Let $M^3$ be a compact $3$manifold with $\partial M=N$ a connected surface. Suppose one has a smooth embedding of $N$ into the interior of $M$ and $N$ bounds a domain $D$ in $M$. Can we show that $D$ is homeomorphic to $M$?

4$\begingroup$ This is certainly not true. One class of examples is given by the complements of satellite knots: en.wikipedia.org/wiki/Satellite_knot. These contain tori that are not parallel to the boundary, and indeed these tori bound a "deeper" knot complement. $\endgroup$– HJRWJun 10, 2021 at 10:23

2$\begingroup$ If you drop the dimensions down to $2$, it is also not true. There is however a way to phrase it for curves. $\endgroup$– Ryan BudneyJun 10, 2021 at 15:23

$\begingroup$ This is true only if M is the 3ball. $\endgroup$– Bruno MartelliJun 11, 2021 at 15:21

$\begingroup$ @BrunoMartelli Why not for other handlebodies? $\endgroup$– Will SawinJun 20, 2021 at 23:57

3$\begingroup$ You can embed a surface of genus >= 1 inside a ball in a strange way, so that it does not bound a handlebody in it. $\endgroup$– Bruno MartelliJun 21, 2021 at 14:02
1 Answer
No, this need not hold.
Here is a concrete example. Suppose that $K \subset S^3$ is the trefoil knot in the threesphere. It is an old result (certainly known to Alexander) that $K$ is not isotopic to the unknot. Let $N(K)$ be a small tubular neighbourhood of $K$, taken in $S^3$. So $N(K)$ is a solid torus. Let $n(K)$ be the interior of $N(K)$. We define $M = S^3  n(K)$. This is a (compact) knot exterior. By the "old result" $M$ is not homeomorphic to a solid torus. Note that $T = \partial M = \partial N(K)$ is a twotorus.
Now choose an embedded loop $\alpha$ in $M$. Then $N(\alpha)$, a small tubular neighbourhood of $\alpha$, is another solid torus. Again $T' = \partial N(\alpha)$ is a twotorus. However, $N(\alpha)$ is not homeomorphic to $M$.
The comments above give many other examples. In general, the ways that a fixed surface embeds into a fixed threemanifold is an interesting question.