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Ok, if one looks at invariants like the fundamental group, the Euler characteristic or orientability, then it is immediate to see that $\mathbb R\mathbb P^2$ is not homeomorphic to $S^2$.

Is there any simple (or maybe not simple but still intersting) proof of this fact that makse no use of sophisticated invariants? (like homology, homotopy etc...)

The purpose is to teach this fact to a class without any of such tools.

user126154
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  • A simple loop always cuts a sphere into two parts. – Thomas Andrews Dec 27 '17 at 17:36
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    this is a big theorem – user126154 Dec 27 '17 at 17:37
  • Will such a class appreciate what $\Bbb RP^2$ is? – Arthur Dec 27 '17 at 17:37
  • @Arthur I hope so :) – user126154 Dec 27 '17 at 17:37
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    Well, I think you have it backwards. The tools you mention are useful partly because they very easily prove things like this. Trying to avoid them seems unpedagogical to me. Of course, if you have some really long proof not using them, and then show how simply it can be proven if you use them, then that's another story. – Arthur Dec 27 '17 at 17:39
  • @user126154 The Jordan curve theorem is a "big theorem," but it is quite intuitive. – Thomas Andrews Dec 27 '17 at 17:40
  • @Arthur You are right, but maybe there is a nice proof that I don't know. – user126154 Dec 27 '17 at 17:42
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    The Jordan Curve Theorem itself is, perhaps, a better example of the phenomenon you are pondering. The JCT does have "simple" proofs that "make no use of sophisticated invariants", but they are very very messy, intricate proofs. So if the first proof of the JCT that you ever saw was one of those "simple" proofs, you would be very pleased when you finally saw it proved using algebraic topology. Take a look at the discussion over here: https://mathoverflow.net/questions/8521/nice-proof-of-the-jordan-curve-theorem – Lee Mosher Dec 28 '17 at 13:55

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It is easy to check that removing a point from the sphere gives you something that continuously retracts to a point, whereas removing a point from the real projective plane gives you something that retracts to a circle. Therefore they can't be homeomorphic.

If you don't want to work with retractions, note that removing a point from a sphere leaves you with a disk, whereas removing a point from the real projective plane leaves you with an open Mobius band.

Mikhail Katz
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    why an open disc is different from an open moebius band? – user126154 Dec 27 '17 at 17:59
  • This follows from the Jordan curve theorem that you mentioned in the comments. Removing the central loop of the Mobius band produces an annulus, but removing a loop from a disk produces two connected components. – Mikhail Katz Dec 27 '17 at 18:09
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    I don't see any way to make either of these approaches work without using any of the big tools OP specifically said they want to avoid. – Eric Wofsey Dec 27 '17 at 20:25
  • Anyway JCT is considerably older than the invariants from algebraic topology mentioned in the OP's question, and also admits a nice hyperreal proof by Kanovei in Real Analysis Exchange. @EricWofsey – Mikhail Katz Dec 28 '17 at 09:41