Let \(x_0\) and \(x_1\) be points of the path-connected space \(X\). Show that \(\pi_1(X, x_0)\) is abelian if and only if for every pair \(\alpha\) and \(\beta\) of paths from \(x_0\) to \(x_1\), we have \(\hat{\alpha} = \hat{\beta}\).
Solution
For two paths \(\alpha,\beta:x_0\to x_1\), set \[ \lambda=\alpha*\bar\beta, \] a loop at \(x_0\). For any loop \(f\) at \(x_0\), a direct path calculation gives \[ \widehat\beta^{-1}\widehat\alpha([f]) =[\lambda^{-1}*f*\lambda], \] so \(\widehat\beta^{-1}\widehat\alpha\) is the inner automorphism determined by \([\lambda]\in\pi_1(X,x_0)\).
If \(\pi_1(X,x_0)\) is abelian, every inner automorphism is the identity. Hence \[ \widehat\beta^{-1}\widehat\alpha=\operatorname{id}, \] so \(\widehat\alpha=\widehat\beta\) for every \(\alpha,\beta\).
Conversely, suppose \(\widehat\alpha=\widehat\beta\) for every pair of paths from \(x_0\) to \(x_1\). Fix one path \(\beta:x_0\to x_1\). For an arbitrary loop \(\lambda\) at \(x_0\), put \[ \alpha=\lambda*\beta. \] Then the displayed calculation shows that the inner automorphism induced by \([\lambda]\) is the identity. Since \([\lambda]\) was arbitrary, every element of \(\pi_1(X,x_0)\) is central. Thus \(\pi_1(X,x_0)\) is abelian.