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Problem:

Let $A_1,A_2 \subset X$ and $X$ be a topological space. In general, if $A_1 \cong A_2$ is homotopy equivalent (or homeomorphic), it does not follow that $X \setminus A_1 \cong X \setminus A_2$ is homotopy equivalent (or homeomorphic).

Question:

I therefore wanted to ask whether there are conditions/theorems that guarantee this if $A_1,A_2,X$ fulfill certain conditions. At least I strongly believe that it is enough (for both hom. equiv. and homeomorphic) if $X$ is hausdorff and $A_1$ and $A_2$ are compact. But that's just a guess, as I haven't been able to show it yet.

Noctis
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    Your guess is wrong: take $X=A_1=[0,1]$ and $A_2=[0,{1\over 2}]$, all with the usual topology. Or even better, $X=[0,1]$, $A_1={0}$, and $A_2={{1\over 2}}$. – Noah Schweber Nov 18 '24 at 18:46
  • @NoahSchweber All right, thank you very much. I hadn't thought of that example. Unfortunately, this makes it more likely that there are no very general theorems if it fails despite such a setting. – Noctis Nov 18 '24 at 18:57
  • @Noctis There are no such theorems, as far as I can tell. Noah’s example perfectly demonstrates why. There are some very specific cases in which you can conclude $X \setminus A_1 \simeq X \setminus A_2$, like, say, $X$ is discrete and $A_1$ has strictly smaller cardinality than $X$; or when $X = \mathbb{R}$, $A_1$ is countable dense, and $A_2$ is dense. But those very specific cases are more or less all you can get. – David Gao Nov 18 '24 at 19:07
  • @Noctis Maybe saying there are no general theorems is not exactly right. Maybe this is true if $X$ is a non-compact connected manifold (without boundary) of dimension at least $2$, and $A_1$ and $A_2$ are compact submanifolds (with boundaries) of dimension the same as $X$? It’s just a guess though, I don’t know if this is actually true, but if it is true it does sound somewhat general. – David Gao Nov 18 '24 at 19:17
  • @DavidGao I don't think that's true: take $X=\mathbb{R}^3$, $A_1$ the 1-ball, and $A_2$ the filled-in Alexander horned sphere. – Noah Schweber Nov 18 '24 at 19:22
  • @Noctis It is essential in knot-theory that $X \setminus A_1 \ncong X \setminus A_2$ need not be homeomorphic. If $A_1,A_2 \subset \Bbb R^3$ are tame knots, then they are both homeomorphic to the circle (and hence to each other). However, $\Bbb R^3 \setminus A_1$ and $\Bbb R^3 \setminus A_2$ are homeomorphic if and only if $A_1$ is an equivalent knot to $A_2$ or the mirror image of $A_2$. – Ben Grossmann Nov 18 '24 at 19:33
  • @NoahSchweber Oh yeah, you’re right. I guess I need to assume the manifolds are smooth and the embedding is smooth. Then I believe it’s at least true in dimension $3$? I don’t know about higher dimensions though. (I was thinking about knots when I added the condition that $A_1$ has the same dimension as $X$, but I guess that’s not sufficient.) – David Gao Nov 18 '24 at 19:45
  • Oh wait no, even that’s not true, tubular neighborhoods around knots should provide counterexamples. I guess the correct thing should have been $A_1$ and $A_2$ are smooth $(\dim(X) - 1)$-dimensional spheres and the embeddings are smooth. (Apparently this is still open for $\dim(X) = 4$, but true in other dimensions?) – David Gao Nov 18 '24 at 19:56

2 Answers2

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Turning my comment into an answer:

It's hard to rigorously state, let alone prove, a negative answer to a question like this; but here's an example which strongly indicates that the answer is indeed negative.

Let $X=[0,1]$, let $A_1=\{0\}$, and let $A_2=\{{1\over 2}\}$. Then $X\setminus A_1$ is connected but $X\setminus A_2$ is not. Lots of analogous things can be done; e.g. if $X$ is a "lollipop" (= a line segment attached to a circle) then removing one point can leave it connected but not simply connected while removing another point can make it contractible. Note that all of the spaces involved here are "nice" (e.g. Hausdorff and compact).

In particular, the OP's guess is incorrect.

To try to turn a negative into a positive, we can ask the following:

Let $X$ be a fixed "nice" space, and define an equivalence relation $\sim$ on the (closed, for simplicity) subsets of $X$ as the smallest equivalence relation with the following two properties:

  • If $A,B$ are homeomorphic subspaces of $X$, then $A\sim B$.

  • If $X\setminus A, X\setminus B$ are homeomorphic subspaces of $X$, then $A\sim B$.

The strong failure of a positive answer to your question might(!) be re-interpretable as a frequent coarseness of $\sim$: it may be the case that $\mathcal{P}_{closed}(X)/\sim$ is small in some sense (= "easily classifiable") even when $X$ has lots of closed subsets. You might try to compute $\mathcal{P}_{closed}/\sim$ in a couple simple cases, e.g. $X=[0,1]$ or $X=S^1$ (although even the case of $X=[0,1]^2$ might be too hard to compute by hand, for reasons related to this old answer of mine).

Noah Schweber
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It should be pointed out that a special case of this problem is the basis of the whole subject of knot theory. Here one takes $X=\mathbf{R}^3$ and $A_1=S^1\subset\mathbf{R}^2\subset\mathbf{R}^3$ and $A_2$ some tangled knot. The famous Gordon–Luecke theorem states that the complements of two knots are homeomorphic if and only if the two knots are equivalent to each other. Another famous related counterexample is when $A_1=S^2$ and $A_2$ is the Alexander horned sphere. These and other counterexamples form the basis of a great deal of modern geometric and algebraic topology

Zbigniew Fiedorowicz
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