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Let $M_K$ be the result of $0$-Dehn surgery along a knot $K$ in $S^3$. Let $m$ be a meridian to $K$, and we view $m$ as a circle in $M_K$ (without changing the notation for it).

The claim I wanted to prove is: Then the fundamental group $\pi_1(M_K)$ is normally generated by the homotopy class $[m]$ of the meridian.

I tried to apply van Kampen with the following notation.

I use $U:=\text{knot complement of }K$, $V:=\text{regular neighbourhood of }K$. Both are meant to be open and thicker than complementary such that they have an intersection $U\cap V$ homeomorphic to the torus crossed with some open interval $T^2\times (a,b)$, homotopy equivalent to $T^2$.

The group $\pi_1(U)$ has some Wirtinger representation, but other than that we cannot say much about it now. Of course, $\pi_1(V)=\mathbb{Z}$, generated by the knot or a longitude, and $\pi_1(U\cap V)=\mathbb{Z}\oplus \mathbb{Z}$, generated by meridian and longitude. Let $\mu$ be the class of the meridian and $\lambda$ the class of the longitude.

Now $0$-Dehn surgery can be realized by choosing $(p,q)=(0,1)$. This means the spanning disc used to fill in the solid torus has the homotopy class $\alpha = \mu$ (instead of $\alpha = \lambda$ for trivial reinserting of the solid torus).

So now the homotopies of fundamental groups induced by inclusions are $i_{UV}:\pi_1(U\cap V)\to \pi_1(U):(k,l)\to k$, because the the longitude becomes null homotopic, and $i_{VU}:\pi_1(U\cap V)\to \pi_1(V):(k,l)\to l$ (this is rather a guess).

Then the generators we have to examine are $i_{UV}(\omega)i_{VU}(\omega)^{-1}$ for $\omega\in \pi_1(U\cap V)$. These are thus $kl^{-1}\in \pi_1(U)\ast \pi_1(V)$ for $(k,l)\in \mathbb{Z}^2$. Let $N$ be the normal subgroup generated by those elements.

Thus $\pi_1(M_K)=\pi_1(U)\ast \mathbb{Z} / N$. But what is this exactly and how can I check the claim?

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