If $K, K' \subset \mathbb{S}^2$ are compact and $K \simeq K'$, is $\mathbb{S}^2 \setminus K \simeq \mathbb{S}^2 \setminus K'$?

Question

One of the most immediate consequences of the Alexander-Pontryagin Duality Theorem is that if $X$ is compact and $f, g: X \rightarrow \mathbb{R}^n$ are two embeddings, then $\mathbb{R}^n \setminus f(X)$ and $\mathbb{R}^n \setminus g(X)$ have the same number of components. The same applies for $\mathbb{S}^2$.

But in the plane, there are no 'wild sets'. So is it in fact the case that $\mathbb{R}^2 \setminus f(X) \simeq \mathbb{R}^2 \setminus g(X)$ when $X$ is connected? A union of two disjoint circles are a counterexample if we don't assume $X$ is connected, depending on whether one is in the interior region of the other.

I thought I had seen some counterexample to this in the plane, some sort of graph, but now I can't think of it and it wasn't in the book I thought I saw it in so now I'm feeling uncertain. Note that $f$ and $g$ would have to be inequivalent embeddings, because every isotopy of compact subsets in the plane extends to an ambient isotopy.

EDIT: As pointed out in a comment below, it's true. The proof outlined here uses shape theory: Homotopy type of the complement of a subspace

So now my question is, is there a more elementary proof? It doesn't have to be for the full-strength "only homotopy type matters" of the above theorem, just when $K, K'$ are homeomorphic would be satisfactory to me.

Answer 1

Here is a proof via the Alexander duality (using the form of duality with the Chech cohomology of $K$):

1. If $K\subset S^2$ is compact and connected, then for $\Omega= S^2\setminus K$ we have $H_1(\Omega)=0$.

2. Thus, each component $\Omega_i$ of $\Omega$ is an oriented surface with $H_1(\Omega_i)=0$. One then proves that such $\Omega_i$ is simply-connected. One proof is that for each noncompact connected surface $S$, $\pi_1(S)$ is free. Then apply Hurewicz theorem in conjunction with the fact that $H_1(\Omega_i)=0$.

3. Thus, assuming that $K$ is nonempty, Hurewicz theorem applied one more time implies that each $\Omega_i$ is contractible.

4. Applying the Alexander duality again, we see that ${H}_0(\Omega)\cong \check{H}^1(K)$. In other words, the homotopy type of $K$ also determines the number of connected components of $\Omega$. Hence, the homotopy type of $K$ determines the homotopy type of $\Omega=S^2\setminus K$.

One can do better, of course, and note that each connected noncompact simply-connected surface $S$ is homeomorphic to the plane. Proving this is not hard once you know the classification of surfaces; in fact, it is enough to know that $S$ is orientable and has $b_1(S)=0$ (thus, simplifying the proof above and avoiding most of the Algebraic Topology arguments):

Exhaust $S$ by compact subsurfaces with (smooth) boundary $S_k$ (find a smooth proper function on $S$ and consider its regular sublevel sets). Use the assumption that $b_1(S)=0$ to conclude that each boundary component of $S_k$ bounds a compact subsurface in $S$. Using this, enlarge each subsurface $S_k$ to a compact subsurface $S'_k$ which has exactly one boundary component and no handles (since $b_1=0$). WLOG, the surfaces $S'_k$ still form an exhaustion: $$ S'_k\subset int(S'_{k+1}). $$ Now, using the classification of surfaces conclude that $S'_k$ is homeomorphic to the disk and each $S'_{k+1} \setminus int(S'_k)$ is homeomorphic to the annulus. Using this information, construct a homeomorphism $S\to R^2$.

See also my answer here.

With this exhaustion argument in mind, you only need algebraic topology to relate the number of connected components of $R^2\setminus K$ to the topology of $K$. One can do this without algebraic topology but it will be long and complicated (since, among other things, you will be reproving Jordan curve theorem) and, as far as I am concerned, pointless.

To conclude: The homotopy type of a connected and compact subset $K\subset S^2$ determines the topology of $S^2 \setminus K$.