Let \(\alpha\) be a path in \(X\) from \(x_0\) to \(x_1\); let \(\beta\) be a path in \(X\) from \(x_1\) to \(x_2\). Show that if \(\gamma = \alpha * \beta\), then \(\hat{\gamma} = \hat{\beta} \circ \hat{\alpha}\).
Solution
Recall that for a path \(\delta\) from \(u\) to \(v\), the change-of-basepoint map is \[ \widehat\delta([f])=[\bar\delta*f*\delta], \] with the convention that products of paths are traversed from left to right.
Let \(\gamma=\alpha*\beta\). Then \(\bar\gamma=\bar\beta*\bar\alpha\). Hence for a loop \(f\) at \(x_0\), \[ \begin{aligned} \widehat\gamma([f]) &=[\bar\gamma*f*\gamma]\\ &=[\bar\beta*\bar\alpha*f*\alpha*\beta]\\ &=\widehat\beta([\bar\alpha*f*\alpha])\\ &=(\widehat\beta\circ\widehat\alpha)([f]). \end{aligned} \] Associativity here is associativity of path multiplication up to the standard reparametrization homotopy, which is enough on homotopy classes. Therefore \[ \boxed{\widehat\gamma=\widehat\beta\circ\widehat\alpha}. \]