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