geometric-topologymanifoldssurfaces
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 $H$?
The formula only holds when there are boundary components, i.e., $b \geq 1$. It doesn't make sense for the sphere, as you remark, but it also isn't true for the torus ($g=1$, $b=0$, but the first homology group has rank $2$). See First homology of a compact connected surface with boundary for more details.
Regarding the disk: indeed, it has no nontrivial cycles. If you take any embedded circle in the disk, you may simply fill it in to get an embedded disk whose boundary is the original circle. Therefore every cycle in the disk is exact.
The first homology group of the torus has rank $2$; a basis consists of two loops going around the torus, one the "short way around" (a meridian) and one the "long way around" (a longitude).
The story for the closed orientable surface of genus $g$ (connected sum of $g$ tori) is similar: the basis consists of $2g$ cycles, namely a meridian and a longitude coming from each torus in the connected sum.
Best Answer
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 $\Sigma$: