I have to solve this one:
Let us consider the sphere $S^3\subset \mathbb{R}^4\cong\mathbb{C}^2$, and the map $\pi:S^3\rightarrow \mathbb{P}^1(\mathbb{C})\cong S^2$ defined as the projection's restriction $(z_1,z_2)\mapsto [z_1:z_2]$. Prove that $\pi$ is a submersion.
Solution:
I identify $\mathbb{P}^1(\mathbb{C})$ and $S^2$ through a diffeomorphism
\begin{array}{crcl} f:& \mathbb{P}^1(\mathbb{C}) & \longrightarrow & S^2 \\ & [z_1,z_2] & \longmapsto & \frac{1}{|z_1|^2+|z_2|^2}\big(|z_2|^2-|z_1|^2,2z_1\overline{z_2} \big) \end{array}
This way we have
\begin{array}{crcl} f\circ\pi:& S^3 & \longrightarrow & S^2 \\ & (z_1,z_2) & \longmapsto & \big(|z_2|^2-|z_1|^2,2z_1\overline{z_2} \big) \end{array}
because $|z_1|^2+|z_2|^2=1$. Now we compute the Jacobian matrix of $f\circ\pi$
\begin{equation} J_{f\circ\pi}=\left( \begin{array}{cc} 2\overline{z_2} & 2z_1\frac{2|z_2|z_2-|z_2|^2}{z_2^2} \\ 2|z_1| & -2|z_2| \end{array} \right) \end{equation} that has full rank, so $f\circ\pi$ is a submersion.
Now the thing is: how can I conclude that $\pi$ is a submersion?
Any other idea to solve it?
Thanks a lot