Calculating homology groups of a generic knot I am trying to calculate the homology groups of a knot embedded in $S^3$. This is what I have so far:
$$H_0(S^3-K)=\mathbb{Z}$$
since $S^3-K$ is path connected; from Alexander duality I have that:
$$H_n(S^3-K)=0 \qquad n\geq3$$
Again from Alexander duality I think it should be
$$H_2(S^3-K)\simeq H^0(K)=\mathbb{Z}$$
I need to find $H_1(S^3-K)$, but I do not know how. I was thinking to use Mayer-Vetoris method as follows:
$$0=H_1(S^3)\rightarrow H_1(S^3,S^3-K)\rightarrow H_1(S^3-K)\rightarrow H_1(S^3)=0$$
but I do not know how to find $H_1(S^3,S^3-K)$
EDIT I just thought I have to use Hurevicz theorem, so that $H_1(S^3-K)\simeq \pi_1(S^3-K)/[\pi_1,\pi_1]$
I am surprised that you know Alexander duality but not $H^1(S^1)$. However, you are right that Alexander duality shows
$$\tilde H_i(S^3 \setminus K) \approx \tilde H^{2-i}(K) \approx \tilde H^{2-i}(S^1). $$
Here $\tilde H_j$ and $\tilde H^j$ denote the reduced homology and cohomology groups. Except for $j = 0$ they agree with the unreduced groups. Moreover, if the reduced group in dimension $0$ is $0$, the unreduced group in dimension $0$ is $\mathbb Z$.
You correctly conclude that $H_i(S^3 \setminus K) = 0$ for $i \ge 3$.
The (reduced) cohomology groups of $S^1$ can be computed by exactly the same methods as the homology groups of $S^1$ which gives $\tilde H^1(S^1) = \mathbb Z$ and $\tilde H^i(S^1) = 0$ for $i \ne 1$. Let us come back to this point later. Anyway, we get
$$\tilde H_0(S^3 \setminus K) \approx  H^2(S^1) = 0, $$
$$H_1(S^3 \setminus K) \approx H^1(S^1) = \mathbb Z, $$
$$H_2(S^3 \setminus K) \approx \tilde H^0(S^1) = 0 . $$
Concerning the cohomology of $S^1$: We can either compute it as indicated above or use Alexander duality. In fact, consider $S^1$ embedded as the set $S = \{(x_1,x_2,0,0) \mid (x_1,x_2) \in S^1\}$. It is well-known that in this special case $S^3 \setminus S \simeq S^1$. Thus you know the homology of $S^3 \setminus S$ and via Alexander duality the cohomology of $S = S^1$.
Classical knots in 3-space have no interesting homology. Instead, Alexander looked at the homology of their 2-fold covering spaces (an easy to see invariant that distinguishes a lot of knots). Reidemeister showed, shortly thereafter, that linking numbers in non-cyclic coverings (just a bit harder to see) fill in all the known gaps (as revealed now through 19 crossings). -- Ken Perko, lbrtpl@gmail.com