Let x0 and x1 be points of the path-connected space X. Show that Pi_1(X,x0) is abelian iff for every pair a and b of paths from x0 to x1, we have a'=b', where a'([f])=[a-]*[f]*[a];( a- means the reverse of a.) and [f] belongs to Pi_1(X,x0). a':Pi_1(X,x0)->Pi_1(X,x1).
If Pi_1(X,x_0) is Abelian, consider two paths alpha and beta from x_0 to x_1.
If now [f] is in Pi_1(X,x_0), then [alpha]*[beta-] is another loop in Pi_1(X,x_0)
and [f] commutes with it: [alpha] * [beta-] * [f] = [f] * [alpha] * [beta-].
(no brackets necessary by the groupoid properties).
So if we premultiply both sides ...
Path-connected Space and Abelian Groups are investigated. The solution is detailed and well presented.