# Fundamental group of 3-manifold obtained by surgery description

Say I have a surgery description of a 3-manifold $M$: an $n$ component link $L$ with $p_i/q_i$ rational surgeries performed on the components (can as well be integer surgeries, $q_i=1$).

Then: $$\pi_1(S^3-L) = <x_1,... | w_1,...,w_m>,$$ where the relations are obtained from the Wirtinger presentation.

And $\pi_1(M)$ can be obtained by adding the 2-discs: $$\pi_1(M) = <x_1,... | w_1,...,w_m, \, l_1^{p_1} m_1^{q_1} \,, ... ,\, l_n^{p_n} m_n^{p_n}>,$$ where $l_i$ and $m_i$ are words that represent the longitudes and meridians of the components in terms of the $x_i$'s.

Is this correct and where can I find a reference to it? I've check Rolfsen's book, but could not find a direct argument for this.

The van Kampen theorem argument goes like this. I'll just do it for a single knot, but the argument extends easily to links. (My motivation is to get this straight myself, so apologies for restating things you already know.) The homology of the torus boundary of a tubular neighborhood of the knot is generated by a longitude $\lambda$ and a meridian $\mu$. The boundary of a solid torus $S^1\times D^2$ has a meridian curve $m$ bounding a solid disk and a corresponding longitude $l$. According to Lickorish exercise 11.9, the $p/q$ surgery is gluing $S^1\times D^2$ to $S^3-N(K)$ along along a homeomorphism $h:\partial(S^1\times D^2)\to\partial(S^3-N(K))$ with $h_*(m)=p\mu+q\lambda$. Since it is a homeomorphism and since the mapping class group of $S^1\times S^1$ is $SL_2(\mathbb{Z})$, the two integers $a,b$ such that $h_*(l)=a\mu+\lambda$ where $a,b\in\mathbb{Z}$ just need to satisfy $pb-qa=\pm 1$.
The fundamental group of the surgery is the quotient of $\pi_1(S^1\times D^2)*\pi_1(S^3-N(K))$ subject to the following additional relations:
• From the inclusion of $m$ into $S^1\times D^2$ and $S^3-N(K)$, $1=\mu^p\lambda^q$.
• From the image of $l$ in each subspace, $l=\mu^a\lambda^b$, where I am identifying $l$ with the generator of $\pi_1(S^1\times D^2)$.
The second relation says that $l$ is already representable in $\pi_1(S^1-N(K))$. Thus, the fundamental group of the $p/q$ surgery is $$\langle \pi_1(S^3-K)\mid \mu^p\lambda^q\rangle$$ which is almost what you said.
To test this out, let's consider the lens space $L(p,q)$, which is $p/q$ surgery on the unknot. Then $\lambda=1$. The relation is just $\mu^p=1$, so $\pi_1(L(p,q))\cong \mathbb{Z}/p\mathbb{Z}$. (Rolfsen Exercise 5, p235.)