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Math Help - Topology Question...

  1. #1
    TTB
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    Topology 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?

    Any help would be appreciated.

    Thanks x
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  2. #2
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    Quote Originally Posted by TTB View Post
    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 B_1, B_2, \cdots, B_k. Then A \times B = \bigcup_{i=1}^{k} (A \times B_i), where each  (A \times B_i) is simply-connected. We see that A \times B is not necessarily simply-connected. However, if we apply the smash product of A and B, A \vee B is identified for A \times B. Now each simply-connected space A \times B_i has a common point. It follows that the smash product 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 A \wedge B is simply connected having a trivial fundamental group.
    Last edited by aliceinwonderland; November 25th 2009 at 09:12 AM.
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  3. #3
    TTB
    TTB is offline
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    It's ok, I've sorted it out now.

    Thanks anyway x
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