# Homology condition - bounding a disk in a handlebody?

Suppose that $$\gamma_1,...,\gamma_n$$ are a set of disjoint simple closed curves on a closed orientable surface $$\Sigma$$ that all bound disks in some handlebody $$H$$ with $$\partial H = \Sigma$$. Let $$\gamma$$ is a simple closed curve on $$\Sigma$$ such that the homology class $$[\gamma]$$ is in the subgroup of $$H_1(\Sigma; \mathbb{Z})$$ generated by $$[\gamma_1],...,[\gamma_n]$$.

Does $$\gamma$$ also bound a disk in $$H$$?

First, it doesn't change anything to suppose each $$[\gamma_i]$$ is nonzero and $$[\gamma_i]\neq [\gamma_j]$$ when $$i\neq j$$. Second, it doesn't hurt to suppose $$n$$ is the genus of $$\Sigma$$ by adding in the rest of the curves.

With this setup, the kernel of $$H_1(\Sigma)\to H_1(H)$$ is generated by $$[\gamma_1],\dots,[\gamma_n]$$. Consider the universal abelian cover $$H'\to H$$, which is the one corresponding to the homomorphism $$\pi_1(H)\to H_1(H)$$, and it can be visualized in $$\mathbb{R}^{n}$$ with a ball at each lattice point $$\mathbb{Z}^{n}$$ and $$1$$-handles connecting the balls in axial directions. Each $$\gamma_i$$ lifts to $$H'$$ on a handle that goes in the $$i$$ direction, and the disk that $$\gamma_i$$ bounds lifts to a disk in that handle.

If $$\gamma\subset\Sigma$$ is a simple closed curve by your hypotheses, then $$[\gamma]$$ is in the kernel, or equivalently that $$[\gamma]$$ is zero in $$H_1(H)$$, which in terms of covering spaces is that $$\gamma$$ lifts to a closed loop in $$H'$$. Furthermore, $$\gamma$$ bounds a disk in $$H$$ if and only if it does in $$H'$$.

So, if we find a homotopically nontrivial simple closed curve in $$H'$$ that downstairs is still an embedded curve, we have a $$\gamma$$ that does not bound a disk. Here's a candidate for a curve in a $$H$$ a genus-$$2$$ handlebody, as seen from the cover $$H'$$: The image of this curve on $$\Sigma\subset H$$ itself is Notice that the curve is actually zero in $$H_1(\Sigma)$$ since it passes over the top of each handle once in both directions.

I personally appreciate seeing these things on a standard handlebody, so here is a $$\gamma$$ with $$[\gamma]\in\ker(H_1(\Sigma)\to H_1(H))$$ and yet $$\gamma$$ does not bound a disk in $$H$$. I was going to give the next picture before the previous one as a sort of joke, but after deforming the above one into this this form (a thickened punctured torus) it becomes manifestly obvious the curve (1) does not bound a disk and (2) is nullhomologous in $$\Sigma$$: 