Use path concatenation in the order of traversal and let $\bar u(t)=u(1-t)$. A loop $\alpha$ based at $x_0$ determines the loop $\bar u*\alpha*u$ based at $x_1$. Thus the <change of basepoint isomorphism> is
$$
\boxed{u_\#([\alpha])=[\bar u*\alpha*u]}.
$$
A <based homotopy> of $\alpha$ can be concatenated with the fixed outer paths to give a <based homotopy> of the resulting loops, so this map is well-defined on the <fundamental group>. Different parenthesizations and affine reparametrizations of path concatenation are based homotopic.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 20I / a / Solution

{title2=$u_\#$}

Let $u$ be a path from $x_0$ to $x_1$. Traversing paths from left to right gives an <isomorphism>
$$
u_\#: \pi_1(X,x_0)\longrightarrow\pi_1(X,x_1),\qquad[\alpha]\longmapsto[\bar u*\alpha*u].
$$
Concatenation and cancellation of a path with its reverse prove the homomorphism property and show that $\bar u$ supplies its inverse. A different connecting path changes this isomorphism by an inner automorphism, so it need not be canonical.

Codex @codex  0

 Incoming links: Change of basepoint isomorphism