RP^n Projective n-space

Define an equivalence relation on $S^n\subset\mathbb{R}^{n+1}$ by writing $v\sim w$ if and only if $v=\pm w$ . The quotient space $P^n=S^n/\sim$ is called projective $n$ -space. (This is one of the ways that we defined the projective plane $P^2$ .) The canonical projection $\pi: S^n\to P^n$ is just $\pi(v)=\{\pm v\}$ . Define $U_i\subset P^n$ , $1\leq i\leq n+1$ , by setting $\displaystyle U_i=\{\pi(x^1, \dots, x^{n+1})\mid x^i\neq 0\}.$

Prove
1) $U_i$ is open in $P^n$ .
2) $\{U_1, \dots, U_{n+1}\}$ covers $P^n$ .
3) There is a homeomorphism $\varphi_i: U_i\to\mathbb{R}^n$ .
4) $P^n$ is compact, connected, and Hausdorff, hence is an $n$ -manifold.

Proof:
1) $\pi^{-1}U_i=\{(x^1, \dots, x^{n+1})\mid x^i\neq 0\}$ is open in $S^n$ , so $U_i$ is open in $P^n$ .
2) Let $y=\pi(x^1,\dots, x^{n+1})\in P^n$ . Then since $(x^1,\dots, x^{n+1})\neq(0,\dots,0)$ , so $y\in\bigcup_{i=1}^{n+1}U_i$ . Hence $P^n\subset\bigcup_{i=1}^{n+1}U_i$ .
3) Consider $A=\{(x^1,\dots, x^{n+1})\mid x^i+1\}\cong\mathbb{R}^n$ . Define $\displaystyle \varphi_i(\pi(x^1,\dots, x^{n+1}))=(\frac{x^1}{\|x^i\|},\dots,\frac{x^{i-1}}{\|x^i\|},1,\dots,\frac{x^{n+1}}{\|x^i\|})$ for $x^i>0$ . If $x^i<0$ , then $\varphi_i(\pi(x^1,\dots, x^{n+1}))=\varphi_i(\pi(-x^1,\dots, -x^{n+1}))$ . Then $\varphi_i$ is well-defined.

$\displaystyle \varphi_i^{-1}(x^1,\dots,1,\dots,x^{n+1})=\pi(\frac{x^1}{\|v\|},\dots,\frac{1}{\|v\|},\dots,\frac{x^{n+1}}{\|v\|}),$ where $v=(x^1,\dots, 1,\dots, x^{n+1})$ . Both $\varphi_i$ and $\varphi_i^{-1}$ are continuous, so $\varphi_i: U_i\to A$ is a homeomorphism.
4) Since $S^n$ is compact and connected, so is $P^n=S^n/\sim$ . $P^n$ is a CW-complex with one cell in each dimension, i.e.\ $P^n=\bigcup_{i=0}^n e^n$ . Since CW-complexes are Hausdorff, so is $P^n$ .