geometric-topologylow-dimensional-topology
Suppose we are given a 2 diagrams for an handle decomposition of two closed, orientable 3-manifold M and $\tilde{M}$. Each of these diagrams consists in an handlebody of genus $g$ (respectively $\tilde{g})$, and closed curves $\{\gamma_1,\dots, \gamma_g\}$ (respectively $\{\tilde{\gamma}_1,\dots, \tilde{\gamma}_\tilde{g}\})$ drawn on them representing the attacching spheres of the 2-handles.
How can we construct a diagram of the connected sum $M\#\tilde{M}$ from the given diagrams?
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 $\Sigma$:
Best Answer
Thanks to Arnaud Mortier's comment. As it can be read in paragraph 3.2 Reducible splittings page 9 of https://arxiv.org/pdf/math/0007144.pdf Martin Scharlemann's survey "Heegaard Splittings of compact 3-manifolds", the diagram for the connected sum is obtained just connecting the two handlebodies with a single 1-handle. Thus the diagram for $M\#\tilde{M}$ consists in an handlebody of genus $g+\tilde{g}$ with the curves $\{\gamma_i\}$ drawn on the first $g$ 1-handles of the handlebody and the curves $\{\tilde{\gamma}_i\}$ drawn on the second set of $\tilde{g}$ 1-handles.
We obtain this by removing cleverly a 3-disk from $M$ that intersects the first handlebody's surface in a 2-disk (thus half of the 3-disk lies in the 3-handle of $M$ and half of it lies in the 1-handlebody induced by the diagram). We remove similarly a 3-disk from $\tilde{M}$, now we have to glue the new $\mathbb{S}^2$ boundary components, to do so we use a 1-handle to glue two hemispheres and a single last 3-handle to cap the resulting $\mathbb{S}^2$ boundary.
We can also decompose this procedure as follows, first we add an isolated 2-handle to the diagram of $M$ and the diagram of $\tilde{M}$ (this accounts to removing the balls). Then we connect the diagrams joining with a 1-handle the disks bounded by these new 2-handles. We end up with 3 boundary components isomorphic to $\mathbb{S}^2$s. We can cap them with 3-handles. Reflecting on this construction one sees that the two 2-handles and two 3-handles cancels out thus we have just added a 1-handle and capped all with a single 3-handle.