I just read a proof stating that $\pi_1(\mathbb{R}^3-K)\cong \langle x,y: x^3=y^2\rangle$ but I am having trouble with some statements.
The proof goes as follows : By using van-Kampen we have that $\pi(S^3-K)\cong \pi_1(\mathbb{R}^3-K)$. Now view $S^3=\partial(D^2\times D^2)=S^1\times D^2\cup D^2\times S^1$ and denote $A:=S^1\times D^2$ and $B:=D^2\times S^1$. We have that $K\subset A\cap B=S^1\times S^1$ and so $A-K$ and $B-K$ are two closed subsets that cover $S^3-K$. They intersect at $A\cap B-K$ which is an annulus . (Why is this true ? I can't seem to visualize it , I would guess the idea would be to flatten what is left of the torus but I am not sure this works).The inclusions $A\cap B-K\rightarrow A-K,B-K$ are cofibrations and so we can apply van kampen's theorem and we get that $\pi_1(S^3-K)$ is isomorphic to the pushout of $\mathbb{Z}\leftarrow^{\times 2} \mathbb{Z}\rightarrow ^{\times 3} \mathbb{Z} $.
Now I don't really understand why we have that $\pi_1(A-K)\cong \pi_1(B-K)\cong \mathbb{Z}$, I guess it would follow that we would make a deformation retract into an annulus . Also I am not sure why we get those induced maps in the fundamental groups. I get that the map that gives the knows is $f(z)=(z^2,z^3)$ but since we are removing I don't see how we can come in with it's influence.
Any help with this is appreciated. Thanks in advance.
To see why the complement of $K$ in $S^1\times S^1$ is a copy of $S^1\times I$, think about what $K$ looks like in the description of the torus as a square with its edges identified - the complement will be easier to visualize there (remember you need to identify edges again to see it properly), and it should be clear that it's an annulus.
As for why $\pi_1(A-K)\cong\pi_1(B-K)\cong\mathbb{Z}$, note that we are now considering the complement of a trefoil knot in a solid torus: the above shows that $(S^1\times S^1)-K$ retracts to an annulus, so have a think about what $(S^1\times D^2)-K$ will retract to.
Finally, for your question regarding the effect on the fundamental groups, it may help to think of it this way: we can think of $S^3$ as the union of two solid tori, as you described in your question. Your curve $K$ is homotopically non-trivial on $S^1\times S^1$, but when we include it into $S^1\times D^2$ (where we think of $K$ as lying first on a hollow donut, then on a solid donut), all the meridional components become nullhomotopic, so the effect on the fundamental group is determined by the longitudinal components. The effect on the "outer" solid torus is the opposite.