Use a Mayer-Vietoris sequence to calculate $H_i(S^3 - K)$ where $K$ is an embedding of $S^1$.
(hint: Let $S^3 = (S^3 - K) \cup O$ where $O$ is an open tubular neighborhood around $K$ that is homemorphic to the solid torus $D^2 \times S^1$.
This is what I have of the proof so far. Will somebody help me correct it, finish it, and just make it better in general? Thanks!
Proof:
Let $A = S^3 - K$ and $B = O \cong D^2 \times S^1$. Then $A \cap B \cong T^2$, the Torus and $A \cup B \cong S^3$.
Note that since $D^2$ is contractible, $B=O$ has the same homotopy type as $S^1$. Thus $H_i(B) = H_i(S^1)$
We thus have the following Mayer-Vietoris Sequence:
$$0 \rightarrow H_3(S^3-K) \oplus H_3(S^1) \rightarrow^z H_3(S^3) \rightarrow^x H_2(T^2) \rightarrow^a H_2(S^3 - K) \oplus H_2(S^1) \\ \rightarrow^b H_2(S^3) \rightarrow^c H_1(T^2) \rightarrow^d H_1(S^3 - K) \oplus H_1(S^1) \rightarrow^e H_1(S^3) \rightarrow^f H_0(T^2) \\ \rightarrow^g H_0(S^3-K) \oplus H_0(S^1) \rightarrow^h H_0(S^3) \rightarrow 0$$
We now make the following known substitutions:
$H_2(T^2) \cong \mathbb{Z}$
$H_2(S^1) \cong 0$
$H_2(S^3) \cong 0$
$H_1(T^2) \cong \mathbb{Z} \oplus \mathbb{Z}$
$H_1(S^1) \cong \mathbb{Z}$
$H_3(S^3) \cong \mathbb{Z}$
$H_1(S^3) \cong 0$
And all the zeroth homologies are $\mathbb{Z}$ as they are all path connected.
Thus our sequences becomes:
$$0 \rightarrow H_3(S^3-K) \oplus 0 \rightarrow^z \mathbb{Z} \rightarrow^x \mathbb{Z} \rightarrow^a H_2(S^3 - K) \oplus 0 \rightarrow^b 0 \rightarrow^c \mathbb{Z} \oplus \mathbb{Z} \\ \rightarrow^d H_1(S^3 - K) \oplus \mathbb{Z} \rightarrow^e 0 \rightarrow^f \mathbb{Z} \rightarrow^g \mathbb{Z} \oplus \mathbb{Z} \rightarrow^h \mathbb{Z} \rightarrow 0$$
And so by exactness of our sequence we have $H_1(S^3-K) = \mathbb{Z}$
I now turn my attention to proving that $H_3(S^3-K) = H_2(S^3-K) = 0$, so I zoom in on this part of our L.E.S.
$0 \rightarrow H_3(S^3-K) \oplus 0 \rightarrow^z \mathbb{Z} \rightarrow^x \mathbb{Z} \rightarrow^a H_2(S^3 - K) \oplus 0 \rightarrow^b 0 \rightarrow...$
or, simplified a bit:
$0 \rightarrow H_3(S^3-K) \rightarrow^z \mathbb{Z} \rightarrow^x \mathbb{Z} \rightarrow^a H_2(S^3 - K) \rightarrow^b 0 \rightarrow...$
note to reader: I can be pretty sloppy when analysing simple sections of a L.E.S. like this one; Here is my attempt. If somebody could show me the simpliest way to think about this that would be great!
Because of exactness, we know that $im(z) = ker(x)$, $ker(a) = im(x)$, and we know that $z$ is injective and $a$ is surjective.
Thus we know that:
$\frac{\mathbb{Z}}{im(x)} = \frac{\mathbb{Z}}{ker(a)} \cong H_2(S^3-K)$
Also, for some reason, perhaps because $\mathbb{Z}$ is free abelian, we know that $\mathbb{Z} = ker(x) \oplus im(x) = im(z) \oplus im(x)$
So if we can prove that something like $im(x)=\mathbb{Z}$ and this would instantly imply that $H_2(S^3-K)$ is trivial, and it would also imply that $im(z)=0$ and since $z$ is injective this means that $H_3(S_3-K)=0$.
Calculating $im(x)$ has to do with understanding the boundary map rather than falling out of the exactness of the sequence.
To understand the boundary map, here is Cheerful_Parsnip, where he use $U$ instead of $O$
"One way to see that is to trace through the definition of the boundary map. Basically, the fundamental class for S3 is a direct sum of simplices in a triangulation of S3. Subdivide these until all of them are contained either in S3∖K or U (or equivalently in S3∖K up to sign). Then look at the direct sum of those in U. The boundary of this is a torus around the knot, and is homologous to the fundamental class of T2. I'm sure there's some fancier way of seeing it."
Thus $x$ is an isomorphism and $H_3(S^3-K) \cong H_2(S^3-K)=0$