 Solution

For loops $\alpha ,\beta$ based at $x_0$, the product of their transported loops is represented byThe middle $u\ast \bar{u}$ contracts relative to its endpoints, leaving $\bar{u}\ast (\alpha \ast \beta )\ast u$. Hence $u_{\#}$ is a group homomorphism. The reverse path induces the inverse: transporting back produces $u\ast \bar{u}\ast \alpha \ast u\ast \bar{u}$, whose outer retraced paths contract. Similarly the other composite contracts at $x_1$. ThereforeThis proves the asserted group isomorphism without needing any global path-connectedness hypothesis beyond the supplied path between the two points.