Understanding the Hopf fibration I'm taking a class in manifolds, and the Hopf fibration recently came up. I'm trying to get a handle on it, so I'm going to try and explain what I think is going on, and hopefully math.stackexchange will either correct me, or have more enlightening/sophisticated ways to explain what's going on.
The Hopf fibration maps circles (these are the 'fibres') of $S^3$ onto points of $S^2$. Identifying $S^3$ with $SU(2)$, is the above description telling us how many copies of $SO(2)$ (that is, $S^1$) fit inside $SU(2)$? I think this makes sense, since $SU(2)/SO(2)$ is isomorphic to $S^2$.
If one takes a 3-dimensional slice of $S^3$, one ends up with a an image that look like this. Are the fibres of $S^3$ that are being mapped to points of $S^2$ the Villarceau circles one gets from slicing these torii?
The fibre bundle map can be written $S^1\hookrightarrow S^3\xrightarrow{\pi}S^2$. So for $p\in S^2$, $\pi^{-1}(p)=S^1$. Wikipedia says that if $U$ is an open neighborhood of $S^2$, then $\pi^{-1}(U)$ should be homeomorphic to $U \times S^1$. How can this homeomorphism be realized? Is this related to the image I linked above, at all?
Sorry if this seems rambling/undirected. Feel free to ask for clarification etc.

If you consider a disk bounded by the red circle in your image, all of the copies of $S^1$ (the fibers) intersect that disk exactly once (with the exception of the the red outer circle).  You can identify this disk with $S^2$ by mapping all the points on that red boundary circle to one point.  Since every point in $S^3$ is on a fiber, you can then define $\pi : S^3 \rightarrow S^2$ by sending that point to where its fiber intersects the disk.   
