questions about fundamental group being abelian Let $X$ be a topological space and let $x\in X$.  Suppose that $\pi_1(X,x)$ is abelian.  Now I know that this means given two closed paths $f,g\in\pi_1(X,x)$ (i.e. start/end at x), then $[f]*[g]=[g]*[f]$.  What I can't see is how this even makes sense physically.  To me this says "first take the loop f, then g" is the same as "taking g, then f". Like if we look at loops on $S^1$, and f=go counterclockwise and g=go clockwise, then going clockwise->counterclockwise is certainly different from counterclockwise->clockwise.  Unless $[f]=[g]$, these seem like two different things and should never be equal.  Can someone give me a little more intuition on where I'm thinking about this incorrectly?
 A: I think the biggest confusion here is over the distinction between the loop $f$ and the corresponding element of the fundamental group, $[f]$. Given two loops, $f$ and $g$, we certainly cannot expect that concatenation in different orders to be the same. Indeed, in the example you gave, the $f\ast g\neq g\ast f$. But if we compare the corresponding elements of the fundamental groups, they $[f\ast g]=[g\ast f]$. This is because the fundamental group doesn't care about the loop. Rather, the fundamental group is concerned only with the homotopy properties of the loop, so if we can find a homotopy from $f\ast g$ to $g\ast f$, then these correspond to the same element of the fundamental group. In your example, $f\ast g$ is the path counterclockwise around the circle to the starting point, and then clockwise back around. We can find a homotopy which basically moves the point where the two paths meet clockwise around the circle tow times, and end up with the path $g\ast f$. Thus, $[g\ast f]=[f\ast g]$.
