$(z_1,z_2)\rightarrow [z_1:z_2]$ is a submersion Consider de map $\phi:\mathbb{S}^3\rightarrow\mathbb{CP}(1)$ given by $\phi(x_1,x_2,x_3,x_4)=[x_1+ix_2:x_3+ix_4]$ (a Hopf map). I want to prove that it is a submersion. By choosing for $\mathbb{CP}(1)$ atlas $A_{\mathbb{CP}}=\{(\varphi_i:U_i)\}, U_i=\{[z_1:z_2] |z_i\neq0\}, \varphi_i[z_1,z_2]=\frac{z_j}{z_i},j\in\{1,2\}-\{i\}$ and for $\mathbb{S}^3$ standart the stereographic atlas I was able to proof that $\phi$ is differentiable.
However I was not able to prove that the differential is surjective. I first tried to use the atlas above and calculate the differential at the basis $\{\partial x_i\}$ but the expression is very hard to get around. I also tried to change the sphere's atlas to $\{(\psi_i^{+},V_i^{+}),(\psi_i^-\,V_i^-)\}$ where $V_i^+=\{(x_1,...,x_4)|x_i>0\}$, $\psi_i^+(x_1,...,x_4)=(x_1,...,\hat{x_i},...,x_4)$ and the ones with the minus are defined in a similar way, but also withouth success. Right now I'm trying to prove that $\dim \ker d\phi = 1$, but without any results yet. If anyone could help me, I would be greateful.
 A: Hint: Things become easier if you view $S^3$ as a submanifold of $\mathbb C^2\setminus\{0\}$ (which is implicitly being done in the description of $\phi$). Then your original atlas immediately shows that $(z_1,z_2)\mapsto [z_1:z_2]$ is smooth. To prove that the map is a submersion, take $U_2:=\{(z_1,z_2):z_2\neq 0\}$. Then compute the derivative of the map $U\to\mathbb C$ defined by $(z_1,z_2)\mapsto \frac{z_1}{z_2}$. This is a linear map $\mathbb C^2\to\mathbb C$, which is obviously surjective and hence has one dimensional kernel. By computation or just by thought, this kernel has to be the complex line consisting of all complex multiples of $(z_1,z_2)$. Now if you take $z=(z_1,z_2)\in U_2\cap S^3$, the tangent space to $S^3$ in that point consits of all vectors $y=(y_1,y_2)$ such that $\langle z,y\rangle$ is purely imaginary. Thus the intersection of your kernel with this tangent space has real dimension one, which implies that the restriction of the derivative to the tangent space is surjective, which is what you want to prove. To deal with the missing point of $S^3$, you have to use $U_1=\{(z_1,z_2):z_1\neq 0\}$ in a similar way.
