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Projective plane - {0} homeomorphic to cyllinder?

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Assuming that [tex]\mathbb S^1 \times (0,1)[/tex] is not homeomorphic to [tex]\mathbb R^2[/tex] show that the sphere [tex]\mathbb S^2[/tex] is not homeomorphic to the projective plane [tex]\mathbb P^2[/tex].


    2. Relevant equations
    N/A


    3. The attempt at a solution
    This asks for a proof by contradiction - assuming [tex]\mathbb S^2 \cong \mathbb P^2[/tex] ("[tex]\cong[/tex]" meaning "homeomorphic"), we remove a point from [tex]\mathbb S^2[/tex] to get [tex]\mathbb R^2[/tex]. However, I cannot prove that by removing a point from the projective plane we get a cyllinder. I have considered [tex]\mathbb P^2[/tex] both as [tex]\mathbb S^2 / \mathbb Z_2[/tex] and producing it from the [tex]\mathbb R^3[/tex]. Any help appreciated!
     
  2. jcsd
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