Is $S^n/(v∼−v)$ diffeomorphic to $\Bbb RP^n$?

Question

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

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

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

Answer 1

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.