Above is in May's book p7. I don't understand the last three sentences:
(i) From "If the group happens to be Abelian...this is just [f]" : $[b\cdot a ] , [(b\cdot a)^{-1}]$ are not necessarily elements of $\pi_1(X,x)$, so why does that imply $\gamma [ b \cdot a ] [f] = [f]?$.
(ii) From "... $\gamma[a]$ is independent of the choice of path class [a]": Does this mean $\gamma[a] [f]= \gamma[b][f]$ for all $a:x \rightarrow y, b:x \rightarrow y$?
Maybe I am completely misinterpreting something... I need some elaboration.
(i) He defines $b$ twice in two sentences; the one directly ahead of the one you're looking at says $b$ is a path from $y$ to $x$. So $[b\cdot a]$ is a path from $x$ to $y$ to $x$, an element of $\pi_1(X, x)$.
(ii)Yes, that is correct. If we think about $\gamma[a][f]$ represents, it means that you go from $y$ to $x$ along $a$ backwards, then loop around in $f$, then go back from $x$ to $y$ along $a$. If you take a different path, along $b$, we want to see that doing $\gamma[a][f]$ and then $\gamma[b][f]^{-1}$ leaves you the same place.
The trip you take is backwards from $y$ to $x$ along $a$, then around $f$, then back to $y$ along $a$, then back to $x$ via $b$, then backwards around $f$, then back to $y$ along $b$. But going around $a$ then backwards around $b$ is an element of $\pi_1(X, x)$, as is $f$; they commute. So this is the same as going backward through $a$ then forward through $a$, then backward through $b$ then forward through $f$ then backward through $f$ then forward through $b$; this is obviously trivial. So this is a more geometric/less symbol-pushing way to see that $\gamma[a][f]$ is independent of $a$.