Find the transition map of the Hopf bundle $p:S^3\to S^2$.
The transition maps in this context is defined as
I know that when $S^2$ is viewed as $\mathbb{C}P^1$, which has a covering by two coordinate charts $U_0=\{[z_0,z_1]|z_0\neq0\}$ and $U_1=\{[z_0,z_1]|z_1\neq0\}$ with homeomorphisms $\phi_0:U_0\to\mathbb{C}$, $[z_0,z_1]\mapsto \frac{z_1}{z_0}$ and $\phi_1:U_1\to\mathbb{C}$, $[z_0,z_1]\mapsto \frac{z_0}{z_1}$. Then the transition maps $\phi_j\circ\phi_i^{-1}:\mathbb{C}\to\mathbb{C}$ are both given by $\lambda\mapsto\frac{1}{\lambda}$.
So, my questions are: Will this covering as defined above work in this case? And, what is the transition map of the Hopf bundle as defined above?
Let $$S^3= \{(z_0,z_1)\in \mathbb{C}^2\mid z_0\overline{z_0}+z_1\overline{z_1}=1\}$$
Then the Hopf bundle $S^3\to S^2 (=\mathbb{C}P^1)$, is given by $p\colon (z_0,z_1) \mapsto [z_0,z_1]$.
Let $U_0,U_1$ be as defined in the question. We have local trivialisations of this bundle: $$\phi_0\colon p^{-1}(U_0)\cong \mathbb{C}\times S^1,\qquad \phi_1\colon p^{-1}(U_1)\cong \mathbb{C}\times S^1,$$ given by: $$\phi_0(z_0,z_1)=\left(\frac{z_1}{z_0},\frac{z_0}{|z_0|}\right), \qquad \phi_1(z_0,z_1)=\left(\frac{z_0}{z_1},\frac{z_1}{|z_1|}\right). $$
The transition maps are then both given by: $$\Psi\colon (\lambda,\mu)\mapsto \left(\frac1\lambda,\mu \frac{\lambda}{|\lambda|}\right)$$
That is, at any point $\lambda\neq 0$ in one of the charts, the homeomorphism from the fibre ($=S^1$) to the corresponding fibre on the other chart, is given by multiplication by $\frac{\lambda}{|\lambda|}$.
Note we have:\begin{eqnarray*} \Psi\phi_0(z_0,z_1)&=&\Psi\left(\frac{z_1}{z_0},\frac{z_0}{|z_0|}\right)\\&=& \left(\frac{z_0}{z_1},\frac{z_0}{|z_0|}\frac{z_1}{z_0}\frac{|z_0|}{|z_1|}\right)\\ &=&\left(\frac{z_0}{z_1},\frac{z_1}{|z_1|}\right)\\&=&\phi_1(z_0,z_1), \end{eqnarray*} as required. By an identical calculation $\Psi\phi_1(z_0,z_1)=\phi_0(z_0,z_1)$ too.