What is the abelianization of $\pi_1(\mathbb{R}^3\setminus k)$, where $k$ is a knot in $\mathbb{R}^3$ I don't understand what the three dimensional plane minus a knot really is. I would like to know this because I am studying for an exam and don't know how to work out these abelianization type of problems.
 A: It is worth writing down some other ways of doing this.
1) If you know homology, you can use Mayer-Vietoris in a number of ways. One would be to take a plane diagram for the knot, and assume it's embedded in a way that makes that diagram reasonably accurate (away from crossings, it's in the $xy$-plane; and at crossings it goes just slightly out of the plane and just slightly in). Then delete a tubular neighborhood of the knot, and use the top and bottom half-spaces for Mayer-Vietoris. Alternatively, let $X$ be the exterior of a tubular neighborhood of the knot, and let $T$ be the solid torus given by a tubular neighborhood of the knot. Then we have an exact sequence $$0 = H_2(S^3) \to H_1(T^2) \to H_1(T)\oplus H_1(X) \to H_1( S^3) = 0.$$ The map $H_1(T^2) \to H_1(T)$ sends the longitude to a generator and the meridian to zero; and the map $H_1(T^2) \to H_1(X)$ sends the longitude to 0.
So because the map $\varphi: H_1(T^2) \to \Bbb Z \oplus H_1(X)$ is an isomorphism, given by $am+b\ell \to (b,a\varphi(m))$, we see in particular that $\varphi(m)$ is a generator of $H_1(X)$ and that $H_1(X) \cong \Bbb Z$.
As suggested by Qiaochu, you could also invoke Alexander duality on the complement of a tubular neighborhood of the knot, which gives $H_1(X) \cong H^{3-1-1}(S^1) = \Bbb Z$. If you try to do this on the knot instead of its complement, you get $$H_1(S^1) \cong H^{3-1-1}(X) = H^1(X) = \text{Hom}(H_1(X),\Bbb Z),$$ which doesn't tell you about possible torsion in $H_1(X)$.
A: As mentioned in the comment of @MikeMiller, the key to this is the Wirtinger presentation for $\pi_1(\Bbb R^3 - k)$. Here's a sketch of the derivation of the presentation:
$k : S^1 \hookrightarrow \Bbb R^3$ be the knot. Isotope $k$ to a knot which mostly lies flat on the $xy$-plane but at the crossings it makes overpasses in the upper half space $\Bbb H^3$. 

$\Bbb H^3 - k$ is just $\Bbb H^3$ with a collection of handles corresponding the overpasses removed. Thus, $\pi_1(\Bbb H^3 - k)$ is free on the generators $x_1, \cdots, x_n$ with corresponding to the loops winding around the handles once clockwise. One can easily prove this by applying the van Kampen theorem.

Now, if one attaches the lower half plane $\Bbb {R}_{-}^3$, one adds extra relators. For example, a loop winding around a handle can now be slided through the lower half plane. Consider the situation that you are staring at a knot overpass, with $x_i$ the homotopy class of the loop winding around that overpass. Homotope a loop winding around some other overpass nearby to bring it along the piece of the knot below that overpass. Call it $x_j$. Similarly bring another one of the generator loops close to $x_i$ by the other branch of the underpass. The picture should look like this -

Now you should be able to see that $x_i x_j x_i^{-1} = x_k$. These are the extra relators, and a little more work and Siefert-van Kampen theorem should tell you that these are all the relators there are. You should try to work out the details here by yourself (warning: care needs to be taken to not forget the basepoint). That said, $\pi_1(\Bbb R^3 - k) \cong \langle x_1, \cdots, x_n | x_i x_j x_i^{-1} = x_k \rangle$. This is the Wirtinger presentation. 
To compute $H_1(\Bbb R^3 - k)$, note that abelianization is the same as attaching commutators as relators in your presentation. We know $x_i x_j x_i^{-1} = x_k$, but we are also adding the relators $x_i x_j x_i^{-1} = x_j$ which implies $x_j = x_k$. Thus all but one generator are redundant in the presentation for the abelianization $\pi_1^{ab}(\Bbb R^3 - k)$. 
Since for path-connected based spaces, $\pi_1^{ab}(X, x_0) \cong H_1(X)$ by the Hurewicz theorem, we conclude $H_1$ is infinite cyclic, that is, $H_1(\Bbb R^3 - k) \cong \Bbb Z$.
