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

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?

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.)

• @PaulFrost So it is not true without the assumption of local contractibility, but a different statement (using Cech cohomology) is true. See, I can use bold too. – Najib Idrissi Feb 21 at 17:03
• @NajibIdrissi there's no need to be edgy, thanks for your answer! Do you know any more elementary approach to my original problem? I'm fine with Alexander duality, I'm just curious. – Javi Feb 21 at 19:11
• Perhaps mention that the compactness assumption is necessary. An infinite line and a finite open segment of it are homeomorphic, but their complements differ in $H_1$ and $H_2$. – Andreas Blass Feb 21 at 19:16
• @AndreasBlass A finite open segment is not closed in $\mathbb{R}^n$. – Paul Frost Feb 21 at 23:00
• @AndreasBlass You are right, but an infinite line and a closed half line are not homeomorphic. In $S^n$ we clearly need $A,B$ compact (which is the same as closed) to apply the Alexander duality theorem. In $\mathbb{R}^n$ we do not need $A,B$ compact, it works with closed $A, B$ provided they are homeomorphic. – Paul Frost Feb 22 at 11:56

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 $$\tilde{H}_i(\mathbb{R}^n \setminus A) = \tilde{H}_i(S^n \setminus A^+) \approx \tilde{H}_i(S^n \setminus B^+) = \tilde{H}_i(\mathbb{R}^n \setminus B) .$$