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$S^{n}/(v∼−v)$ is homeomorphic to $\mathbb{RP}^n$.

But, is $S^{n}/(v∼−v)$ diffeomorphic to $\mathbb{RP}^n$?

If $n=1$, the answer is yes.

Lee Mosher
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Clin
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  • Yes, although for my definition of the standard smooth structures on both of these it is obvious that they are diffeomorphic (because the smooth structure on $\mathbb{R}P^n$ is inherited from the quotient map from $S^n$). What definition do you have in mind? – hunter Aug 17 '19 at 15:50
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    (note that if you don't define a smooth structure on source and target then the question is a little broken.) – hunter Aug 17 '19 at 15:51

1 Answers1

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In fact, consider the smooth map:

$$f: \mathbb R^{n+1}-\{0\}\longrightarrow S^n, x\longmapsto \frac{x}{|x|}.$$

Let $\pi_2: S^n\longrightarrow S^n/(v\sim -v)$ be the canonical projection. This is also smooth. In particular, the composite is smooth $$\pi_2\circ f: \mathbb R^{n+1}-\{0\}\longrightarrow S^n/(v\sim -v).$$

This induces a smooth map at the level of quotient:

$$\overline{f}: \mathbb R\mathbb P^n\longrightarrow S^n/(v\sim -v)$$

This is the unique map such that $$\pi_2\circ f=\overline{f}\circ \pi_1$$ where $\pi_1: \mathbb R^{n+1}-\{0\}\longrightarrow \mathbb R\mathbb P^n$ is the canonical projection.

The map $\overline{f}$ will be the desired diffeomorphism.

PtF
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