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 $\displaystyle B_1, B_2, \cdots, B_k$. Then $\displaystyle A \times B = \bigcup_{i=1}^{k} (A \times B_i)$, where each $\displaystyle (A \times B_i)$ is simply-connected. We see that $\displaystyle A \times B$ is not necessarily simply-connected. However, if we apply the smash product of A and B, $\displaystyle A \vee B$ is identified for $\displaystyle A \times B$. Now each simply-connected space $\displaystyle A \times B_i$ has a common point. It follows that the smash product $\displaystyle A \wedge B$ 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 $\displaystyle A \wedge B$ is simply connected having a trivial fundamental group.