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In exercise 4-5 from John M. Lee's "introduction to smooth manifolds" I have proven that there is a surjective smooth submersion from $\mathbb C^2\setminus\{0\}$ to $\mathbb{CP}^1$, namely the map $\pi(z_1,z_2) = [z_1:z_2]$. The next exercise asks to show that $\mathbb{CP}^1$ is diffeomorphic to the sphere $\mathbb S^2$.

Theorem 4.31 states that if $M, N_1$ and $N_2$ are smooth manifolds, and $\pi_1 \colon M\to N_1$ and $\pi_2\colon M\to N_2$ are surjective smooth submersions that are constant on each other's fibers, then there exists a unique diffeomorphism $F\colon N_1\to N_2$ such that $F\circ \pi_1 = \pi_2$.

If possible I would like to use this theorem, because I already have a surjective smooth submersion from $\mathbb C^2 \setminus \{0\}$ to $\mathbb{CP}^1$. However, I need help with finding a good candidate map from $\mathbb{C}^2\setminus \{0\}$ to $\mathbb S^2$. Do you have any suggestions? Or do you have an argument why this would not work. Thanks in advance!

Math
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    Maybe this will help. – Kelvin Lois Oct 04 '20 at 12:21
  • @SiKucing thanks! Unfortunately they take another route, I was hoping to do it as described above because I already put much effort in showing that $\pi(z_1,z_2) = [z_1:z_2]$ is a surjective smooth submersion. – Math Oct 04 '20 at 15:52
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    Do you already know about the Hopf map $S^3\rightarrow S^2$? – Jason DeVito - on hiatus Oct 04 '20 at 16:19
  • @JasonDeVito I am not familiar with the Hopf map. Just started Lee's book. – Math Oct 04 '20 at 18:15
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    @Math: Well, it's a relatively famous (i.e., has its own wikipedia page) submersion $S^3\rightarrow S^2$. If you compose it with the map $\mathbb{C}^2\setminus 0 \rightarrow S^3$ given by $x\mapsto \frac{x}{|x|}$, you get what you want. I don't know how to motive discovering the Hopf map at this point.... – Jason DeVito - on hiatus Oct 04 '20 at 18:44
  • @JasonDeVito: In the Lee's book, problem 21-3 the author called this map $\Bbb S^{2n+1}\to \Bbb Cp^n$ a Hopf map. – C.F.G Oct 04 '20 at 18:48
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    @C.F.G: Sure, and under the diffeomorphism $\mathbb{C}P^1\cong S^2$, we are essentially talking about the same map. – Jason DeVito - on hiatus Oct 04 '20 at 18:49

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As Jason DeVito states one can use the map $\mathbb C^2 \setminus 0\to \mathbb S^3:x\mapsto \frac{x}{|x|}$ followed by the map $$ \mathbb S^3 \to \mathbb S^2 : (w,z) \mapsto \left( z\overline{w}+ w\overline{z}, iw\overline{z}-iz\overline{w}, z\overline{z}-w\overline{w} \right) $$ as defined in exercise 2-3(c) of Lee's introduction to smooth manifolds.

Math
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