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? (Worried)

Any help would be appreciated.

Thanks x

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- Nov 24th 2009, 08:00 PMTTBTopology Question...
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? (Worried)

Any help would be appreciated.

Thanks x - Nov 24th 2009, 10:10 PMaliceinwonderland
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. - Nov 25th 2009, 03:29 AMTTB
It's ok, I've sorted it out now. :)

Thanks anyway x