TITLE: Obtain any 3-manifold from repeating surgeries on knots in $S^3$
QUESTION [14 upvotes]: In Witten's “QFT and Jones Polynomials” paper, page 383, it states that: "It is a not too deep result that every 3-manifold can be obtained from or reduced to $S^3$ (or any other desired 3-manifold) by repeated surgeries on knots.
What are the methods to show this? In the simplest intuitive level?
I understand this may be a relevant post, but I hope there are better illuminations. Thanks.
p.s. I am a QM/QFT theorist trying to understand the topology better.

REPLY [5 votes]: This is really an extended comment. Firstly, as pointed out by Neil Hoffman, the result was proved by Wallace two years before Lickorish, and Lickorish was well aware of this (Wallace's paper is cited in Lickorish's). I really can't understand why Lickorish's name is attached to this result at all. Lickorish's reasonable claim was that his argument was more elementary than Wallace's. Indeed, this is true, since Lickorish's entire paper is eight pages (Wallace's is 28), BUT Lickorish (as suggested in my comment re Sam Nead's answer) does NOT prove that a finite number of Dehn twists suffice (that result is proved in a 1964 paper - two years after, and if you look at the math review, the reviewer is not so sure that it is really proved). Of course, had Lickorish been aware of Dehn's work, his 1962 paper would be down to around two pages. So, another (somewhat complicated) example of Arnold's principle at work.

REPLY [4 votes]: Math Reviews says of the paper "A type of homology-preserving surgery on 3-manifolds", J. London Math. Soc. 17 (1978), 183–185 by Norris Weaver:
"Let $M$ be a 3-manifold and let $T \subset {\rm int}\,M$ be a tame submanifold homeomorphic to $S^1\times D^2$. Let $g: \partial T \to \partial T$ be an orientation-preserving homeomorphism, and let $n$ be an integer. Glue $T$ or $M - {\rm int}\, T$ by $g^n$ to obtain a manifold $M_1$. Call this construction an $n$th power surgery. The following result is obtained. Theorem: If $n\equiv 0 \, ({\rm mod}\,6)$ and $0\leq n\leq 36$, then $H^1(M_1:\mathbb{Z}_n) \cong H^1(M:\mathbb{Z}_n)$. If $n \not\equiv 0\,({\rm mod}\,6)$ then every closed, orientable 3-manifold can be obtained from $S^3$ by successively performing a finite number of $n$th power surgeries."