# Can every manifold with torus boundary be cut?

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.

• The question is unanswerable until you define what you mean by "cut." An example is not a substitute for a definition. Oct 26, 2022 at 10:45

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.