Is it possible to have a properly embedded annulus $A$ in a genus two handlebody $V$ such that the two boundary curves of $A$ in $\partial V$ represent different isotopy classes of curves on $\partial V$? (I should also mention that in the case I care about, the curves bounding $A$ cannot be meridians, i.e. cannot bound embedded disks in $V$). My intuition says that this is not possible, but maybe I'm just not visualizing things correctly. And I haven't yet thought up a good way to prove this either way.


If the answer is yes, such an annulus exists, can we say anything about whether the boundary curves of $A$ in $\partial V$ are separating/non-separating? For instance, could you have both boundaries be non-separating (as is possible in genus 3)? Or could you have one boundary separating and one non-separating?


I should mention that I mean the following by "properly embedded": $X$ is properly embedded in $Y$ via $f:X \to Y$ if $f(\partial X) = f(X) \cap \partial Y$, and $f(X)$ is transverse to $\partial Y$ at any point of $f(\partial X)$.



Properly embedded annuli in genus two handlebody?
Choose a non-trivial loop $a\subset \partial V$ and a disjoint meridian $b\subset \partial V$. Let $D\subset V$ be a disk with $\partial V = b$ and let $A'$ be an annulus in $V$ with $\partial A'$ equal to two parallel copies of $a$. Define $A$ to be the result of taking a boundary connected sum of $A'$ and $D$ parallel to an embedded arc $p\subset \partial V$ joining $a$ to $b$. Then one boundary component of $A$ is homologous (in $\partial V$) to $a$ and the other boundary component is homologous to $a \sqcup b$.


If $a$ is separating and $b$ is non-separating then one boundary component of $A$ is separating and the other is not.


