I am trying to prove the following statement.
If $A$ and $B$ are closed homeomorphic subsets of $\mathbb{R}^{n}$, then show that the homology group of $H_i(\mathbb{R}^{n},\mathbb{R}^{n} - A)$ and $H_i(\mathbb{R}^{n},\mathbb{R}^{n} - B)$ are isomorphic.
I tried using long exact sequence for pairs twice. I got
$$\dots \to H_i(\mathbb{R}^{n} - A) \to H_i(\mathbb{R}^{n}) \to H_i(\mathbb{R}^{n},\mathbb{R}^{n} - A) \to H_{i-1}(\mathbb{R}^{n} -A)\to \dots$$ and $$\dots \to H_i(\mathbb{R}^{n} - B) \to H_i(\mathbb{R}^{n}) \to H_i(\mathbb{R}^{n},\mathbb{R}^{n} - B) \to H_{i-1}(\mathbb{R}^{n} -B)\to \dots$$
I tried to use the following diagram.
\begin{matrix} \dots \to H_i(\mathbb{R}^{n} - A)& \to H_i(\mathbb{R}^{n}) &\to H_i(\mathbb{R}^{n},\mathbb{R}^{n} - A) &\to H_{i-1}(\mathbb{R}^{n} -A)\to \dots \\ \downarrow & \downarrow &\downarrow &\downarrow \\ \dots \to H_i(\mathbb{R}^{n} - B)& \to H_i(\mathbb{R}^{n}) &\to H_i(\mathbb{R}^{n},\mathbb{R}^{n} - B) &\to H_{i-1}(\mathbb{R}^{n} -B) \to \dots \end{matrix}
I thought diagram chasing would help, but it doesn't as I am not getting any vertical maps.
Can anyone help me with this problem. I read few answers but I cannot use Alexander's Duality. Thank you.
Edit 1
As RJP suggested, I got
$ \dots \to 0 \to H_i(\mathbb{R}^{n},\mathbb{R}^{n} - A) \to H_{i-1}(\mathbb{R}^{n} -A)\to 0 \to \dots $ is exact implies $H_i(\mathbb{R}^{n},\mathbb{R}^{n} - A) \cong H_{i-1}(\mathbb{R}^{n} -A)$ and simmilarly $H_i(\mathbb{R}^{n},\mathbb{R}^{n} - B) \cong H_{i-1}(\mathbb{R}^{n} -B)$.
Now how do I show that $H_{i-1}(\mathbb{R}^{n} -A) \cong H_{i-1}(\mathbb{R}^{n} -B)$?
Note- I tried and refered this
But it didn't help. Can anyone give a simpler reasoning?
