Surface with boundary cross an interval is a handlebody I was reading an argument and it seemed to suggest that given an orientable surface $\Sigma$ with nonempty boundary, $\Sigma \times I$ is a handlebody (i.e. a bunch of closed balls glued together along disks embedded the balls boundaries). Where are the disks in $\Sigma \times I$ that I can cut along in order to get a bunch of closed balls?

Let $\alpha$ be any embedded essential arc intersecting the boundary at two points (essential here means non-trivial in $H_1(\Sigma,\partial \Sigma)$). Consider the curve given by $\alpha$ on both $\Sigma \times 0$ and $\Sigma \times 1$, and the obvious vertical arcs between $\alpha(i) \times 0$ and $\alpha(i) \times 1$ (where $i= 0$ and $1$). This curve bounds an obvious disk. To get a complete system of disks take a basis for $H_1(\Sigma,\partial \Sigma)$ so that the corresponding arcs do not intersect (it should be easy to come up with at least one such basis).  
