# Condition for a curve to bound disks on the surface of a handlebody

Let $\Sigma$ be a closed orientable surface of genus $g$ and let $\alpha = \{\alpha_1,...,\alpha_g\}$ be a set of pairwise nonintertersecting nonseparating simple closed curves on $\Sigma$. Then $\alpha$ determines a way of identifying the boundary of a genus $g$ handlebody $H$ with $\Sigma$. If $\gamma$ is a simple closed curve in $\Sigma$ that does not intersect any of the $\alpha_i$, does it follow that $\gamma$ will bound a properly embedded disk when we identify $\Sigma$ with $\partial H$?

• Not sure if I understand. The curves $\alpha_j$ can be contractible? Further, if you have $\alpha_1$, a nontrivial circle on a torus, and $\gamma$ be just close to $\alpha_1$, going "paralelly", then $\gamma$ doesn't bound any disc... – Peter Franek Apr 24 '17 at 20:12
• Whoops - forgot to say nonseparating. Parallel is fine - those will bound disks. – user101010 Apr 25 '17 at 1:17

It is probably assumed that the curves are also pairwise non-isotopic (since otherwise (1) the curves do not determine the handlebody attachment uniquely and (2) the answer would be "no" since you could take $$g$$ parallel copies of the same nonseparating curve in a genus-$$g$$ surface, and for $$g>1$$, given any handlebody attachment there are plenty of disjoint curves that don't bound disks).
Identify $$\Sigma$$ with $$\partial H$$, and let $$D_1,\dots,D_g\subset H$$ be a system of disjoint disks such that $$\partial D_i=\alpha_i$$ for each $$i$$. If $$\gamma$$ does not intersect any of the $$\alpha_i$$, then $$\gamma$$ is an embedded loop in the boundary of $$B=H-\bigcup_i \nu(D_i)$$. Since $$B$$ is a ball, $$\gamma$$ bounds a disk $$D\subset B$$, which is also a disk in $$H$$.