Let $K\subset S^3$ be a knot, $N(K)$ be a tubular neighborhood of $K$ in $S^3$, $M_K$ to be the exterior of $K$ in $S^3$, i.e., $M_K=S^3-\text{interior of }{N(K)}$.
Now, it is clear that $\partial M_K=\partial N(K)=T^2$, the two dimensional torus, and when using the (M-V) sequence to the triad $(S^3, N(K)\cup M_K, \partial M_K)$ to calculate the homology group of $H_i(M_K, \mathbb{Z})$, $$H_3(\partial M_K)\to H_3(M_K)\oplus H_3(N(K))\to H_3(S^3)\overset{\partial_3}{\to}H_2(\partial M_K)\to H_2(M_K)\oplus H_2(N(K))\to H_2(S^3)\overset{\partial_2}{\to}\cdots$$
I have to figure out what the boundary map $\mathbb{Z}=H_3(S^3)\overset{\partial_3}{\to}H_2(\partial M_K)=\mathbb{Z}$ looks like.
I think it should be the $\times 1$ map, but I do not know how to deduce it geometrically, so my question is:
Q1, Is it true that in the above (M-V) sequence $~ \mathbb{Z}=H_3(S^3)\overset{\partial_3}{\to}H_2(\partial M_K)=\mathbb{Z}$ is $\times 1$ map? Why?
Q2, If the above $K$ is a link with two components, then $\partial M_K=T^2\sqcup T^2$, do the same thing as above, what does $\mathbb{Z}=H_3(S^3)\overset{\partial_3}{\to}H_2(\partial M_K)=\mathbb{Z}\oplus \mathbb{Z}$ look like now?
The generator of $H_3(S_3)$ can be given by taking the closures of $M_K$ and $N(K)$ and triangulating them so the triangulation agrees on the boundary, then taking the union of all the simplices as your cycle. The boundary map takes this cycle and sends it to the common boundary of its two chunks, which is exactly the torus, so the map is indeed 1. The map in the second question should be the diagonal map $(\times 1,\times 1)$.