If I have a connected manifold and I poke a hole in the interior of the manifold, it seems obvious that the manifold is still connected. But how would you prove this?

More precisely, let $M$ be a connected $n$-manifold with $n>1$ and let $B$ be a subset of $\operatorname{Int} M$ homeomorphic to the unit closed ball in $\mathbb{R}^n$. Is $M \setminus B$ connected? Can $B$ be any simply connected closed set?


I think the result is true if $B$ is any connected $n$-dimensional submanifold whose boundary is also connected. That includes a closed $n$-ball. I'm assuming the manifolds and embedding are piecewise linear (PL), and that we're talking about path-connectedness rather than general connectedness.

Here's a proof. Consider a path $\gamma$ with endpoints in $M - B$. The intersection of $\gamma$ with $B$ is a disjoint union of paths $\sigma$, each with its endpoints in the boundary Bd $B$. Since Bd $B$ is connected, we can replace each $\sigma$ with a $\sigma'$ with the same endpoints which lies entirely in Bd $B$.

Now we want to push $\sigma'$ slightly away from $B$ so that it doesn't meet $B$ at all. Choose a small regular neighborhood $N$ of Bd $B$. $N$ is an $I$-bundle over Bd $B$, in which Bd $B$ embeds as a section and each fiber has one end inside $B$ and one end outside $B$. So the bundle is trivial, and we have an embedding $f: N \times [0, 1] \rightarrow M$ with $f(N \times \{0\})$ in the interior of $B$, $f(N \times \{1/2\}) =$ Bd $B$, and $f(N \times \{1\})$ outside $B$. In particular the obvious projection $f(N \times [1/2, 1]) \rightarrow f(N \times \{1\})$ is a retraction which pushes $\sigma'$ out of $B$.

Doing this for each $\sigma$, we get a path $\gamma'$ which doesn't meet $B$. QED.

It's not actually necessary for $B$ to be connected, as long as each of its connected components has connected boundary. So you can poke multiple holes in the original manifold and it will still be connected. Nor is it necessary for $B$ to be simply connected: it can be a solid torus, for example.

However, we do need that connected-boundary condition on $B$. For example, $\mathbb{R} \times [0, 1]$ as a subset of $M = \mathbb{R}^2$ is closed and simply connected, and it lies in the interior of $M$ because $M$ has no boundary, but removing that subspace divides $\mathbb{R}^2$ into two connected components.

I don't know about closed subsets in general, but I would bet that the result holds for any compact PL subset $B$.

  • $\begingroup$ Nice proof! If we're considering the general case, things get a little tricky if $B$ is something like a line segment in $\mathbb R^2$. $\endgroup$ – Rahul Nov 22 '12 at 17:37
  • $\begingroup$ Can you be a bit more explicit with how you modify the path so it doesn't enter $B$? $\endgroup$ – wj32 Nov 22 '12 at 21:12
  • $\begingroup$ Good point. I added a bit about how to use that regular neighborhood of Bd $B$ -- see if you believe it. $\endgroup$ – Hew Wolff Nov 23 '12 at 19:40
  • $\begingroup$ @HewWolff: I don't know anything about fiber bundles yet. Give me a few months to think about this... $\endgroup$ – wj32 Nov 24 '12 at 1:36
  • $\begingroup$ OK. Intuitively, I just want to argue that there's a well-defined direction for us to push $\gamma'$, because every point in Bd $B$ has an "inward" and an "outward" direction. There may be an easier way to do that. $\endgroup$ – Hew Wolff Nov 24 '12 at 19:16

Take local charts around each point in the boundary $\partial B$. By compactness of $\partial B$, finitely many local charts suffice. Then, if $\gamma$ was a path that ran through $B$ and connected two points not in $B$, we should be able to modify it to $\gamma'$ by pushing it around the boundary of $B$ by working in these finitely many local charts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.