \(k\) have the same endpoints.
Since \(X\) is path connected, all fundamental groups at different basepoints are isomorphic, so it is enough that \(\pi_1(X,x_0)\) is abelian.
For \([f]\in\pi_1(X,x_1)\), set \[ a=[h\cdot\bar k]\in\pi_1(X,x_0), \qquad b=\beta_k([f])=[k\cdot f\cdot\bar k]. \] Then \[ a^{-1}=[k\cdot\bar h]. \] Using associativity of path classes and cancellation of \(\bar k\cdot k\) and \(\bar h\cdot h\) up to homotopy, \[ \begin{aligned} a b a^{-1} &=[h\cdot\bar k]\,[k\cdot f\cdot\bar k]\,[k\cdot\bar h]\\ &=[h\cdot f\cdot\bar h]\\ &=\beta_h([f]). \end{aligned} \] Since \(\pi_1(X,x_0)\) is abelian, \[ a b a^{-1}=b. \] Therefore \[ \beta_h([f])=\beta_k([f]) \] for every \([f]\), so \(\beta_h=\beta_k\).
2. Conversely, suppose every basepoint-change homomorphism depends only on the endpoints of its path.Then \(\pi_1(X,x_0)\) is abelian for every \(x_0\in X\).
Fix \(x_0\in X\) and let \(h\) be any loop based at \(x_0\). Let \(c\) be the constant path at \(x_0\). The paths \(h\) and \(c\) have the same endpoints, so by hypothesis \[ \beta_h=\beta_c. \]
For any \([f]\in\pi_1(X,x_0)\), \[ \beta_h([f])=[h]\,[f]\,[h]^{-1}, \] while \[ \beta_c([f])=[f]. \] Hence \[ [h]\,[f]\,[h]^{-1}=[f]. \] Multiplying on the right by \([h]\) gives \[ [h]\,[f]=[f]\,[h]. \] Both \([h]\) and \([f]\) were arbitrary, so \(\pi_1(X,x_0)\) is abelian.
3. Therefore \(\pi_1(X)\) is abelian if and only if the maps \(\beta_h\) depend only on the endpoints of \(h\).
The forward implication is 1 and the reverse implication is 2.