A is path-connected &
B is not empty & all path components of B are open in B.
Show that the suspension of the smsh product of A & B has trivial fundamental group.
Any ideas?
Any help would be appreciated.
Thanks x
OK. My idea is as follows:
Theorem (Munkres p161). A space X is locally path connected if and only if for every open set U of X, each path component of U is open in X.
B is locally path-connected by the above theorem. Let path components of B be . Then , where each is simply-connected. We see that is not necessarily simply-connected. However, if we apply the smash product of A and B, is identified for . Now each simply-connected space has a common point. It follows that the smash product is simply-connected. We see that a suspension of a simply-connected space is simply connected.
Thus we concluce that the suspension of a smash product is simply connected having a trivial fundamental group.