Showing that the Hopf fibration is a non-trivial fibre bundle
The easier way to show that is to remark that $\pi_1(S_3)=1$ and $\pi_1(S_2\times S_1)=Z$.
Showing that the Hopf fibration is a non-trivial fibre bundle
The easier way to show that is to remark that $\pi_1(S_3)=1$ and $\pi_1(S_2\times S_1)=Z$.