Representation of $S^{3}$ as the union of two solid tori 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.
 A: 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. 
A: The easiest way I know to do this is to think of $S^3$ as the boundary of $D^4$. This is the same as the boundary of $D^2\times D^2$, which is $S^1\times D^2 \cup D^2\times S^1$. 
By a similar trick you can cook up decomposition of spheres of other dimensions too. 
A: I'm not going to give you an actually formal proof, but a way to think about this, which anyway I find quite satisfactory. What you are asking for is known as a Heegaard splitting of a 3-manifold (in your case it's $S^3$), i.e. a decomposition of the manifold in two copies of the same "easier" 3-manifold with boundary, such that gluing these two copies along their boundary gives you the original manifold. The easy pieces are the so called handlebodies, which roughly speaking are the compact 3-manifolds bounded in $\mathbb{R}^3$ by closed orientable surfaces (if you restrict to the orientable case). So for example a 3-ball is a handlebody of genus $0$, a solid torus is a handlebody of genus $1$, and so on...You say that a 3-manifold has Heegaard genus $g$ if it admits a splitting made of handlebodies of genus $g$ (or higher) but not of genus $g-1$ (or lower).
In general you can Heegaard-split your manifold in several different ways, and conversely if you start from two handlebodies of the same genus, then you can glue them in several different ways and obtain several different (non homeomorphic) 3-manifolds. This correspondence between manifolds and Heegaard splittings is thus a little subtle and has something to do with the map you choose to do the gluing and the structure of the Mapping class group (=homeomorphisms modulo isotopy) of your 3-manifold and that of the boundary surface of the handlebodies, and the way they interact via the inclusion map.
I don't know if you are familiar with this things or not, so I'm being a little sketchy here; please let me know if this is clear or if you want me to elaborate more.
Anyway, some basic results tell you that:


*

*if you take two genus $g$ handlebodies and glue them via two maps which are isotopic (as homeomorphism of the genus $g$ surface which bounds the handlebodies ), then you get homeomorphic 3-manifolds. This can tell you why the mapping class group of the surface gets involved in these facts.

*the sphere $S^3$ has Heegaard genus $0$, i.e. has a splitting made of 3-ball. Put in another way you can take two 3-balls, glue them and obtain $S^3$. Let's see how this can be done. Choose two 3-balls $B_1$ and $B_2$ in $\mathbb{R}^3$ and a plane which stands between them as a mirror (I mean in such a way that you can obtain one ball from the other via a reflection in this plane). Now define a gluing of the boundary 2-spheres simply identifying a point on the first with its mirror image on the second. To visualize this, you can proceed like this: remove a point from each sphere, for example the farther from the mirror; then you are left with two half-spaces and the identification you are doing tells you to glue their boundary plane; so the result is just $\mathbb{R}^3$. Now remind of the points you previously took away: the chosen identification tells you to identify them in a single point. If you glue this single point to the $\mathbb{R}^3$ you have just obtained, then you get $S^3$ as the Alexandrov compactification of $\mathbb{R}^3$. Moreover, since the mapping class group of the 2-sphere is trivial, this is really the only thing you could obtain: whatever gluing you choose for your 3-balls you always end with a 3-sphere.

*another general result is this: if you have constructed a genus $g$ splitting for your manifold, then you can get a genus $g+1$ splitting of your manifold by removing a solid cylinder $I\times D^2$ from a handlebody and gluing it to the other. 


Let's see how this works on the genus $0$ splitting we constructed before for $S^3$. Now that we know $S^3$ is obtained from to 3-balls, we can represent them in an easy way: just think of $S^3$ as $\mathbb{R}^3$ plus a point at infinity, take the unit 3-ball as the first ball and it's complement in $S^3$ as the other ball (notice that according to 2. this is indeed a 3-ball). Now take a solid cylinder $I\times D^2$ inside the unit ball, such that its boundary 2-disks lie on the 2-sphere which bounds the ball; it's just an inscribed cylinder the basis of which have been pushed to the boundary of the ball. If you remove this cylinder from the unit ball, then the result is homeomorphic to a solid torus. But if you do this inside $S^3$, this cylinder is necessarily and automatically glued to the complement of the unit ball, which is another ball: so you get a ball with a solid cylinder glued to its boundary, and this is again homemorphic to a torus. So you have to solid tori in $S^3$ which are "linked somehow like a Hopf link" (ok, I know this is not easy to visualize, and is just some handwaving, but I think it helps). This is a genus $1$ splitting for $S^3$. Notice that these two tori are glued in such a way that a meridian of the first gets identified with a parallel of the other, so this is not the identity map of the torus boundary. If you did the gluing with the identity map then you would obtain $S^1 \times S^2$ and, as Neal pointed out in a comment, as you vary the gluing map you obtain all the lens spaces (and, as a matter of fact, nothing else).
A: EDIT: this was worked out for me by the same team of gerbils that does Jordan Normal Form of matrices when I need that. Amazingly versatile.
Take the coordinates as named $(w,x,y,z)$ to reduce subscripts. The sphere is $$  w^2 + x^2 + y^2 + z^2 = 1. $$ The common boundary is the Clifford Torus, $$ w^2 + x^2 = y^2 + z^2 = \frac{1}{2}.   $$ One solid torus is
$$ w^2 + x^2 \leq \frac{1}{2}, \; y^2 + z^2 = 1 - w^2 - x^2. $$ The other solid torus is $$ y^2 + z^2 \leq \frac{1}{2}, \; w^2 + x^2 = 1 - y^2 - z^2. $$
Or you could just write
$$ w^2 + x^2 \leq \frac{1}{2}, \; y^2 + z^2 \geq \frac{1}{2}  $$ and
$$ w^2 + x^2 \geq \frac{1}{2}, \; y^2 + z^2 \leq \frac{1}{2}.  $$
