How to prove: Real projective plane is a manifold I need to correct and fill the gaps in my following home work problem
To show $\mathbb{R}P^2$ is a manifold, I defined the map $f:\mathbb{R}P^2\to\mathbb{D}=\{z\in\mathbb{C}:|z|\le1\}$ by $$f([x])=\dfrac{x_1+ix_2}{\sqrt{x_1^2+x_2^2+x_3^2}},$$ where $x=(x_1, x_2, x_3)\in\mathbb{R}^3.$ 
 As I calculated, for any $re^{i\theta}\in\mathbb{D},$ $$re^{i\theta}\to\left[\left(r\cos\theta, r\sin\theta, \sqrt{1-r^2}\right)\right]$$ gives the inverse map $f^{-1}:\mathbb{D}\to\mathbb{R}P^2.$
But I could not show that this is a homeomorphism between these two spaces. In fact, I doubted  that whether this a bijection or not.
Can someone help me to figure out this. 
Also still I have no fine feeling about projective planes and manifolds. Do you know a reference on these topics?
 A: We know that the real projective space $\mathbb{R}P^2$ is the quotient space of
$ \mathbb{R}^3-\{0\}$ by the equivalence relation $ x\sim y$ iff $y=tx$ for some arbitrary choice of $t$. We denote the equivalence class of a point in $ \mathbb{R}^3-\{0\}$ by $[a^0, a^1, a^2]$. To show that  $\mathbb{R}P^2$ is a manifold you need to define an atlas of homeomorphic charts i.e. $(U_i,\phi_i) $.
Let $U_0:=\{[a^0 , a^1 , a^2]\in \mathbb{R}P^2 : a^0\not=0\}$ then $\phi_0 \colon U_0\to \mathbb R^2$ such that $[a^0, a^1, a^2] \to(a^1/a^0,a^2/a^0)$. You can check it is continuous and has a continuous inverse $\phi_0^{-1} \colon (b^1,b^2)\to[1,b^1,b^2]$
and is therefore a homeomorphism.
More generally $\phi_i : U_i\to \mathbb R^2$ for $i \in \{ 0, 1, 2 \}$ such that $[a^0, a^1, a^2]\to(a^0/a^i,\hat a^i/a^i,a^2/a^i)$ where the $\hat a_i$ means the $i^{th}$ value is omitted. This shows that $\mathbb RP^2$ is locally Euclidean. The real projective space is second countable and Hausdorff. It is a manifold.
