I've read in a couple of different places (a paper and a blog) the following fact:
if $F$ is a surface, properly embedded in a three-dimensional handlebody of genus at least two, then $F$ is either compressible or boundary compressible.


Neither of these sources included a reference or a proof and I am a bit of a three-manifold newbie. Could somebody please direct me to reference for this fact? Or provide a proof?



Properly embedded surfaces in handlebodies are compressible or boundary compressible?
Suppose that the surface $F$ is properly embedded in the handlebody $V$. We induct on the genus of $V$.


In the base case $V$ has genus zero and so is a three-ball. In this case the desired result follows from a theorem of Alexander. (Briefly, suppose that $F$ is not a sphere or a disk. Sweepout the ball $V$ by planes $(P\_t)$ and track the pattern of intersections with the surface $F$. Then there is at least one time $t'$ where $F \cap P\_{t'}$ contains a compressing curve or a boundary compressing arc.)


Suppose that $V$ has genus greater than zero. Suppose that $(D, \partial D) \subset (V, \partial V)$ is a non-separating compressing disk for $V$. Properly isotope $F$ to minimise the intersection between $F$ and $D$. If $F$ is disjoint from $D$ we are done by induction. Suppose instead that $F \cap D$ contains a component $\alpha$ which is either an innermost curve or an outermost arc. If $\alpha$ is not a (boundary-)compression for $F$ in $V$ then there is an isotopy that further reduces $|F \cap D|$, a contradiction.


