$\mathbb{C}\mathbb{P}^1$ is homeomorphic to $S^2$ I just read on wikipedia that the the complex projective line is homeomorphic to the riemann sphere. How do I prove this? But, before that I have an extremely silly doubt that has been eating me. In the complex projective line, antipodal points are identified, but in the Riemann sphere they are clearly distinct. Then how can the two be homeomorphic. 
I found out the following map from the complex projective line to $S^2$. But how do I prove this is a homeomorphism? Finding the inverse of this functions seems complicated, and then how do I prove that the inverse map of a open subset is a open subset, and vice versa?
$f:\mathbb{C}\mathbb{P}^1 \to S^2 = [z:w] \to \frac{(2Re(w\bar{z}), 2Im(w\bar{z}),|w|^2-|z|^2)}{(|w|^2+|z|^2)} $
 A: Sorry I'm late to the party; I would comment on Berci's post but I'm not allowed to comment just yet.
While I'm sure you can actually prove that's a homeomorphism by computing inverses and showing that everything's continuous/smooth, I think most people would balk at actually carrying out such computations, and would simply accept that it can be done without actually doing it out.
But if you want, there is another way to see this fact using CW-complexes: on pages 6-7 of Hatcher's book he describes the standard CW-structure for $\mathbb{CP}^n$.
In this structure you get one $2k$-cell for each $k$ between $0$ and $n$, and no other cells.
This helps here, because it means that the CW-structure of $\mathbb{CP}^1$ is just a $2$-cell glued onto a $0$-cell ($=$ a point).
But there is only one way to do that, and that's to map all of the boundary of the $2$-cell to the point, which gives you the $2$-sphere.
A: The real projective line is basically a real line plus one point in the infinity.
The complex projective line is the complex plane plus one point in the infinity.
Have you heard about the stereographic projection?
A: This is actually a very interesting/useful formalism used in physics, usually mentioned as the Bloch sphere.
I will only provide a sketch. Indeed, $\mathbb{C}\mathbb{P}^1$ can be regarded as the set of rank-1 projections on $\mathbb{C}^2$, which can be denoted as $|\hat{n}\rangle \langle \hat{n}|$ (physicist notation) where $|\hat{n}\rangle$ is the unit vector in $\mathbb{C}^2$ being projected on. Then by the Bloch sphere isomorphism, one can show that
$$
|\hat{n}\rangle \langle\hat{n}| =\frac{1}{2}(I+\hat{n}\cdot\sigma)
$$
For some unique unit vector $\hat{n}\in \mathbb{S}^2$, where $\sigma=(\sigma_1,\sigma_2,\sigma_3)$ is the vector of Pauli matrices. The usual proof of this statement basically follows from the fact that
$$
|0\rangle \langle 0| = \frac{1}{2} (I+\sigma_3)
$$
Where $\sigma_3 |0\rangle = |0\rangle$, and that if $U\in SU(2)$ corresponds to a rotation $R\in SO(3)$ which take $e_3$ to $\hat{n}$, then
$$
U|0\rangle \langle 0|U^* =\frac{1}{2}(I+\hat{n}\cdot\sigma)
$$
One can also easily check that this is indeed the isomorphism the OP mentioned, i.e., if $\hat{n}$ is the unit vector in $\mathbb{S}^2$ specified by the spherical coordinates $(\theta,\phi)$, then the state $|\hat{n}\rangle$ is uniquely defined (up to a global phase) as
$$
|\hat{n}\rangle =(\cos{(\theta/2)}e^{-i\phi/2},\sin{(\theta/2)}e^{+i\phi/2})
$$
