Let $M$ be a connected closed surface. Suppose $N$ is a connected closed surface embedded in the interior of $M \times [0,1]$ such that $N$ separates $M \times \{0\}$ and $M \times \{1\}$. Can we prove the region bounded by $M \times \{0\}$ and $N$ is homeomorphic to $M \times [0,1]$?
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This is not true in general.
One way to build examples is via "stabilisation". Take $N_0 = M \times \{1/2\}$. Suppose that $\alpha$ is an arc in $M \times [0,1]$ with the following properties.
- $\alpha$ is simple (does not self-intersect).
- $\alpha \cap N_0 = \partial \alpha$.
- $\alpha$ is isotopic, relative to its boundary, to a simple arc embedded in $N_0$.
We take a regular neighbourhood of $N_0 \cup \alpha$. Let $N$ be the boundary component of the neighbourhood that is on the same side of $N_0$ as $\alpha$ is. The genus of $N$ is one higher than that of $M$. Also, $N$ separates $M \times \{0\}$ from $M \times \{1\}$.
Of course the arc $\alpha$ could also be "knotted". Also, we could stabilise more than once. Finally, we could combine these operations in various ways. It is a pretty theorem, appearing early in the theory of three-manifolds that fibre over the circle, that the above operations are all that can happen.
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$\begingroup$ Is the original conclusion true if we assume $N$ is homeomorphic to $M$? $\endgroup$– AdterramCommented Jun 10, 2021 at 8:36
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$\begingroup$ Yes. If $N$ separates the boundary components, and is homeomorphic to $M$, then there is an ambient isotopy taking $N$ to $M$. $\endgroup$– Sam NeadCommented Jun 10, 2021 at 19:49
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$\begingroup$ Could you give a reference for this fact? $\endgroup$– AdterramCommented Jun 11, 2021 at 1:14
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$\begingroup$ I poked around, and could not find an obvious place to point to on-line. I'll guess that this appears in either Jaco's book or Hempel's - unfortunately I don't have them to hand. The proof is not so difficult, however. Suppose that $M$ has genus $g$. Fix a cell structure on $M$ with one vertex, $2g$ edges, and one $4g$-gon (as the two-cell). Cross this with the closed interval to get a cell structure on $M \times [0, 1]$. Isotope $N$ to be in general position with respect to the one- and two-cells of the cell structure on $M \times [0, 1]$. Since $N$ separates the boundary components, $\endgroup$– Sam NeadCommented Jun 11, 2021 at 2:26
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$\begingroup$ it meets the internal one-cell an odd number of times. We now perform isotopies of $N$ to ensure that all intersections of $N$ with the "vertical" rectangles (two-cells in the interior of $M \times [0, 1]$) are horizontal arcs. Also, we can arrange that intersections of $N$ with the three-cell are disks. We deduce that $N$ meets the internal one-cell exactly once. A final isotopy gives the result. Here is a more algebraic proof: compress $N$ to be incompressible, appeal to Dehn's lemma to deduce that $N$ is essential, and then use the classification of subgroups of surface groups. $\endgroup$– Sam NeadCommented Jun 11, 2021 at 2:32