Homology groups of $\mathbb{R}^3$ relative to a disjoint union of two copies of $S^1$

Question

Problem #5: Let $C_1$, $C_2$ be two copies of $S^1$ disjointly embedded in $\mathbb{R}^3$. Compute $H_i(\mathbb{R}^3,C_1\cup C_2)$ for all $i\in\mathbb{N}$.

If I am understanding this correctly, this problem seems really easy. However, I'm not sure that I am. To start things I will present my final answer:
$$H_0=\mathbb{Z},\quad H_1=\mathbb{Z},\quad H_2=\mathbb{Z}\oplus \mathbb{Z},\quad H_n=0\text{ for }n>2.$$

My reasoning for these calculations are as follows: Let $A$ be the set consisting of the two disjoint circles embedded into $\mathbb{R}^3$. $\widetilde{H}$ represents reduced homology.

First, I know that $\widetilde{H}_n(\mathbb{R}^3)=0$ for all $n$. Also, I know $\widetilde{H}_0(A)=\mathbb{Z}$, $\widetilde{H}_1(A)=\mathbb{Z}\oplus\mathbb{Z}$ and $\widetilde{H}_n=0$ for all $n>1$ because it has it's two path components as copies of $S^1$, so the homology groups of $A$ are the direct sum of the homology groups of two copies of $S^1$. Changing to reduced homology yields what I put.

At this point I used the long exact sequence and then converted from the reduced homology groups back to the usual.

If someone has a way to make those direct sums look sexier, feel free to edit. I'm still quite new to the latex world — user82668, Jun 18 '13 at 07:14
You can see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. — Zev Chonoles, Jun 18 '13 at 07:28
The long exact sequence involves relative homology groups... how did you compute those? — Mariano Suárez-Álvarez, Jun 18 '13 at 07:32
@MarianoSuárez-Alvarez Those were what I was using the long exact sequence to compute. — user82668, Jun 18 '13 at 07:33
Sorry, I meant to ask about the maps in the sequence... Too late here :-) — Mariano Suárez-Álvarez, Jun 18 '13 at 07:34
@MarianoSuárez-Alvarez The way the goups turn out, unless I did something wrong, the maps don't turn out to be that important (i.e. just that there are such maps is enough). — user82668, Jun 18 '13 at 07:39

Homology groups of $\mathbb{R}^3$ relative to a disjoint union of two copies of $S^1$

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