Homology of the complement of a compact subset inside ball

Question

Let $A$ be a abelian group and let $\mathbb{B}^m=\{|x|<1\}$ and let $L\subseteq \mathbb R^m$ be compact such that $K:=L\cap \mathbb B^m\ne\emptyset$ is connected. Clearly we have an iso $$H_{m-1}(\mathbb{D}^m\setminus \mathbb{B}^m;A) \to H_{m-1}(\mathbb{D}^m\setminus K;A) \to H_{m-1}(\mathbb{D}^m\setminus 0;A),$$ so $H_{m-1}(\mathbb{D}^m\setminus K;A)\to H_{m-1}(\mathbb D^m\setminus 0;A)$ is an epimorphism. Can we also conclude that it is an isomorphism?

Answer 1

Here is the argument I was referencing in the comments:

We might as well assume that the coefficients are taken to be $\mathbb{Z}$ since the Universal Coefficient Theorem plus the Five Lemma imply the equivalence of the statement with coefficients in $\mathbb{Z}$ and with an arbitrary abelian group. As well, we may as well use reduced homology.

With mild additional restraints the answer is yes. Assume that $L$ is a subset of $B^n$ and it is locally contractible. These assumptions allow us to apply Alexander duality:

Let $X$ be a ball at the origin strictly containing the unit ball. Let $\bar X$ denote its one point compactification. If we define $\bar L$ to be $L \cup \{*\}$ (the point at infinity) we see that $\bar X \setminus \bar L \simeq D^m \setminus L $.

This means we can calculate $\tilde H_{n-1}(D^m \setminus L)$ by calculating $\tilde H_{n-1}(\bar X \setminus \bar L)$ which by Alexander duality equals $\tilde H^0 (\bar L)$ which by assumption of path-connectedness is $\mathbb{Z}$.

This means the map $\tilde H_{m-1}(D^m \setminus L) \rightarrow \tilde H_{m-1}(D^m \setminus 0)$ is a surjective map $\mathbb{Z}\rightarrow \mathbb{Z}$, so must be an isomorphism.