Follow-up on $H_n(\mathbb{R}^3 - S^1)$ I'm trying to apply Mayer-Vietoris to compute $H_n(\mathbb{R}^3 - S^1)$ as asked by me here. Let $A := X -$"z axis" and $B:= B(0, 0.5) \times \mathbb{R}$ where $B(0, 0.5)$ is the open ball around $0$ with radius $0.5$.
The case $n=0$ is clear to me because $X$ is path connected so $H_0(X) = \mathbb{Z}$.
To compute $H_1(X)$ I wrote down the MV sequence as follows:
$$ \dots H_1(A \cap B) \rightarrow H_1(A) \oplus H_1(B) \xrightarrow{i_\ast} H_1(X) \xrightarrow{\partial_\ast} H_0(A \cap B) \xrightarrow{j_\ast} H_0(A) \oplus H_0(B) \xrightarrow{g} H_0(X) \rightarrow 0$$
I know 
$H_1(A \cap B) = \mathbb{Z}$, 
$ H_0(A) \oplus H_0(B) = H_1(A) \oplus H_1(B) = \mathbb{Z} \oplus \mathbb{Z}$ and 
$H_0(A \cap B) = \mathbb{Z}$.
I also know that $j_\ast$ is injective so $ker j_\ast = 0 = im \partial_\ast \implies$ $\partial_\ast = 0 \implies$ $ i_\ast$ is surjective
But here I think I'm stuck. What am I missing? Thanks for your help.
Edit
Another question: it's not possible to compute $H_0$ using Mayer-Vietoris. I have to use that $X$ is path-connected. Right? Because the only information I can gain from MVS is that $g$ is surjective.
 A: I used what I learned from this question here to compute this and I would appreciate it if you could tell me if my reasoning is correct:
$n=0$: 
$$H_0(X) = \mathbb{Z}$$ because $X$ is path-connected.
$n=1$:
$$ \dots \rightarrow H_1(A \cap B) \xrightarrow{(i,j)} H_1(A) \oplus H_1(B) \xrightarrow{k-l} H_1(X) \xrightarrow{\partial} H_0(A \cap B) \xrightarrow{(i^\prime , j^\prime)} H_0(A) \oplus H_0(B)\dots$$
Observations:
(i) $im (k-l) = H_1(X)$ because $ker (i^\prime , j^\prime) = 0 = im (\partial)$.  $ker (i^\prime , j^\prime) = 0$ because $(i^\prime , j^\prime)((p,q)) = (0,0) \implies (p,q)=(0,0)$.
$\implies k-l$ is surjective.
(ii) $ker (k-l) = im ((i,j)) \cong \mathbb{Z} \oplus 0 \cong \mathbb{Z}$ because the inclusion of $H_1(A) = H_1(S^1)$ maps the generator to a generator.
(iii) $im (k-l) = H_1(X) \cong (H_1(A) \oplus H_1(B))/ker(k-l) = (\mathbb{Z} \oplus \mathbb{Z})/ \mathbb{Z}$. Using the splitting lemma it follows that $H_1(X) \cong \mathbb{Z}$
$n=2$:
$$  0 \xrightarrow{(i,j)} H_2(A) \oplus H_2(B) \xrightarrow{k-l} H_2(X) \xrightarrow{\partial} H_1(A \cap B) \xrightarrow{(i^\prime , j^\prime)} H_1(A) \oplus H_1(B) \dots $$
Observations:
(i) $im (k-l) = H_2(X)$ because $ker ((i^\prime , j^\prime)) = 0$ $\implies (k-l)$ is surjective.
(ii) $ker(k-l) = im((i,j)) = 0$ $\implies (k-l)$ is injective
(iii) $\implies H_2(X) \cong H_2(A) \oplus 0 = \mathbb{Z}$. 
Thanks for your help!
