If $A$ and $B$ are homeomorphic proper closed subsets of $\mathbb{R}^n$, do their complements have the same homology?

Question

Let $f_1,f_2:[0,1]\to S^3$, $g:M\to S^3$ and $h:\mathbb{R}P^2\to S^3$ be inyective maps, where $M$ is the Möbius strip. Assume that $\mathrm{Im}f_1\cap \mathrm{Im}f_2=\emptyset$.

I want to compute $H_*(S^3- \mathrm{Im}f_1\cup \mathrm{Im}f_2)$, $H_*(S^3-\mathrm{Im}g)$ and $H_*(S^3-\mathrm{Im}h)$.

Sice all maps are injective and continuous between compact Hausdorff spaces, they are homeomorphisms onto their images. My idea is to compute these homologies using excision surrounding each image by an open ball, and then using Mayer-Vietoris.

In order to use Mayer-Vietoris I'd like to be able to say that $H_*(S^3-A)=H_*(S_3-B)$ whenever $A$ and $B$ are homeomorphic proper closed subsets, or at least $H_*(\mathbb{R}^3-A)=H_*(\mathbb{R}^3-B)$ (and therefore I would use the excision part before). That's because I would be able to easily decompose the spaces, for example, I could choose any obvious embedding of the unit interval.

From Madsen's From Calculus To Cohomology (theorem 7.2) I know that $H^p(\mathbb{R}^n-A)=H^p(\mathbb{R}^n-B)$ for the deRham Cohomology. I would like to mimic the proof, but it uses the fact that the the deRham cohomology has coefficients on the field $\mathbb{R}$, namely, he says that

$H^0(\mathbb{R}^n-A)/\mathbb{R}\cong H^0(\mathbb{R}^n-B)/\mathbb{R}$

implies $H^0(\mathbb{R}^n-A)=H^0(\mathbb{R}^n-B)$, which is not always true for $\mathbb{Z}$ coefficients, since one of the groups could have non-trivial torsion subgroup. In addition, it is a statement about cohomology, not about homology, though I think that's not essential in the proof.

So my question is

Is any of the isomorphisms $H_*(S^3-A)\cong H_*(S^3-B)$ or $H_*(\mathbb{R}^3-A)\cong H_*(\mathbb{R}^3-B)$ true? If not, how can I approach this problem?

Answer 1

If both $A$ and $B$ are compact and locally contractible, then Alexander duality tells you that: $$\tilde{H}_i(S^3 \setminus A) \cong \tilde{H}^{3-i-1}(A) \cong \tilde{H}^{3-i-1}(B) \cong \tilde{H}_i(S^3 \setminus B).$$ and the result you seek is true. This is the case in your example, since your $A$ and $B$ are the images of compact spaces. (For a reference, this is e.g. Theorem 3.44 in Hatcher's Algebraic Topology.)

Answer 2

This is only an extended comment to Najib Idrissi's answer.

The Alexander duality theorem in its most genearal form says that if $A \subset S^n$ is compact, then $\tilde{H}_i(S^n \setminus A) \approx \check{H}^{n-i-1}(A)$. Here $\tilde{H}_*$ denotes reduced singular homology and $\check{H}^*$ denotes Cech cohomology. Hence $$\tilde{H}_i(S^n \setminus A) \approx \tilde{H}_i(S^n \setminus B)$$ if $A, B$ have the same homotopy type.

It is even more generally true if $A,B$ have the same shape. For an idea what this means see my answer to Homotopy type of the complement of a subspace.

The above result answers your question for arbitrary closed $A, B \subset \mathbb{R}^n$. Let us regard $S^n$ as the one-point compactifaction of $\mathbb{R}^n$, i.e. $S^n = \mathbb{R}^n \cup \{\infty\}$. Then $A^+ = A \cup \{\infty\}$ and $B^+ = B \cup \{\infty\}$ are compact subsets of $S^n$ and if $A, B$ are homeomorphic, then so are $A^+, B^+$. Therefore $$\tilde{H}_i(\mathbb{R}^n \setminus A) = \tilde{H}_i(S^n \setminus A^+) \approx \tilde{H}_i(S^n \setminus B^+) = \tilde{H}_i(\mathbb{R}^n \setminus B) .$$