# Proving that $\mathbb{R}P^n$ is a manifold

Consider $\mathbb{R}P^n$ as the quotient space of $S^n$ with antipodal points identified. Prove that $\mathbb{R}P^n$ is a manifold of dimension $n$.

(I'd like to clarify that I've seen the solution to this exersice when we see $\mathbb{R}P^n$ as the quotient of $\mathbb{R}^{n+1}$ with lines identified. I'd like to know if the same answer would solve the problem when it is quotient of $S^n$.)

I already proved $\mathbb{R}P^n$ is $T_2$ and second countable. By doing the same thing when $\mathbb{R}P^n$ is quotient of $\mathbb{R}^{n+1}$, I consider the open set:

$$V_i=\{x\in S^n:x_i\neq 0\},\quad \quad i=1,...,n+1,$$

and $$F_i:V_i\to\mathbb{R}^n,\quad F_i(x_1,...,x_{n+1})=\dfrac{1}{x_i}(x_1,..,x_{i-1},x_{i+1},...,x_n).$$

And then one should prove that $\phi_i:\pi(V_i)\to\mathbb{R}^n$ given by $\phi_i(\pi(x))=F_i(x)$ is a homeomorphism (where $\pi$ is the projection, which is open).

I already proved $\phi _i$ is injective and continuous, but I can't prove that it is surjective. If we take any $(x_1,..,x_n)\in\mathbb{R}^n$, the natural choice would be: $$F_i(x_1,...,x_{i-1},1,x_{i},...,x_n)=(x_1,...,x_n).$$

But $(x_1,...,x_{i-1},1,x_{i},...,x_n)$ need not be in $S^n$. Also, what would be $\phi^{-1}$? (In order to prove the inverse is continous...)

Or maybe $\phi_i$ is not surjective and we need another function. Any help?

Thank you.

To work with the sphere, it may be convenient to introduce the hemispheres $$V_{i}^{\pm} = \{x \in S^{n} : \pm x_{i} > 0\}.$$
An inverse $G_{i}^{\pm}:\Reals^{n} \to V_{i}^{\pm}$ of $F_{i}^{\pm}:V_{i}^{\pm} \to \Reals^{n}$ is $$G_{i}^{\pm}(x_{1}, \dots, x_{n}) = \pm\frac{(x_{1}, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_{n})} {\|(x_{1}, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_{n})\|}.$$ Surjectivity of $\phi_{i}$ follows from invertibility of $F_{i}^{\pm}$.