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.


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

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    $\begingroup$ Hint: $f^{-1} \circ f \circ \pi= \pi$ $\endgroup$ – Saal Hardali Dec 4 '17 at 10:35
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    $\begingroup$ $f$ is a diffeomorphism. $\endgroup$ – B. Pasternak Dec 4 '17 at 11:14
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    $\begingroup$ $f$ is a diffeomorphism so $f_*$ is an isomorphism, and then $f$ is both an immersion and a submersion. The same can be said for $f^{-1}$. I get it, thank you $\endgroup$ – Andrew_Paste Dec 4 '17 at 11:17
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    $\begingroup$ Yes, but I guess it depends on your prerequisites if you like the proof. You can obtain $\mathbb{P}^1(\mathbb{C})$ equally well as the quotient of $\mathbb{S}^3$ by the multiplication action of $\mathbb{S}^1$ on $\mathbb{S}^3$. This action is free and proper (freeness is easy to check, proper because all spaces are compact), hence there exists a unique smooth structure on $\mathbb{S}^3/\mathbb{S}^1$ s.t. the projection is a submersion. You can find the proof of this general fact (any proper, free group action of a Lie group on a manifold) in Lee's book for example. $\endgroup$ – B. Pasternak Dec 4 '17 at 12:42
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    $\begingroup$ Your derivatives don't make much sense and you should have a $2\times 3$ matrix for the derivative, once you work in charts. Here's an alternative suggestion. Consider the map $g\colon\Bbb C^2\to\Bbb C$ given by $g(z_1,z_2)=z_2/z_1$. Show that this is a submersion (officially you need to think about this in real coordinates — can you justify getting around that?). Now what does it take to see that the restriction to $S^3$ is still a submersion? How is $\ker dg_z$ related to $T_zS^3$? $\endgroup$ – Ted Shifrin Dec 4 '17 at 18:44

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