The inside of a closed compact surface $\sum_g$ It would be nice to have the following fact: The surface $\sum_g\subset S^3$ bounds a compact "solid $\sum_g$", call it $M_g$, on one side and some other unbounded manifold $U_g$ on the other. Note that $U_g$ depends on more than just $g$: if a torus is embedded as the boundary of a thickened knot, the unbounded component here is different from if the knot were the unknot.
I thought it would be nice to show this by induction. The statement for $g=1$ is "A torus bounds a solid torus on one side". I have had a very hard time seeing how to do this in the method of Alexander's Theorem. I know that, after setting up an appropriate Morse function, we can surger a torus into a disjoint collection of spheres that all bound balls, but what does this tell us about the reverse surgery? It tells us that not every reverse surgery joins two spheres into one... but that isn't much.
Now, even if the result were true for $g=1$, I don't think there's an easy way to do this inductively in the method of Alexander's theorem for similar reasons. How do we know where to do surgery so that we obtain $\sum_{g'}$ and $\sum_{g''}$ and then apply the inductive hypothesis?
Any insight about how to tackle this is appreciated.
 A: This is actually false in general.  There are genus $g$ surfaces which do not bound a solid genus $g$ torus, which are usually called handlebodies, on either side in $S^3$.  It is true for $g=1$ in $S^3$, though.  
To make the counter-example, take any nontrivial knot and thicken it to create a genus 1 surface surface, call this surface $S$.  So the inside is a solid torus and the outside is NOT a solid torus (which I did not prove here).  Then on the "inside" of that torus (the side that the meridian bounds a disk on) draw a nontrivially knotted arc which has its endpoints on the surface.  Thicken this new arc and cut that out of $S$, and call the new genus 2 surface $S'$.  This makes the inside of $S'$ no longer a solid torus and the outside still is not a solid torus. (Again, this needs proof)
This is exercise 4 on page 108 of Rolfsen, Knots and Links.

What you really want here is called a Heegaard splitting of $S^3$ or whatever 3-manifold you want.  You may assume you have a Heegaard splitting, and then each side of your genus $g$ surface is a handlebody, for some $g$.  See the wikipedia page I linked or Jesse Johnson's notes for more info on this.
