# 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?

• This is a great question here, and I hope you find Nick L's answer useful :)
– user98602
Dec 6, 2018 at 13:27

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
1. $$U+V = [S^{3}]$$.
2. $$\partial U = - \partial V = [T]$$.
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.