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$f$ is a mapping: $(\mathbb{R}^2\setminus{0}) \rightarrow \mathbb{R}P^1$, which assigns to a point of the plane the line joining that point to the origin. The imbedding of $(\mathbb{R}^2\setminus{0})$ in $\mathbb{R}^2$ imbeds the graph of $f$ in the product $M= \mathbb{R}^2 \times \mathbb{R}P^1$. The closure of the graph in $M$ is called $S$. Prove that $S$ is diffeomorphic to a Möbius band.

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    It might help to think of M as a solid torus.2012-10-21

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You can take the open annulus $1 diffeomorphic to the punctured plane, for ease of seeing things. Suppose one draws an arrow from $(1,0)$ to $(2,0)$ and sees where it goes. After rotating through the angle $\pi$ the arrow ends up pointing from $(-1,0)$ to $(-2,0)$, and since you're in projective space you are identifying the "ends" of the annulus at this point. So that gives the twist in the usual Möbius band.

It's a good thing the overall dimension of $R^2$ X $RP^1$ is three, since I recall one cannot embed a strip into the plane. :-)

Not really sure this is rigorous, but it seems intuitive to me.

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    @azimut -- Any idea about a rigorous answer? I just dashed this off (many moons ago) and didn't think again about it.2013-05-06