When I take some arbitrary compact and connected $d$-manifold $\mathcal{M}$ without boundary and I cut out the interior of some (sufficient nicely) embedded closed $3$-ball $B^{3}$ inside $\mathcal{M}$, does the result depend on the chosen ball? In other words, are the manifolds obtained by cutting two different balls from $\mathcal{M}$ homeomorphic?
Naively I would say they are homeomorphic, however, I am not totally sure. Maybe it is also related to the Annulus Theorem, because if the results would be non-homeomorphic, also the connected sum would be not well-defined, I guess.
Best Answer
If the ball is embedded nicely enough, it will have a neighborhood which is contractible (ambiently contractible, even). Let me pretend each ball is in a single coordinate chart. If the manifold is connected, take a path connecting the balls, which we can take to be embedded. Take a small normal neighborhood.
So we have two balls with a little tube connecting them. By the usual look-at-the-complement-of-a-slightly-smaller-neighborhood, the manifold is this, plus stuff glued to the boundary. But then it's obvious that there is a homeomorphism which swaps the two balls, by going through the tube.