Boundary Map in Mayer Vietoris and Homology of Knot Complement Let $K$ be a knot in $S^3$, and N(K) be its tubular neighborhood. I want to compute the homology of $S^3-N(K)$ using Mayer-Vietoris. Let $A$ be $N(K)$ minus a small neighborhood, and $B$ be the complement of N(K) with a small neighborhood added. Note that $A\cup B$ is $S^3$, and $A\cap B$ deformations retracts onto a torus. $H_1(S^3-N(K))$ and $H_n(S^3-N(K))$, $n>2$, pop out of Mayer Vietoris applied to A and B. However I am struggling with $H_2$. According to Mayer-Vietoris we have the following exact sequence $$H_3(S^3) \to H_2(T) \to H_2(A)\bigoplus H_2(B) \to 0$$ 
The part I don't understand is the map $H_3(S^3) \to H_2(T)$.  Let $\alpha$ be the generator of $H_3(S^3)$.  Under the boundary map of Mayer-Vietoris, $\alpha$ is subdivided into a part residing in $A$ and a part residing in $B$.  Then we take the boundary of the part residing in $A$.  Thinking of the boundary map like this makes me think that it is an isomorphism.  However, I am not sure of this.
Are there other ways to see what this map does to the generator of $H_3(S^3)$?  More generally, how does one work with the boundary map in the MV sequence in higher dimensions, where things cannot be simply visualized?  
 A: You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map. 
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = \partial N(K)$) is contained in the $2$-skeleton.  Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} \setminus T$ the interior of each tetrahedron is contained in.  This gives us our two cycles $U,V$, with 


*

*$U+V = [S^{3}]$.

*$\partial U = -  \partial V = [T]$.
Then, by definition, the boundary map maps $U+V \in H_{3}(S^{3},\mathbb{Z})$ to  $\partial U \in H_{2}(T,\mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.
