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.

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    $\begingroup$ This is not an answer to your question. I've seen the approach that you are taking before, so I know that it will work, but I have always viewed it as a bit of a pain. There is an alternative way: 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.) $\endgroup$ – Andres Mejia Feb 25 '17 at 9:57
  • $\begingroup$ @AndresMejia how do you conclude it is smooth (in the differentiability sense)? $\endgroup$ – Vim Feb 25 '17 at 10:02
  • $\begingroup$ Lee's book theorem 9.19 should do it, I searched it on SE, and here is a post that references it: math.stackexchange.com/questions/496571/… $\endgroup$ – Andres Mejia Feb 25 '17 at 10:05
  • $\begingroup$ @AndresMejia that would be definitely too advanced for me, but thanks still. I now think it might be best to introduce new compatible charts into the stereographic atlas of $S^n$. (Verification won't be so hard, since only orthogonal transformations would be involved, i suppose). $\endgroup$ – Vim Feb 25 '17 at 10:17
  • $\begingroup$ Okay, I hope that somebody gives you a good answer, I would try, but I think that this is a little too hard for me :). $\endgroup$ – Andres Mejia Feb 25 '17 at 10:18

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