Why is the complex projective line homeomorphic to a $2$-sphere? Define an equivalence relation $\sim$ on $X={\bf C}^2\setminus \{(0,0)\}$ by 
$(x_1,y_1)\sim(x_2,y_2)$ if and only if there exists $t \in C\setminus\{0\}$ such that $(x_1,y_1)=(tx_2,ty_2)$
show that $X/{\sim}$ is homeomorphic to 2-sphere $S^2$
In this problem, $\bf C$ is complex plane as usual.
Can you help me please?
I have tried about 2 h. but failed..
 A: Hints (intuitive approach, develop it as much as you need): if $ 0 \neq \beta$ then you can kill $\beta$ by multiplying for $\beta^{-1}$ and you get $(\alpha \beta^{-1},1)$ so that you have one degree of freedom running over $\mathbb{C}$. You need to add all the points of the type $(\alpha,0)$ which are all indentified under your equivalence relation, thus you can think of them as a point at infinity: you've obtained the one point compactification of $\mathbb{C}$, or $\mathbb{R}^2$ for visual simplicity (which is topologically the same). 
A: Define the surjection 
$$q:\Bbb C^2\setminus\{0\}\ \to\ \hat{\Bbb C},\quad q(z,v)=z/v$$
 where $\hat{\Bbb C}$ is the one-point compactification of the complex numbers. We show that it is open and continuous:
The continuity of $q$ is clear on $\mathbb C \times (\mathbb C\setminus\{0\})$ since it can be composed as $(z,w)\mapsto (z,w^{-1})\mapsto zw^{-1}$, the last map being multiplication on $\mathbb C$.
It remains to show continuity in $(z,0)$. To this end, let $B_M(\infty)$ be a neighborhood with $\infty>M>1$. Let $\varepsilon=\lVert z\rVert/2M$. Now if $(x,y)\in B_\varepsilon(z,0)$, then $\lVert x\rVert >\lVert z\rVert (2M-1)/2M$ and $\lVert y\rVert <\lVert z \rVert /2M$, so $(2M-1)\lVert y\rVert  < \lVert x\rVert $, implying that $\lVert x/y \rVert > 2M-1 > M$.
For the openness of $q$, let $U$ be an open set and $(z,w)$ a point in $U$.
If $w\ne0$, then the map $z\mapsto z/w$ is a homeomorphism on $\mathbb C$, so $q[U]$ contains the open neighborhood $q[ U \cap (\mathbb C\times \{w\}) ]$ of $z/w$. 
In the case $w=0,z\ne 0$, the image $q[U]$ contains $q[\{z\}\times B_\varepsilon(0)\}]$ for some $\varepsilon>0$, which is equal to $\{z/y\ ;\ ||y||<\varepsilon\}$. This set is a neighborhood of $z/0=\infty$ since it contains $B_{||z||/\varepsilon}(\infty)$.
It follows that $q$ is open. Note that this also implies that the map $\mathbb C \to \mathbb C^*, w\mapsto z/w$ is an open map when $z\ne 0$.
You can then proceed by showing that $q$ respects the equivalence relation and thus induces a continuous bijection $\tilde q$ between $\mathbb CP^1$ and $\hat{\Bbb C}$, and this bijection is a homeomorphism because $q$ is an open map. Since $\hat{\Bbb C}$ is homeomorphic to $S^2$, you are finished.
A: In fact complex projective line biholomorphic to the Riemann sphere $S^2 $.
The idea is to construct a meromorphic function on $\Bbb{C}P^1$ which has single simple pole. Then use the fact there is one to one correspondence between the meromorphic function $f:\Bbb{C}P^1 \to \Bbb{C}$ and holomorphic function $F:\Bbb{C}P^1 \to S^2$.Therefore it's enough to prove the corresponding $F$ is biholomorphic.
First $F$ must be surjective since $\Bbb{C}P^1$  is compact and $S^2$ is connect.
Secondly the degree of $F$ is 1, in particular, it must be injective .Therefore as bijective holomorphic map is biholomorphic.


*

*the meromorphic function $f:\Bbb{CP}^1 \to \Bbb{C}$ such that $f([z_1:z_2]) = z_1/z_2$ has single simple pole.

*the degree of $F$ is 1 since $\text{deg}F = \text{multi}_{[z_1:0]} F = -\text{ord}_{[z_1:0]}f = 1$
