TITLE: Representation of $S^{3}$ as the union of two solid tori
QUESTION [26 upvotes]: Well, I'm trying to prove that you can express the 3-dimensional sphere $S^{3}$ as the union of two solid tori. I tried first use that a solid tori is homeomorphic to $S^{1}$$\times$$D^{2}$ and use this to obtain some quotient space which would be homeomorphic to $S^{3}$, but I couldn't go any further. Is this the way or I need a little more to prove that? Thanks in advance.

REPLY [6 votes]: The easier way is to think of $S^3$ as follows:
$S^3$ ={$ (z,w)\in  \mathbb{C}: |z|^2 +|w|^2=2$}. Take $V_1$={$ (z,w)\in  S^3: |z|\geq |w|$} and $V_2$={$ (z,w)\in  S^3: |z|\leq |w|$}. It is not difficult to show that $V_1$ and $V_2$ are both solid tori glued along their common boundary namely $\delta V_1$=$\delta V_2$={$ (z,w)\in  S^3: |z|=|w|$}, which is a torus.
More generally,
There are different ways of gluing two solid tori to get different 3-manifolds. One way is to glue a meridian curve on the ist torus boundary to a longitude curve on the second. If this has to be injective, you have to glue each meridian on the ist to a longitude of the second. This gives you $S^3$. Hopf fibration helps you visualize this: See a short video here movie. (However, you might need some skills to notice what is going on. You can ask me of course).
If on the other hand you glued a meridian on the ist torus boundary to a meridian on the second, what you get is an $S^1$-worth of 2-spheres i.e $S^2$ $\times$ $S^1$.
In the two cases above, we got irreducible 3-manifolds. But in general, to obtain a new 3-manifolds, you will have to glue the meridian of the ist to any non-trivial simple curve on the second. A fancy name for such a curve is $(p,q)$-curve, where $p$ and $q$ denote the number of meridians and number of longitudes respectively. The manifold formed is called a Lens space $L(p,q)$.
Something similar to Heegaard splitting discussed before is  Dehn surgery on knots. Here you can do the 3 procedures that I have described. It involves removing a regular neighbourhood of a knot in $S^3$ and gluing it back in a different way.