I know that similar question were asked and answered but I want to finish this very construction. In other answers different approaches are used (like construction of sections). I would like to know if there exists the inverse to the map defined below.
Fix the notation.
$$S^3 = \{(z,w) \in \mathbb{C}^2 \;| \; |z|^2 + |w|^2 = 1 \},$$
$$p: S^3 \to \mathbb{C}P^1, \; (z, w) \mapsto (z:w) \; (\text{homogeneous coordinates})$$
I am showing local triviality. I choose an open set
$$U_w = \{ (z:w) \; | \; w \neq 0 \} \cong \mathbb{C},$$
take its preimage
$$p^{-1}(U_w) = \{ (z,w) \subset S^3 \; | \; w \neq 1 \} $$
and construt a trivialization map
$$ \phi: p^{-1}(U_w) \to U_w \times S^1, \; (z,w) \mapsto ( (z:w), \frac{w}{|w|}). $$
Obviously $p = p_1\phi$, where $\pi_1: U_w \times S^1 \to U_w$ the canonical projection.
So it is almost it. But I fail to construct the inverse map to $\phi$.