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 $M \times [0,1]$?



Surface in a product domain
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.


