Diffeomorphism de $\mathbb{R}^n$ with open dense of $\mathbb{RP}^n$

Question

Define $F: \mathbb{R}^n \to \mathbb{RP}^n$ by $F(x^1,...,x^n) = [x^1,..., x^n,1]$. Show that $F$ is a diffeomorphism onto a dense open subset of $\mathbb{RP}^n$.

$F$ is smooth, since exist charts $Id_\mathbb{R}^n$ and $\phi: U_{n+1}\subset\mathbb{RP}^n \to \mathbb{R}^n$, with $U_{n+1} = \{[x^1, \dots, x^{n+1}] \in \mathbb{RP}^n \mid x_{n+1} \neq 0\}$, and $\phi([x^1,...,x^{n+1}]=(x^1/x^{n+1},..., x^n/x^{n+1})$, such that $\phi\circ F\space \circ Id_{\mathbb{R}^n}(x^1,...,x^n)=\phi([x^1,..., x^n,1])=(x^1,...,x^n)$, that clearly smooth.

The inverse $F^{-1}([y^1,...,y^n,y^{n+1}])=(y^1/y^{n+1},...,y^n/y^{n+1})$. This are my suggestions for exercise. But as i show that the image is a dense open in $\mathbb{RP}^n$?

Answer 1

Let $[x_{1},...,x_{n},0]$ be a point in $\mathbb{RP}^{n}\setminus F(\mathbb{R}^{n})$. Since the sequence $(x_{1},...,x_{n},1/k)$ converges to $(x_{1},...,x_{n},0)$ in $\mathbb{R}^{n+1}$ and the canonical projection is continuous, we have $[x_{1},...,x_{n},1/k]=[kx_{1},...,kx_{n},1]=F(kx_{1},...,kx_{n})\to[x_{1},...,x_{n},0]$, and thus $[x_{1},...,x_{n},0]\in \overline{F(\mathbb{R}^{n})}$.

Hope this helps!

Answer 2

HINT:

Consider the mappings: $$ G : \Bbb{R} ^n \to S^n \qquad (x_1,x_2,\dots x_n)\mapsto \frac{1}{(1+\sum_{i=1}^nx_i^2)^{1/2}}(x_1,x_2,\dots x_n, 1)$$ $$ \pi : S^n \to \Bbb{PR} ^n \qquad (y_1,y_2,\dots y_n, y_{n+1})\mapsto [y_1,y_2,\dots y_n, y_{n+1}]$$ Now, notice that $F=\pi \circ G$.