Why the projective space $ \mathbb{P}^{1} $ is homeomorphic to $ S^{3}/S^{1} $?

Question

Define $ \mathbb{P}^{1} $ to be the space of lines in $ \mathbb{C}^{2} $ passing through the origin, i.e. $ \mathbb{P}^{1} = \{ \mathbb{C}^2 - \{ (0, 0) \} \} / {\mathbb{C}^{*}} $ where $ \mathbb{C}^{*} $ acts on $ \mathbb{C}^{2} $ by $ k(x, y) = (kx, ky) $ for $ k\in \mathbb{C}^{*} $.

Take $ S^{3} = \{ (z_0, z_1) \in \mathbb{C}^{2} | |z_0|^{2}+|z_1|^{2} = 1 \} $. $ S^{1}=\{e^{i\theta}\} $ the unit circle in the complex plane acts on $ S^{3} $ by the natural action. Then how to deduce that $ \mathbb{P}^{1} \cong S^{3}/S^{1} $?

Answer 1

This is an interpretation of the Hopf fibration. Define $f:S^3\rightarrow P^1$ by $f(z_1,z_2)=[z_1,z_2]\in P^1$. It is surjective and a fibre of an element is $S^1$.