Let $f: S^1 \times D^2 \to S^3 $ be an embedding, and let $g=f|_{S^1 \times 0}$ denote the induced embedding of a circle, and let $K=g(S^1)$. I want to compute the homology groups of $A:=S^3-K$ using the Mayer-Vietoris sequence.
Since $A$ is path-connected (by, for example, the Jordan curve theorem), $H_0(A)=\Bbb Z$.
Let $B=f(S^1 \times D^2)$. Since $f$ is an embedding, $B$ is homeomorphic to a solid torus, hence homotopy equivalent to a circle, so we know the homology groups of $B$. Also, we have $A \cup B =S^3$. Now we consider the Mayer-Vietoris sequence for the decomposition $S^3 = A \cup B$ (Actually, we have to consider an open neighborhoods of $B$ that deformation retracts onto $B$, instead of $B$). Note that $A \cap B$ is homotopy equivalent to a torus. Then it is easily shown that $H_k (A)=0$ for all $k >3$. So we only need to consider $k=1,2,3$.
For $k=1$, it is quite easy. We have:
$H_2(S^3)=0 \to H_1(T^2)= \Bbb Z^2 \to H_1(A) \oplus H_1(S^1)=H_1(A) \oplus \Bbb Z \to H_1(S^3)=0.$
Thus, $H_1(A)=\Bbb Z$.
For $k=2,3$ we need to consider the following:
$H_3(T^2)=0 \to H_3(A) \oplus H_3(S^1)=H_3(A) \to H_3(S^3)=\Bbb Z \to H_2(T^2)=\Bbb Z \to H_2(A)\oplus H_2(S^1)=H_2(A) \to H_2(S^3)=0$.
If we can examine the map $H_3(S^3) \to H_2(T^2)$, we will be done, but I have no idea here. How do I have to proceed?
On the other hand, is there another method, other than the Mayer-Vietoris, to compute the homology groups of $A$?