Can every manifold with torus boundary be cut?

Question

Let $\mathcal{M}$ be a compact, oriented and connected $3$-manifolds, whose boundary satisfies $\partial\mathcal{M}\cong T^{2}$, where $T^{2}:=S^{1}\times S^{1}$ denotes the $2$-torus. If I "cut" through the manifold $\mathcal{M}$, do I always end up with a well-defined manifold $\mathcal{M}^{\prime}$ whose boundary is a $2$-sphere? For example, if $\mathcal{M}$ is the solid torus (the genus 1 handlebody), then we can just cut along an embeded disk, whose boundary circle lies purely in $\partial\mathcal{M}$. If $\mathcal{M}$ is a manifold obtained by performing the connected sum of the solid torus with some closed $3$-manifold, we can apply the same logic. But it is also true more generally? In general, manifolds with torus boundary can have a highly non-trivial bulk topology, for example manifold with incompressible boundary, etc.

Answer 1

I'm not sure quite what you mean by "cut". If your question is whether there is always a properly embedded surface $S$ in $M$ such that $M\backslash \backslash S$ has boundary consisting of a single 2-sphere, this is not the case: if $\delta S$ is contractible in $\delta M$, then it cuts $\delta M$ into a once-punctured torus and a disc, so the boundary of $M\backslash \backslash S$ has genus. Otherwise, $\delta S$ cuts $\delta M$ into a single annulus, so $S$ would need to be a disc to cap this off to form a 2-sphere. But, as you note, many 3-manifolds have incompressible boundary, which precisely implies that no non-trivial curve in the boundary bounds a disc in the manifold.