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I'm trying to show that the square where the opposite edges are glued together in the inverse direction is homemorphic to the real projective plane $ℝP^{2}$. I proofed that $ℝP^{2}$ is homemorphic to $S^²/\tilde{}$ with $x\tilde{}(-x)$, so finding a Homeomorphism to $S^2/\tilde{}$ would also be enough. My problem is that I don't really know where to start because I can't even imagine how the square would look after being glued together.

fundamental polygon

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cocreature
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  • You can think to the map that sends $BA$ to the equator and the interior of the square to a hemisphere... – Dario Nov 18 '14 at 13:50
  • Ah that makes sense. Is it possible to write this down explicitely? – cocreature Nov 18 '14 at 14:00
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    I think that what you need can be found here: http://math.stackexchange.com/questions/365920/prove-a-square-is-homeomorphic-to-a-circle – Dario Nov 18 '14 at 14:12
  • @Dario That's square to circle. What I was having trouble with was square to hemisphere or circle to hemisphere, but I just realized that I can just vertically project the points on the hemisphere, so I think I got it know. Thanks for your help. – cocreature Nov 18 '14 at 14:18

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