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 $\partial H$?

  • $\begingroup$ 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... $\endgroup$ – Peter Franek Apr 24 '17 at 20:12
  • $\begingroup$ Whoops - forgot to say nonseparating. Parallel is fine - those will bound disks. $\endgroup$ – 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 $H$.


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