Hopf fibration is fibration indeed

In my topology course, we defined Hopf fibration as the map $$p\colon S^3\to \mathbb{C}P^1$$, $$(z_0,z_1)\mapsto [z_0:z_1]$$, where $$S^3$$ is considered as the unit ball in $$\mathbb{C}^2$$: $$S^3=\{(z_0,z_1)\in\mathbb{C}^2|\,|z_0|^2|+|z_1|^2=1\}$$. Hometask question is to generalize the map on higher dimensions and prove that results are fibrations.

Howerever, I'm confused in this low-dimensional case already. To prove that Hopf fibration is truly fibration one has to find neihgbourhood $$U$$ for every point $$[z_0:z_1]$$ in $$\mathbb{C}P^1$$ that is homeomorphic to $$U\times S^1$$. I can show that $$p^{-1}[z_0:z_1]\cong S^1$$: namely, $$p^{-1}[z_0:z_1]=\{(zz_0,zz_1)|\,|z|=1/\sqrt{|z_0|^2+|z_1|^2}\}\cong S^1.$$ Further, let $$U_1=\{[z:1]| z\in\mathbb{C}\}\cong\mathbb{C}$$ and $$U_2=\{[1:z]| z\in\mathbb{C}\}\cong\mathbb{C}$$. My question is how do I show that $$p^{-1}(U_1)=\bigcup_{z\in\mathbb{C}}p^{-1}[z:1]$$ is homeomorphic to $$U_1\times S^1$$?

• How about $U_1\times S^1 \to p^{-1}(U_1)$ defined by $([z:1], \lambda)\mapsto \frac{\lambda}{\sqrt{1+|z|^2}} (z, 1)$ ? – Max Sep 25 '18 at 22:09