Trivialisation of the Hopf bundle Consider  the Hopf fibre bundle. $
\pi: S^3 \to S^2 $
and let $S^{3}=
\{(z_1,z_2) \in \mathbb C^2:  \left|z_{1}\right|^{2}+\left|z_{2}^{2}\right|=1\}$ and $ S^{2}=\{
(z_0,x)\in \mathbb C\times \mathbb R : \left|z_{0}\right|^{2}+x^{2}=1\}.
$
$\pi$ is defined by $\pi\left(z_{1}, z_{2}\right)=(\underbrace{2 z_{1} z_{2}^{*}}_{\mathbb{C}}, \underbrace{\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}}_{\mathbb{R}})$
I want to show that $\pi: S^3 \to S^2$ is a bundle with fibre $U(1)$, so we have to find local trivialisations. For this, we will use "stereographic coordinates" on $S^{2}$. These are defined on the open subsets $U_{N / S}$ of $S^{2}$ given by
$$
U_{N / S}=\left\{\left.\left(z_{0}, x\right)|| z_{0}\right|^{2}+x^{2}=1 \text {, and } x \neq \pm 1\right\},
$$
where $z_{0} \in \mathbb{C}$ and $x \in \mathbb{R}$. Thus $U_{N / S}$ cover all of the $S^{2}$ except for the north / south pole, respectively. We define $$
\phi_{S / N}: \pi^{-1}\left(U_{S / N}\right) \longrightarrow U_{S / N} \times \mathrm{U}(1)
$$
by
$$
\begin{aligned}
&\phi_{S}\left(z_{1}, z_{2}\right)=\left(\frac{z_{1}}{z_{2}}, \frac{z_{2}}{\left|z_{2}\right|}\right), \\
&\phi_{N}\left(z_{1}, z_{2}\right)=\left(\frac{z_{2}}{z_{1}}, \frac{z_{1}}{\left|z_{1}\right|}\right) .
\end{aligned}
$$

But the problem is that $U_{N/S}$ are opens of $S^2$ but here $z_1/z_2$ and $z_2/z_1 \not \in S^2$ ! I don't know if I misunderstand something or what ! any help please !
 A: Let me first observe that $U(1) = S^1 = \{ z \in \mathbb C \mid \lvert z \rvert= 1\}$. Moreover, for $(z_1,z_2) \in S^2$ we have $\lvert z_1 \rvert^2 +  \lvert z_2 \rvert^2 = 1$, thus $\lvert z_1 \rvert^2 -  \lvert z_2 \rvert^2 = 1 - 2 \lvert z_2 \rvert^2$. Hence we can write
$$\pi(z_1,z_2) = (2z_1z_2^*,1- 2 \lvert z_2 \rvert^2) = (2z_1z_2^*,1- 2 z_2 z_2^*) .$$
Let us consider $U_N$ ($U_S$ can be treated similarly). What you call "stereographic coordinates" is described by stereographic projection
$$p_N : U_N \to \mathbb C, p_N(z,x) = \frac{z}{1-x} .$$
It is well-known that this is a homeomorphism, its inverse being $F_N(w) = \dfrac{1}{\lvert w \rvert^2+1}(2w,\lvert w \rvert^2-1)$. See e.g. The range of the function $F$ is $S^2\setminus \{\textbf{n}\}$.
Your map $\phi_N$ is given by
$$\phi_N : \pi^{-1}(U_N) \to U_N \times S^1, \phi_N(z_1,z_2) = \left(F_N\left(\frac{z_1}{z_2}\right),\frac{z_2}{\lvert z_2 \rvert}\right) .$$
We have
$$\pi^{-1}(U_N) =  \{ (z_1,z_2) \in S^3 \mid 1- 2 \lvert z_2 \rvert^2 \ne 1 \}.$$
The condition $1- 2 \lvert z_2 \rvert^2  \ne 1$ is equivalent to $z_2 \ne 0$. Thus
$$\pi^{-1}(U_N) =  \{ (z_1,z_2) \in S^3 \mid z_2 \ne 0\} .$$
This shows that $\phi_N$ is well-defined. To see that $\phi_N$ is a trivialization over $U_N$, we have to check

*

*$q \circ \phi_N = \pi \mid_{\pi^{-1}(U_N)}$, where $q  : U_N \times S^1 \to U_N$ denotes projection.


*$\phi_N$ is a homeomorphism.
Proof of 1. :
We have to show that $F_N\left(\frac{z_1}{z_2}\right) = \pi(z_1,z_2)$. This is equivalent to $\frac{z_1}{z_2} = p_N(\pi(z_1,z_2))$. And in fact we compute
$$p_N(\pi(z_1,z_2)) = \frac{2z_1z_2^*}{1 - (1- 2 z_2 z_2^*)} = \frac{z_1}{z_2} .$$
Proof of 2. :
This is equivalent to showing that
$$\psi_N:   \pi^{-1}(U_N) \to \mathbb C  \times S^1, \psi_N(z_1,z_2) = \left(\frac{z_1}{z_2},\frac{z_2}{\lvert z_2 \rvert}\right) $$
is a homeomorphism. Define
$$\rho_N : \mathbb C  \times S^1 \to \pi^{-1}(U_N), \rho_N(w_1,w_2) = \frac{1}{\sqrt{\lvert w_1 \rvert^2 + 1}}(w_1,w_2) .$$
Then by definition $\rho_N(w_1,w_2) \in S^3$. Since $w_2 \in S^1$, we have $w_2 \ne 0$, thus the second coordinate of  $\rho_N(w_1,w_2)$ is $\ne 0$. This means that $\rho_N(w_1,w_2) \in \pi^{-1}(U_N)$.
It is now a routine exercise to verify $\rho_N \circ \psi_N = id$ and $\psi_N \circ \rho_N = id$.
