In an arcwise connected topological space $X$, we can show that the two groups $\pi(X,x)$ and $\pi(X,y)$ are isomorphic for $x,y \in X$ by defining a mapping $u: \pi(X,x) \to \pi(X,y)$ by $\alpha \mapsto \gamma^{-1} \alpha \gamma$. $\gamma$ is a path class with inital point $x$ and terminal point $y$. This is how William Massey does it in his book A Basic Course in Algebraic Topology.
My question is this: Why is $\gamma^{-1} \alpha \gamma$ an element of $\pi(X,y)$? As I see it, $\gamma$ takes $x$ to $y$, $\alpha$ makes a loop at $y$ and $\gamma^{-1}$ takes it back to $x$. So why is it an element of $\pi(X,y)$?
Hope you can help!
I feel this situation is better understood as a basic fact on groupoids. Recall that a groupoid $G$ is a small category in which ever arrow is an isomorphism. Also $G$ is connected, or transitive, if for all objects $x,y$ of $G$, $G(x,y)$ is nonempty.
If $a \in G(x,y)$, and we write $G(x)$ for $G(x,x)$, then $a$ induces by conjugation an isomorphism $G(x) \to G(y)$, which is independent of the choice of $a$ in $G(x,y)$ if and only if $G(x)$ is abelian.
I don't write this out explicitly as there is a question of notation for composition: if $a \in G(x,y)$ and $b \in G(y,z)$ the their composite will lie in $G(x,z)$. Higgins book Categories and Groupoids writes it one way and my book Topology and Groupoids writes it the other!
The reasons for writing an abstract argument are as usual: 1) It applies to several known examples. 2) It will apply to new examples. 3) It simplifies the proof by concentrating on the relevant aspects.