Suppose that $A,B$ are subsets of $\Bbb{R}^2$. If $A$ and $B$ are homeomorphic and $A$ and $B$ are compact and connected, are their complements homeomorphic?
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4$\begingroup$ The complement is a disjoint union of open sets where all but the unbounded one are simply connected, i.e., homeomorphic to $\mathbb R^2$, and the unbounded one is homeomorphic to $\mathbb R^2\setminus \{O\}$. Hence any differences can only come from different number of components. $\endgroup$– Hagen von EitzenJul 8, 2015 at 17:18
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4$\begingroup$ It is interesting that this clearly fails in $\Bbb R^3$. Take a circle and a non-trivial knot. $\endgroup$– SpenserJul 8, 2015 at 17:18
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1$\begingroup$ @HagenvonEitzen And it's obvious that the number of components must be the same. Where as always "obvious" means I have no idea how to prove it... ? $\endgroup$– David C. UllrichJul 8, 2015 at 17:40
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1$\begingroup$ We can prove the very special case when $A$ and $B$ are topological manifolds with boundaries. A homeomorphism $\varphi:A\to B$ always restricts to a homeomorphism $\varphi:\partial A\to\partial B$ so they have the same number of components and @HagenvonEitzen arguments applies. $\endgroup$– SpenserJul 8, 2015 at 17:46
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2$\begingroup$ @HagenvonEitzen: Alexander daulity finishes your argument. I don't know how to write the upside-down-hat so I'm just going to write $\hat H$. $\hat H^0(X) = \prod_{qC(X)} \mathbb Z$, where the product is taken over the number of quasicomponents of $X$. Quasicomponents of a manifold are the same as components. $\prod_{A} \mathbb Z \cong \prod_{B} \mathbb Z$ if and only if $A$ is in bijection with $B$. We're done. $\endgroup$– user98602Jul 8, 2015 at 17:59
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1 Answer
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Yes. Let's work in the sphere $S^2$ for convenience. Let $A$ be a connected compact proper subset of $S^2$. Its complement is an open (noncompact without boundary) surface $M'$; its complement in $\mathbb R^2$ is $M' - \{\infty\}$. This is a disjoint union of connected noncompact surfaces without boundary; each of these must have free fundamental group.
For manifolds $M$, the Cech cohomology $\check H^*(M)$ is isomorphic to the singular cohomology $H^*(M)$. By Alexander duality, and the fact that $\check H_0(A) = 0$ (everything is happening in the reduced world!), we see that $H^1(M') = 0$. (The $0$th Cech homology more precisely counts quasicomponents, but for compact Hausdorff spaces, quasicomponents are the same as components.) Because for a disjoint union $H^1(X \sqcup Y) \cong H^1(X) \times H^1(Y)$, we see that every component of $M'$ is simply connected. By the uniformization theorem, each simply connected component is homeomorphic to $\mathbb R^2$. Now delete the point at infinity; this makes one of your copies of $\mathbb R^2$ into an annulus $S^1 \times (-1,1)$.
So your $M$ is homeomorphic to $S^1 \times (-1,1) \sqcup \kappa \mathbb R^2$, where by $\kappa \mathbb R^2$ I mean the disjoint union of $\kappa$-many copies of $\mathbb R^2$. What's left is to determine $\kappa$. But again we just need to use Alexander duality, because $H^0(M') \cong \prod_{\kappa+1} \mathbb Z$, and this is isomorphic to $\prod_{\kappa'+1} \mathbb Z$ iff there's a bijection $\kappa \to \kappa'$.
So $M$ is determined by something even coarser than the homeomorphism type of $A$: all one needs is the 0th and 1st Cech homologies $\check H_*(A)$.
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1$\begingroup$ you work hard and proof it for me, i have to say tnx mr Miller. $\endgroup$– JalilJul 12, 2015 at 18:42