TITLE: Does every torus $T\subset S^3$ bounds a solid tours $S^1\times D^2\subset S^3$?
QUESTION [9 upvotes]: I want to show this by using Alexander's Theorem's proof method.
So here's what I thought. 
As I surger $T$, I have 2 $S^2$. So one bounds $S^1$ and the other bounds $D^2$. 
By reversing the surgery, I have that $T$ bounds a solid torus $S^1\times D^2$ in $S^3$.
Is it correct?

REPLY [6 votes]: "There's also a proof of this that involves basically following along with what Hatcher did for spheres and... doing it for tori. I don't remember the details, but that's how I did this argument when I looked at the book first."
Let $T$ be an embedded torus in $\mathbb{R}^3$. Then isotope $T$ so that $h:T\to \mathbb{R}$, the height function, is a Morse function. Nothing fancy here, push $T$ around so that none of the saddle points, maximums, and minimums are at the same height. Then slice $T$ along horizontal planes at heights which trap each critical value of $h$ between two planes. Since $T$ is closed and $2$ dimensional, the intersections of $T$ with the horizontal planes will be circles. By cutting $T$ along innermost circles in the intersection with the horizontal planes, pushing the circles off of the planes, and capping the wounds with disks, we surger $T$ into a bunch of components which are isotopic to spheres.
If all of our surgery loops, pre-surgery, bound disks in $T$, then it turns out $T$ is a sphere. This can be seen by induction: The base case is after all of the surgeries. Our manifold is a collection of spheres. The inductive step is a single reverse surgery: by hypothesis will connect a two spheres together, which is always a sphere.
Thus there is a loop $\gamma$ which doesn't bound a disk in $T$ that does bound a disk in $\mathbb{R}^3$. This is the loop OP wanted.
Remark: This is the general proof that no surface $S\subset S^3$ is incompressible