Submersion surjective on the complex projective space $\mathbb{C}P^1$. 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 $\pi$?
 A: 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 $n$.
A: 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. 
