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If we define the Möbius strip by a relation, $R$, by $(1,y)R(-1,-y)$ on the space $I^2$ how can one prove that it is homeomorphic to a subset of $\Bbb R^3$?

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up vote 1 down vote accepted

There are many realizations of the Möbius strip into $\mathbb{R}^3$ as an image of $I^2$. For example, here.

You could pick one and show that it is one-to-one, except on edges where points related by your relation are mapped to the same place. The smoothness and overall one-to-oneness of these maps gives you a homeomorphism. If you need details of continuity proofs, you could hash them out.

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I'm having trouble demonstrating this is a homeomorphism, certainly it is continuous but the other direction is giving me problems – user71839 Apr 17 '13 at 5:55
Try looking for a theorem that proves that such functions have continuous local inverses when their Jacobian has full rank everywhere. I was thinking about the gluing along the edges, and it might be less tedious to map from $\mathbb{R}^2$ to $\mathbb{R^3}$ factoring through $\mathbb{R}^2/I^2$. – alex.jordan Apr 17 '13 at 15:26

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