Here, $S^n\subset\Bbb R^{n+1}$ has the canonical stereographical projection $\phi_{N,S}$ (from north and south pole respectively) as its atlas, and $\Bbb RP^n$ has the following charts as its atlas: $$\psi_i: U_i\ni [x]\mapsto (x_1/x_i,\cdots,x_{i-1}/x_i,x_{i+1}/x_i,\cdots, x_{n+1}/x_i)\in\Bbb R^n$$ where $U_i=\{[x]\mid x\in\Bbb R^{n+1}\setminus\{0\}\,\text{and}\,x_i\ne 0\}$.
I have already done the case where $x$ is not on the "equator" i.e. $x_{n+1}\ne 0$: because we just have to check that, say when $x$ is on the upper hemisphere, the map $\psi_{n+1}\circ p\circ \phi_S(x)=\dfrac{2x}{1-\|x\|^2}$ is a local diffeomorhphism. Some calculations lead to its smooth inverse $$y\mapsto \frac12 y(1-(\sqrt{1+\frac1{\|y\|^2}}-\frac1{\|y\|})^2)$$ (perhaps there can be some glitches at $x = N$ i.e., $y=0$; I only check the differentiability, but not $C^1$ or higher.)
When $x$ is at the equator, i.e., $x_{n+1}=0$, my thought is to change the chart $\psi_{n+1}$ to some $\psi_i$ such that $x_i\ne 0$. This is of course possible, but would give rise to an asymmetric form of $\psi_{i}\circ p\circ \phi_S(x)$ and hence make it complicated to compute its inverse.
Perhaps I could also consider introducing new stereographic projections whose poles are different from $N,S$ into the atlas $\{\phi_{N,S}\}$ to $S^n$, but it will take some non-trivial effort to show the compatibility, I guess.
Anyway, is there a computationally economic way to complete the proof? Thanks.
The equivalence relation on $\mathbb R P^n$ restricts to the $n$-sphere under the action of the two element group $G={1,-1}$ (the antipodal map) so that $\mathbb R P^n := S^n/G$. The action is smooth, proper, and free (of course), so we see that projection $\pi:S^n \to \mathbb R P^n$ is a smooth submersion (and with a bit more effort, we can see a local homeomorphism.)
– Andres Mejia Feb 25 '17 at 09:57