It is obvious that the product of a 2-disk with $n$ holes with the unit interval is a handlebody of genus $n$. I have read that this generalizes to the product of an (orientable) surface of genus $g$ with $n$ holes is a handlebody of genus $2g + n - 1$. I'm trying to visualize this but so far I'm failing. Can you please help me?
PS. I know there is a question here that is pretty much exactly like what I'm asking but I can't quite understand the answer there. Also I'm interested in more of a geometric understanding of the fact.
EDIT. Can someone please verify if my thinking is correct? Let $S_g^n$ be an orientable surface of genus $g$ and with $n$ boundary components, i.e. $S_g^n$ is a sphere with $g$ handles and $n$ holes. Then the $3$-manifold $M=S_g^n \times [0,1]$ has boundary that is two disjoint copies of $S_g^n$ with $n$ hollow tubes connecting the boundary components of the copies of $S_g^n$. By thinking of one of the tubes as being the the tube that connects the two $S_g^n$ into one connected sum, we lose one tube and what we have is a sphere with $2g+n-1$ handles. But that is the boundary of $M$ and therefore $M$ is a handlebody of said genus.
Your solution is okay as far as identifying the surface that bounds $M$, but it does NOT follow that $M$ is a handlebody of the same genus. A fixed compact surface can be homeomorphic to the boundary of many different, non-homeomorphic 3-manifolds.
Let me explain a method which works in general by first specializing to the surface $S^1_g$, the one-holed torus. Define an essential arc to be an embedded arc which intersects the boundary solely in its endpoints, and which is not homotopic rel endpoints into the boundary. Two essential arcs are equivalent if they are homotopic by a homotopy which keeps the endpoints in the boundary.
You can choose two inequivalent essential arcs $\alpha_1,\alpha_2 \subset S^1_g$ such that when you cut along $\alpha_1,\alpha_2$ the result is a disc. To see this, start with the standard square gluing diagram for a torus, and then remove a small quarter-circle neighborhood of each vertex of the square. Now consider $S^1_g \times [0,1]$. Each of $\alpha_i \times [0,1]$ is a properly embedded disc. When you cut open along these 2 discs, what's left is a single 3-ball. Therefore, $S^1_g \times [0,1]$ is handlebody of genus $2$.
In general, using that $\chi(S^n_g) = 2-2g-n$, you can cut $S^n_g$ along a non-separating set of $2g+n-1$ pairwise inequivalent essential arcs $\alpha_1,\ldots,\alpha_{2g+n-1}$ so that what's left is simply connected. In $S^n_g \times [0,1]$, each of $\alpha_i \times [0,1]$ is a properly embedded disc. When you cut $S^n_g \times [0,1]$ along these $2g+n-1$ discs, what's left is a single 3-ball. Therefore $S^n_g \times [0,1]$ is a handlebody of genus $2g+n-1$.