TITLE: What is an intuitive explanation of the Hopf fibration and the twisted Hopf fibration?
QUESTION [10 upvotes]: As the question suggests, what is an intuitive explanation of the Hopf fibration and the twisted Hopf fibration? Many thanks in advance.

REPLY [4 votes]: Circle bundles $E \mapsto S^2$ are indexed by their Euler class, an element of $H_2(S^2;\mathbb{Z}) \approx \mathbb{Z}$. One can characterize this integer using obstruction theory, as the obstruction to the existence of a section of the bundle map $E \mapsto S^2$.
"The" Hopf fibration is (up to oriented bundle equivalence) the circle bundle of Euler class $+1$. 
The circle bundle of Euler class $0$ is simply the projection map to $S^2$ of the Cartesian product $E = S^2 \times S^1$. In general, the total space of the circle bundle of Euler class $n$ is the Lens space $L(n,1)$ (thanks to @PVAL-inactive for this correction)). I believe (the OP might want to confirm that this is the intent of the question) that these are the "twisted Hopf fibrations". Perhals, just as for the Hopf fibration itself (as explained in the answer of PVAL), one might be able visualize the Euler class in each of these cases as the linking number in the total space $L(n,1)$ of the fibers over two points of the base space $S^2$.