Let $M^3$ and $N^3$ be compact orientable manifolds with connected boundary. When we perform the connected sum operation, will the resulting manifold $X = M \#N$ have a connected or disconnected boundary?
For instance, if $M$ and $N$ are two solid tori, their connected sum will be a genus $2$ handlebody, and hence have a connected boundary, or will it be another $3$-manifold whose boundary has $2$ connected components?
Best Answer
The trouble is that for manifolds with connected boundary $M_1, M_2$ there are two notions of connected sum:
The ordinary connected sum, denoted $M=M_1\# M_2$ where you take out open balls $B_1, B_2$ from the interiors of $M_1, M_2$ and glue $M_i-B_i$ via a homeomorphism of the new boundary spheres $\partial B_1, \partial B_2$. Then, clearly, $M$ has two boundary components, homeomorphic to $\partial M_i, i=1, 2$.
The boundary connected sum, denoted $N=M_1\#_{\partial} M_2$, where you pick two closed (tame) disks $D_i\subset \partial M_i$ and glue $M_1, M_2$ via a homeomorphism $D_1\to D_2$. Then $N$ will have connected boundary, homeomorphic to $\partial M_1 \# \partial M_2$.
(In both cases, there are some issues with the choice of gluing homeomorphisms, I will ignore these. Separately, there are issues with the category in which everything is done: TOP, PL or DIFF, in dimension 3 this does not matter.)
Thus, the answer to your question is: It depends on your notion of connected sum.