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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 $U_1\times S^1$?

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

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