Let $\Sigma_g$ be a smoothly embedded surface with genus $g$ in $\mathbb R^3$. Can we prove that $\Sigma_g$ bounds a solid $g$-holed torus?
If $g=0$, the conclusion follows from the famous Alexander's Theorem. I was wondering if the higher-genus version holds.
On
$\text{The following is incorrect; see Moishe's counterexample}$
Yes, let's proceed by induction. You already remark the base case holds. So suppose we have proven it true in the case of $g$. Now take an embedding $\Sigma_g \rightarrow \mathbb{R}^3$. We may perform surgery on $\Sigma_{g+1}$ to change it to $\Sigma_{g}$. This amounts to excising an embedded cylinder in $\Sigma_{g+1}$ and gluing on two disks to close it up.
Clearly, we can restrict the embedding to the manifold with the cylinder excised. The obstruction to extending to the surgered surface is if the link given by the two boundary circles is knotted, but this cannot happen because the links are joined by an embedded cylinder. Hence, we can extend the embedding to the $\Sigma_{g}$ obtained by surgery, moreover we can assume it contained in the interior of the embedded $\Sigma_{g+1}$ holed surface.
Now by induction, this embedding bounds a solid g-holed torus. Now the interior of the sphere formed by gluing the excised cylinder and the attached disks bounds a 3-disk, so we can identify the interior of the embedding of the $\Sigma_{g+1}$ with a solid g-holed torus with a handle attached, i.e. it is diffeomorphic to a solid (g+1)-holed torus.
To convert my comments into an answer:
Assume that $g=1$. Then the following holds:
Lemma. For a (smooth/tame/locally flat) torus $S$ embedded in $S^3$ let $S_\pm$ denote the closures of the two components of $S^3 \setminus S$. These are compact manifolds with boundary equal to $S$. Then either $S_+$ or $S_-$ is homeomorphic to the solid torus.
Proof. If both inclusion maps $S\to S_\pm$ were to induce injections on the fundamental groups, then by Seifert-Van-Kampen's theorem, $\pi_1(S_+\cup S_-)=\pi_1(S^3)$ would be nontrivial, which is, of course, nonsense. Therefore, WLOG, the map $\pi_1(S)\to \pi_1(S_+)$ has nontrivial kernel. Hence, by the Loop Theorem, there is a simple loop $\alpha\subset S$ which bounds a disk $D$ properly embedded in $S_+$. By doing the surgery, as suggested by Connor's answer, along $\alpha$, we attach to $S_-$ a 2-handle $H$ in $S_+$ which is a "thickening" of $D$. The result is a manifold $S_-\cup H$ whose boundary is a 2-sphere $F$ tamely embedded in $S^3$. Thus, by the Jordan-Scheoflies theorem in $S^3$, the sphere $F$ bounds closed 3-balls $B_\pm$ on both sides. From this, one concludes that $S_+$ is a solid torus. (It is $B_+$ with a 1-handle attached.) qed
Now, your actual question was about surfaces in ${\mathbb R}^3$, not $S^3$. If you have a smooth/tame/locally flat torus in ${\mathbb R}^3$, you also obtain one in $S^3$ by taking the 1-point compactification $S^3= {\mathbb R}^3\cup \{\infty\}$. The point $\infty$ could be either in the interior of $S_+$ or of $S_-$ defined above. If $\infty\in int(S_-)$ then we see that $S$ bounds the solid torus in ${\mathbb R}^3$, as you requested. However, it can as well be in the interior of $S_+$, in which case $S$, obviously does not bound a solid torus in ${\mathbb R}^3$ unless $S_-$ is also a solid torus. To find such examples, take a knotted solid torus $S_+$ in $S^3$ and choose a point $p$ in the interior of $S_+$. Then $S^3\setminus \{p\}$ is homeomorphic to ${\mathbb R}^3$ and, thus you obtain an example of a torus in ${\mathbb R}^3$ which does not bound a solid torus in ${\mathbb R}^3$.
One can ask what happens in the case of higher genus surfaces $S$ in $S^3$. One can find examples where neither side of $S$ is a handlebody, so the situation is even worse than in the genus 1 case. The idea is to take a knotted solid torus $M\subset S^3$ and "drill out" a knotted arc from $M$, producing a genus 2 surface so that both sides are not handlebodies. I will leave it to you to work out the details.