Fibre Hopf Fibration diffeomorphic to $S^1$ I have been given the map 
$
h: S^3\to S^2
$
Given by $h(z_0,z_1)=(|z_0|^2-|z_1|^2,2i\overline{z_0}z_1)$.
I have proved that this map is a well-defined smooth submersion. Next is to show that the fibres of $h$ are an embedded submanifold of $S^3$ diffeomorphic to $S^1$. Now, the first part worked out well with the regular value theorem, but I get stuck with proving that the fibres are diffeomorphic to $S^1$. What I have already proved is that for $(a_1,b_1),(a_2,b_2)\in h^{-1}(\{q\})$, there is a $z\in S^1$ such that $(a_1,b_1)=(za_2,zb_2)$, i.e. two points in the fibre have the same norm. Can anybody help me please with proving the actual diffeomorphism?
 A: Fix $(a_o,b_o)\in h^{-1}(q)$.  Since at least one of $a_0, b_0$ is non-zero, we assume $a_0\neq 0$.
Now, define a map $f:h^{-1}(q)\rightarrow S^1$ by $f(a,b) = \frac{a}{a_0}$.
We first check that $f$ really does map into $S^1$.  But you have already shown that $a = za_0$ for some $z\in S^1$, so $\left|\frac{a}{a_0}\right| = |z| =1$.  Smoothness is clear since $z\mapsto \frac{z}{a_0}$ is a linear function on $\mathbb{C}$.
We now define an inverse $g:S^1\rightarrow h^{-1}(q)$ by $g(z) = (za_0, zb_0)$.  Does this really hap into $h^{-1}(q)$?  Yes, since $h(za_0, zb_0) = h(a_0, b_0) = q$.  Note that $g$ is obviously smooth since its a linear function when extended to $\mathbb{C}\rightarrow \mathbb{C}^2$.
Let's check compositions.  $(f\circ g)(z) = f(za_0, zb_0) = \frac{za_0}{a_0} = z$.  Also $(g\circ f)(a,b) = g\left(\frac{a}{a_0}\right) = \left(\frac{a}{a_0} a_0, \frac{a}{a_0} b_0\right) = \left(a, \frac{a}{a_0} b_0\right).$
So, if we can show that $\frac{ab_0}{a_0} = b$, then we're done.  But, as you showed above, we can write $(a,b) = (za_0, zb_0)$ for some $z\in S^1$.  Then $\frac{ab_0}{a_0} = \frac{z a_0 b_0}{a_0} = z b_0 = b$. 
