Is Swiss cheese homeomorphic to a ball? In topology, I understand how most solids are homeomorphic to $n$-holed donuts, but I've never seen anyone mention internal holes (or should I say "bubbles", because holes are something else).
If you take a block of Swiss cheese, how can you get rid of the bubbles inside without cutting or gluing parts that were not close to each other? 
I've never heard anyone talk about it so I assume it's not a problem but I don't understand how you can remove internal holes without breaking the rules of topology.
 A: To give a more explicit view and for the sake of reducing the number of unanswered questions.
Assume that there $n$ "bubbles" i.e. removed $n$ balls from the whole space. Then you can always arrange them in a line without breaking the rule of topology. Note that I'm talking about the solid shape instead of just the surface.
Let $X$ denotes the resulting space (swiss cheese). We can regard this as many $K=D^2\setminus B(x_0,r=1/2)$ joining together, where $D^2$ and $B$ centered at the same point, since the outer boundary of $X$ is homeomorphic with $ \partial D^2$. So, $$X\cong\bigsqcup_{i=1}^n K_i/\sim$$
where $\sim$ is generated by $x_i\sim x_j$ iff $x_i,x_j$ belongs to the outer boundary of $K_i,K_j$ respectively, and $j=i+1$, for each $K$ there is only one point that can be identified.
For each point $x_j\in K_j$, we have a vector between $x$ and its final position (those vectors intersect at the center), so then define a family of maps $f_{i,t}:K_i\to \partial D^2$ by
$$f_{i,t}(x)=t\dfrac{x}{||x||}+(1-t)x$$
(We're sliding each point through a vector from the center and map it to the boundary, and this is well-defined and continuous, you can try to prove it，note that the center can be regarded as the orgin of $K_i$)
and the inverse which is an natural inclusion map $g_i:\partial D^2\to K_i$.
So $X\simeq \bigsqcup S^2_i/\sim$ (i.e. is homotopy equivalent to $n$ spheres joining together) It is not homeomorphic with a Torus because $X$ is simply connected but a torus isn't. And it is not homeomorphic with a ball, because a solid ball doesn't have any bubble in it. Hope this will help.
