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Why is P1 homeomorphic to S1?

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove P1 is homeomorphic to S1

    3. The attempt at a solution

    P1 is the set of all lines through the origin. I have shown for myself that this is homeomorphic to S1/R, where R is given by: (x,y) in R iff x=y or x=-y.
    This also shows there is a continuous surjective map f from S1 to S1/R. To prove S1 and S1/R are homeomorphic I need to show that f is injective, and that it has continuous inverse.

    However, I think there is something fundamentally wrong with my understanding of the problem, because I cannot imagine the map f being injective. To me it feels as though S1/R is the same as the 'northern hemisphere' of S1, because it sends -x and x to the same line through the origin (where x is an element of the real plane on the circle S1). So wouldn't that imply that f(x)=f(-x) while x is not equal to -x, so f is not injective, so P1 is not homeomorphic to S1?

    Anyone who can tell me why I'm fundamentally off here, and give me a push in the right direction, thank you.

    Jacob
     
  2. jcsd
  3. Sep 26, 2008 #2
    Think of the origin as a point at the top (north pole?) of the circle. As the lines pass through the origin they project on to the circle like a stereographic projection.
     
  4. Sep 26, 2008 #3
    Ah I was taking the centre as origin... it confused me deeply.

    But then again, isn't S1={x in the real plane | ||x||=1}... so why do we get to say the origin is on the top of the circle instead of in the middle?

    This new view will probably help me take a better shot at the problem later this evening.
     
    Last edited: Sep 26, 2008
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