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If $S^3=\{ (z_1,z_2)\in\mathbb{C}^2\mid \vert z_1\vert^2+\vert z_2\vert^2=1\}$ and $\pi:S³\rightarrow\mathbb{C}P^1$ for $(z_1,z_2)\mapsto [(z_1,z_2)]$ since $[(z_1,z_2)]=\{ (w_1,w_2)\in\mathbb{C}^2-\bar{0}\mid (z_1,z_2)=\lambda(w_1,w_2),\lambda\in\mathbb{C}-\{0 \}\}$.

I need to proof that $\pi$ is a submersion surjective and $\pi$ induces a fibre bundle with fibre $S^1$.

The surjectivity is almost immediate since the equivalence relation can restrictes a $S^3$ and for the fibre for the fact that $S^1$ act on $S^3$, $i.e.$ $(w,(z_1,z_2))\mapsto w\cdot(z_1,z_2)\in S^3$ and then $\mathbb{C}P^1=S^3/S^1$. For Ehresmann's Lemma $\pi$ induces a fibre bundle since the $S^3$ is compact (and then $\pi$ is proper function). But I had a trouble to prove $\pi$ is a submersion because I dont know how to express the derivative $D\pi$ since that is a projection for the equivalece relation. How am I supposed to derive the function $\pi$?

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For $i=1,2$ let $V_i\subset \mathbb P^1$ be the subset defined by $z_i\neq 0$ and $U_i=\pi^{-1}(V_i)\subset S^3$.
We can write down explicitly a diffeomorphism (commuting with the projections onto $V_i$) between $U_i$ and the trivial bundle $V_i\times S^1$ as follows:

Define $$f_i:U_i\stackrel \cong\to V_i\times S^1: (z_1,z_2)\mapsto ([z_1:z_2],\frac {z_i}{ |z_i|})$$ and $$g_i:V_i\times S^1 \stackrel \cong\to U_i : ([w_1:w_2],s)\mapsto \frac {1}{||w||}(w_1\frac { |w_i|}{w_i}s,w_2\frac { |w_i|}{w_i}s)$$ Then $f_i,g_i$ are mutually inverse diffeomorphisms proving that $\pi$ is a fiber bundle with fiber $S^1$.

Remark
I have set up the notation in a way that makes it trivial to generalize the result to obtain a fibration $ S^{2n+1}\to \mathbb P^n(\mathbb C)$ with fiber $S^1$ just by allowing the index $i$ to run from $1$ to $n$.

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  • $\begingroup$ How does it follow (if it does at all) that $\pi$ is a submersion? $\endgroup$ – user500094 Feb 7 '18 at 19:10
  • $\begingroup$ I see, every fiber bundle is a submersion. But what is $w$ and $||w||$ in your proof? $\endgroup$ – user500094 Feb 8 '18 at 3:57
  • $\begingroup$ @user500094: $w$ is a point of $\mathbb C^2$ satisfying $w_i\neq0$ and $\vert\vert w\vert\vert$ is its euclidean norm. $\endgroup$ – Georges Elencwajg Feb 8 '18 at 11:51
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You could treat all of $CP^1$ except $z_2 = 0$ as being "parametrized" by $C$ via $z \mapsto (z, 1)$. Then the map you've defined just takes $(z_1, z_2) \mapsto z_1/z_z \in \mathbb C$. Now it's not so hard to compute the derivative.

You can then make a second coordinate chart for where $z_1 \ne 0$, via $z \mapsto (1, z)$, and do more or less the same thing.

The map you're looking at is called the Hopf Map, and is described in many topology books.

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  • $\begingroup$ Dear John, you are avoiding the difficulty because you don't take any account of the equation of the sphere. $\endgroup$ – Georges Elencwajg Nov 11 '14 at 18:30
  • $\begingroup$ If $(g,\mathbb{C})$ ($g[(z_1,z_2)]=z_1/z_2$) is the chart for $z_2\not= 0$ then $D\pi$ is surjetive if the derivative of $g\circ\pi$ is surjective. It is right? $\endgroup$ – Donyarley Nov 11 '14 at 19:22
  • $\begingroup$ I don't think so, @Georges. He still needs to prove that the derivative of map $K: S^3 \to \mathbb C : (z_1, z_2) \mapsto z_1/z_2$ has maximal rank on each tangent space, rather than on all of $\mathbb R^4$. I was just giving him a way to avoid thinking about equivalence classes in the codomain. $\endgroup$ – John Hughes Nov 11 '14 at 21:35
  • $\begingroup$ Yes, Donyarley, that's mostly correct...as long as you regard the domain of $D(g \circ \pi)$ as the tangent space to $S^3$ at $(z_1, z_2)$ rather than the tangent space to $\mathbb R^4$ at $(z_1, z_2)$. (The first is 3-dimensional; the second is 4-dimensional.) $\endgroup$ – John Hughes Nov 11 '14 at 21:37

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