I am trying to do the following exercise and I wanted to make sure if my thought process is correct :
Let $L$ be the $2$-component link consisting of two unknotted circles $C\cup D$ in which $C$ wraps around $D$ $k$ times. Where we can think of $C$ being in the torus and $D$ be the generator of the fundamental group of the solid torus.We wanna compute $\pi_1(\mathbb{R}^3-L)$.
So first we note that using Van-Kampen we have that $\pi_1(\mathbb{R}^3-L)=\pi_1(S^3-L)$. Then we have that $S^3=\partial (D^4)=S^1\times D_2 \cup D_2\times S_1$. Let's call $A=S^1\times D^2$ and $B=D^2\times S_1$ where $B$ is the torus that contains the point at infinity. Now we will have that $C\subset A\cap B$, $L\subset A$ and $C\subset B$. So that $A-L$ and $B-L$ are closed sets such that $A-L \cup B-L=S^3-L$. Now we note that $(A\cap B-L ,A-L)$ and $(A\cap B-L,B-L)$ are cofibrations. And so we can apply the van-kampen theorem for closed subspaces. Now we have that $\pi_1(A\cap B-L)\cong \mathbb{Z}$ since $A\cap B-L$ is essentially an annulus, $\pi_1(A-L)\cong \mathbb{Z}\bigoplus \mathbb{Z}$ and $\pi_1(B-L)\cong \mathbb{Z}$. Now we need to find out what are the induced maps. We have that $\pi_1(A\cap B-L)\rightarrow \pi_1(A-L)$ is the map that sends $1$ to $(1,k)$ and that the map $\pi_1(A\cap B-L)\rightarrow \pi_1(B-L)$ is the map that sends $1$ to $k$. And so by the Seifert–Van Kampen theorem we get that $\pi_1(S^3-L)$ will be the group $\langle x,y,z:x^k=zy^k, zy=yz\rangle$.
What do you guys think ? Any insight is appreciated. Thanks in advance.
This seems correct to me: I've been studying these links as part of my dissertation, and I had to check you got the same fundamental group I did. Indeed, you can see from your presentation that $z=x^ky^{-k}$, so we can eliminate it from the generators to get $$\pi_1(S^3-L)\cong\langle x,y\mid (x^ky^{-k})y=y(x^ky^{-k})\rangle\cong\langle x,y\mid x^ky^{1-k}=yx^ky^{-k}\rangle$$ and multiplying by $y^{k-1}$ on both sides of the relation, we get $\pi_1(S^3-L)\cong\langle x,y\mid x^k=yx^ky^{-1}\rangle$, which is the Baumslag-Solitar group $BS(k,k)$.