# Real projective plane and Möbius strip

Let $\mathbb{S}^2$ be the unit sphere in $\mathbb{R}^3$. Define the quotient space $$\mathbb{R}P^2:=\mathbb{S}^2/(x\sim -x).$$

How can we show that $\mathbb{R}P^2$ contains an embedding of the Möbius strip

$$M:=([0,1]\times (-1,1))/((0,y)\sim (1,1-y))$$

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## 1 Answer

The natural way to show this would be to explicitly write down the embedding.

For example, the Möbius strip can map to a thin belt around the equator of the sphere. Then the relation between this belt (in spherical coordinates!) and your defining parameterization of the Möbius strip is a simple affine transformation of each coordinate separately. Remember that $[0,1]$ should map to only one hemisphere of longitude due to the identification of antipodes.

Then there remains some slightly tedious fiddlework proving that your map respects the two relations you quotient out, and you're set.

But actually there's no reason to make it a thin belt. If instead you make the belt extend all the way to the poles, you'll have proved the stronger result that the real projective plane minus one point is homeomorphic to the Möbius strip without boundary.

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I posted a question following up on your last remark: math.stackexchange.com/q/830511/706. – Gerard Jun 11 '14 at 20:51