Finding $H_3$ and $H_2$ for $S^3 \setminus K$ where $K$ is a knot I'm trying to compute (reduced) $H_n(S^3 \setminus K)$ where $K$ is a knot, an embedding of $S^1$ into $S^3$. I'm following a suggestion with the problem that if I let $O$ be a small open tube around $K$ then I can use the Mayer-Vietoris sequence where by homotopy equivalence $H_n(K) = H_n(S^1)$, $H_n((S^3 \setminus K) \cap O) = H_n(S^1 \times S^1)$, and this works fine except around $n = 2, 3$.
My issue is this: Assuming I've done everything else right, I have the exact sequence
$0 \to H_3(S^3 \setminus K) \oplus 0 \to \mathbb{Z} \to \mathbb{Z} \to H_2(S^3 \setminus K) \oplus 0 \to 0$ and I can't figure out the nature of any of the maps in the middle beyond what the sequence already makes obvious. That is, I don't understand how $H_3(S^3)$ maps to $H_2(S^1 \times S^1)$ here (these are the two $\mathbb{Z}$'s in the middle), and I really can't figure out anything about the relationship between the outer groups and the inner ones.
 A: The boundary operator $\partial \colon H_k(X)\to H_{k-1}(A\cap B$ in Mayer Vietoris for $X=A\cup B$ is defined as follows. Given a $k$-cycle $x$ in X, write it as $[x]=[a+b]$ where $a$ and $b$ are chains supported in $A$ and $B$ respectively. Then $\partial a$ is contained in $A\cap B$, and is the image of the MV boundary operator. For the sphere, the generator of $H_3(S^3)$ is represented by the sum of tetrahedra in any triangulation of $S^3$. One can choose such a triangulation to contain a triangulation of $S^1\times S^1$, and let $a$ be the sum of tetrahedra on the outside of the $S^1\times S^1$, while $b$ is the sum of tetrahedra on the inside. $\partial a$ is then the sum of triangles in the triangulation of the torus, so represents the generator of $H_2(S^1\times S^1)$. Therefore the MV boundary map is an isomorphism in this case.
A: Alexander duality says that:

If $K$ is a compact, locally contractible, nonempty, proper subspace of $S^n$, then $$\tilde{H}_i(S^n-K;\mathbb{Z})\cong \tilde{H}^{n-i-1}(K;\mathbb{Z})$$ for all $i$. 

For a proof see Hatcher Theorem 3.44.
