# Connected sums where the separating "2"-sphere bounds a punctured ball

I have a question about connected sums in 3-dimensional spaces in relation to normal surfaces. I was reading the paper https://arxiv.org/abs/math/9712269 by Joel Hass, which at one point outlines a proof of "Knesers Theorem", which states that a triangulated $$3$$-dimensional manifold (without boundary) $$M$$ can be decomposed non-trivially along $$2$$-spheres (in terms of the connected sum) up to a number of times bounded by a constant $$k(M)$$ dependent on $$M$$ and its triangulation.

For the proof, the notion of a normal surface was introduced, which is a surface embedded in $$M$$ which is a disjoint union of normal triangles or normal quadrilaterals, where the latter are intersecting a tetrahedron of $$M$$ in 3 or 4 edges and faces.

At Page 7 in the proof, the notion of a "punctured ball" is introduced, which is just defined as "a ball with some open balls removed". At Page 8 in the proof it is mentioned, that if we use a separating (normal) $$2$$-sphere $$S$$ to decompose $$M$$ into a connected sum and $$S$$ bounds a punctured ball, then one of the summands in the connected sum is trivial, meaning its homeomorphic to $$S^3$$. I can't get my head around the following things:

1. Do we have specific extra conditions regarding the definition of a punctured ball? For example that we only allow for finitely many open balls cut out or that they have to be completely enclosed in the main ball?
2. I can't really see why a $$2$$-sphere bounding a punctured ball leads to a $$S^3$$-summand. Is it a simple trick I don't see (maybe having to do with normal $$2$$-sphere) or is this maybe done by calculation of homology groups? I think I am missing something important here.

I appreciate any kind of help and ideas, thanks in advance for your time!