## 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 $P^n$.