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Math Help - Topology: Fund Grp of wedge of two real projective planes

  1. #1
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    Topology: Fund Grp of wedge of two real projective planes

    How can I compute the fundamental group of the wedge of two real projective planes? I would guess that it is (a,b | a^2 = 1 = b^2} with a,b each loops in each respective real projective plane, but I'm not seeing a way to prove it. I'd like to do it without using covering spaces if possible, maybe with SVK, but any suggestion is very welcome at this point.
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  2. #2
    Senior Member Tinyboss's Avatar
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    SVK will give you that the fundamental group of the wedge product is the free product of the fundamental groups, which agrees with your guess.
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  3. #3
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    I think this works but there is still a gap.

    \mathbb{P}^2 can be considered a closed disk with boundary identified antipodally. In the proof below I have assumes that the wedge point is taken in the open disk from each projective plane. If I wedge on the boundary of \mathbb{P}^2 I get confused - not clear how to choose the open sets.

    Here's the gap: It must be true that any two wedges of projective planes are homeomorphic, but how do you prove it.

    If I assume the above is true, then without losing generality we can wedge two projective planes at a point interior to \mathbb{D}^2 - call this wedge space X at the point q. Let A = \mathbb{P}^2 \vee \mathbb{D}^2 and B = \mathbb{D}^2 \vee \mathbb{P}^2. Then A, B are open, A \cup B = X, and A \cap B = \mathbb{D}^2 \vee \mathbb{D}^2 which has a trivial fundamental group. So SVK gives that \pi_1(X,q) = \pi_1(A,q) * \pi_1(B,q) \cong \{ a,b | a^2 = 1 = b^2 \} where * denotes the free group.
    Last edited by huram2215; May 13th 2011 at 05:57 AM.
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