Is there a way to classify incompressible surfaces in $\Sigma \times [0,1]$ where $\Sigma$ is any closed surface? I know of the [Hatcher-Thurston](http://www.math.cornell.edu/~hatcher/Papers/2-bridgeKnots.pdf) classification of incompressible surfaces in 2-bridge knot exteriors,
 but I wonder if a classification of incompressible surfaces can be carried out in this other simple setting.



Is there a way to classify incompressible surfaces in $\Sigma \times [0,1]$ ?
An incompressible and boundary incompressible connected surface is isotopic to either (a) a vertical annulus or (b) a horizontal surface. A vertical annulus is of the form $\alpha \times I$ where $\alpha$ is an essential simple closed curve. A horizontal surface is of the form $\Sigma \times \{t\}$. 


Here is a sketch of the proof. Let $F$ be the given incompressible, boundary incompressible, connected surface. Suppose that $\beta \subset \Sigma$ is essential simple closed curve. Let $B = \beta \times I$ be the corresponding vertical annulus. An innermost disk/outermost bigon argument simplifies the intersection between $F$ and $B$ until it is a disjoint union of either vertical arcs or horizontal curves (ie, copies of $\beta \times \{t\}$). 


Now cut $\Sigma \times I$ along $B$ to get a handlebody with a product structure. Repeat the above argument, replacing the vertical annulus with a sequence of vertical rectangles. 


I believe that you can find all of the tools you need for this kind of thing in Gordon's lecture notes on normal surfaces. <http://homepages.warwick.ac.uk/~masgar/Articles/Gordon/normal.pdf>


There is also a proof of a similar fact by Scharlemann and Thompson in their paper "Heegaard splittings of (surface)×I are standard." 


Allowing boundary compressible surfaces makes the classification more annoying. I haven't thought that through. 


