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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?

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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 $\frac{ab_0}{a_0} = \frac{z a_0 b_0}{a_0} = z b_0 = b$.

Jason DeVito - on hiatus
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    Incidentally, another proof that the fibers are $S^1$: The fibers are closed $1$-manifolds, and the only closed $1$-manifold is $S^1$. – Jason DeVito - on hiatus Sep 27 '18 at 14:32
  • Jason, I think you'd better put connected in there! You certainly would need to argue that the fibers are connected. – Ted Shifrin Sep 27 '18 at 17:32
  • @Ted: Oops. I often forget that disconnected manifolds exist! Thanks for the correction. – Jason DeVito - on hiatus Sep 27 '18 at 18:56
  • @TedShifrin What do you mean? Why do the fibres need to be connected? –  Sep 27 '18 at 20:57
  • @James: I think Ted was referring to my comment. My answer provides an explicit diffeo between $h^{-1}(q)$ and $S^1$. But my "easy" proof in the comments only works if you aleady know (a priori) that the fibers are connected. – Jason DeVito - on hiatus Sep 27 '18 at 21:01
  • @JasonDeVito Ahh okay! That makes more sense now. I was already wondering where your extensive proof assumed connectivity but I couldn't find it. So I assume connectivity is not needed for the extesnive proof you have written? –  Sep 27 '18 at 21:05
  • @James: Correct, I just wrote down an explicit diffeo. (Well, we all make mistakes, so I guess I'm not 100% confident... but I'm reasonably confident in my answer) – Jason DeVito - on hiatus Sep 27 '18 at 21:05