Hopf Fibration in Local Coordinates I have the following task: 
Consider the unit sphere $\mathbb S^3$ in $\mathbb R^4$. We know $\mathbb CP^1\simeq \mathbb S^2$ (homeomorphic). Identifying $\mathbb R^4$ with $\mathbb C^2$, we have a canonical inclusion $\imath: \mathbb S^3\rightarrow \mathbb C^2-\{0\}$. Composing this inclusion with the canonical projection $\mathbb C^2-\{0\}\rightarrow \mathbb CP^1$ we have a map $h:\mathbb S^3\rightarrow \mathbb S^2$ called Hopf Fibration. Write down $h$ in local coordinates. Identify $h^{-1}(p)\subseteq \mathbb S^3$ topologically.
I wrote down the following proof:
Possible Proof: We have a homeomorphism $\phi:\mathbb CP^1\rightarrow \mathbb S^2$ given by,
$$ \phi([z_1, z_2])=(2\textrm{Re}(\overline{z}_1z_2), 2\textrm{Im}(\overline{z}_1z_2), |z_2|^2-|z_1|^2),$$ and an inclusion, $\imath:\mathbb S^3\rightarrow \mathbb C^2-\{0\}$ given by.
$$\imath(x_1, x_2, x_3, x_4)=(x_1+ix_2, x_3+ix_4). $$ If $z=(x_1, x_2, x_3, x_4)\in \mathbb S^3$ then we can write $z=(z_1, z_2)\in \mathbb C^2-\{0\}$ where $z_1=x_1+ix_2$ and $z_2=x_3+ix_4$ with $|z_1|^2+|z_2|^2=1$. Hence, $$h(z)=(\phi\circ \pi)(z_1, z_2)=\phi([z_1, z_2])=(2\textrm{Re}(\overline{z}_1z_2), 2\textrm{Im}(\overline{z}_1z_2), |z_2|^2-|z_1|^2).$$
Problem: I don't know if what I've done was writing in local coordinate, was it? As to the identification $h^{-1}(p)$ I wasn't able to do it. Can anyone help me?
 A: Yes. What you wrote is correct. Instead of viewing $S^2\subset\Bbb{R}^3$ as you did, let
$$S^3=\{(\alpha,\beta)\in\Bbb{C}^2: |\alpha|^2+|\beta|^2=1\}$$ 
$$S^2=\{(x,z)\in\Bbb{R}\times\Bbb{C}: x^2+|z|^2=1\}$$
$$S^1=\{z\in\Bbb{C}: |z|^2=1\}$$
Then $h:S^3\rightarrow S^2$ can be written slightly more compactly as
$$h(\alpha,\beta)=(|\alpha|^2-|\beta|^2,2\alpha\bar{\beta})$$
Now let's come to the fibers $h^{-1}(p)$.
Claim: $\, h(\alpha_1,\beta_1)=h(\alpha_2,\beta_2)\iff (\alpha_1,\beta_1)=(z\alpha_2,z\beta_2) \ \mbox{ for some } z\in S^1$.
Consequence: each fiber $h^{-1}(p)\simeq S^1$ is a circle!
Proof of the Claim: If $z\in S^1$ then $|z|^2=z\bar{z}=1$, so
$$h(z\alpha,z\beta)=(|z|^2|\alpha|^2-|z|^2|\beta|^2,2z\bar{z}\alpha\bar{\beta})=h(\alpha,\beta)$$
Viceversa, suppose $h(\alpha_1,\beta_1)=h(\alpha_2,\beta_2)$. In polar form we can write
$$\alpha_k=r_ke^{i\theta_k}, \ \beta_k=s_ke^{i\phi_k} \quad (k=1,2)$$
Imposing the two conditions $(\alpha_k,\beta_k)\in S^3$ and $h(\alpha_1,\beta_1)=h(\alpha_2,\beta_2)$ we get
$$\begin{cases} r_1^2+s_1^2=r_2^2+s_2^2=1 \\ r_1^2-r_2^2=s_1^2-s_2^2 \end{cases} \Rightarrow \begin{cases} r_1=r_2 \\ s_1=s_2 \end{cases}$$
Moreover, it has to be $\theta_1-\theta_2=\phi_1-\phi_2$, which implies $\phi_1-\theta_1=\phi_2-\theta_2=\vartheta$ for some fixed $\vartheta$. Conclusion: $(\alpha_1,\beta_1)=(z\alpha_2,z\beta_2)$, where $z=e^{i\vartheta}$.
Edit: Some fibers stereographically projected on $\Bbb{R}^3$, as explained in the comments below:

