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Hopf Fibration is $F:S^3\to S^2$ given by formula $F\left(z_1,z_2\right)=\left(\left(\phi^+\right)^{-1}\left(\frac{z_1}{z_2}\right)\right)$ for $z_2 \ne0$ and $F\left(z_1,0\right)=\left(1,0,0\right)$ where $\phi^+$ is stereographic projection from $\left(1,0,0\right)$ of $S^2$.

Now to prove this I tried to compute the rank of the Jacobian matrix of $\psi\circ F\circ\phi^{-1}$ with $\psi$ being the stereographic projection from North of $S^3$. But is this the easy way?

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up vote 3 down vote accepted

It depends on what you consider easy to be. First of all, since $\phi^+$ is a diffeomorphism between the $2$-sphere and the Riemann sphere, you can throw it away and consider only the map $G : S^3 \to \overline{\mathbb C}$ given by $G(z_1,z_2) = z_1/z_2$. And $G$ extends to a scale-invariant map $G : \mathbb C^2 \setminus \{(0,0)\} \to \overline{ \mathbb C }$ via the same formula. So $F$ is a submersion if and only if $G$ (with domain $\mathbb C^2 \setminus {(0,0)}$) is a submersion. That this latter map is a submersion is easy-enough to check -- you'll have to use two charts on the Riemann sphere $\overline{\mathbb C}$ to see it, but that's fine. You could also eliminate that step by making some symmetry observations.

A totally different way to make the argument would be to write $S^3$ as the union of two solid tori, and define the map on those solid tori, I believe Hatcher does this in his 3-manifolds notes. This gives not only a quick-and-dirty proof but it also gives you a strong intuition for how to visualize the map.

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