# Why surgery produce a new 3-manifold?

I was studying a proof of the fact that any closed orientable 3-manifold is obtained by integer surgery along a link.

I read the several proofs but I don't understand well.

A proof is as follows.

First, consider a Heegaard decomposition of a 3-manifold $M$. Assume first that gluing homeomorphism of this Heegaard decomposition and a gluing composition of these handle body to produce the $S^3$ differ only by a twist along a curve $c$ on a handlebody.

We push the curve into a handlebody and cut out the neighborhood $N$ of it. Then we twist the complement along $c$.

Then there is a homeomorphism from $S^3\setminus N$ to $M\setminus N$. Thus $M$ is obtained from $S^3$ by cutting out a solid torus and glue it back with a different homeomorphism.

What I do not understand is as follows. In the proof above, we cut out a torus $N$ and we twist the complement along $c$, which is a longitude. So the meridian is mapped to a meridian plus/minus a longitude.

However, this is a description of $1$-surgery, which produce again $S^3$. So how come can we get a new manifold by this procedure?

I think you might be using the term "longitude" incorrectly. Given a solid torus $N \subset S^3$ with boundary 2-torus denoted $T = \partial N$, a "meridian" on $T$ is a curve that bounds a disc in $N$, and a "longitude" on $T$ is a curve which is homologically trivial in $S^3-N$ (equivalently it bounds a Seifert surface in $S^3 - N$).
It is certainly true that if you do surgery using any map of $T$ which takes the meridian to the meridian, that is any power of a Dehn twist around the meridian, then the result of that surgery is $S^3$.
But if you do surgery using a map of $T$ that takes the meridian to some curve other than the meridian, such as a Dehn twist around the longitude, that's what is called a "nontrivial surgery", the result of which is never $S^3$. That's Property P, which has been proved.
Having said that, I will add that in the context of the piece of proof that you show, there is no guarantee that $c$ even represents a longitude in $\partial N$ (which is what made me suspect you are using "longitude" incorrectly).
• Maybe a more precise /less ambiguous way of describing the meridian 'c' is as the inclusion of {pt}$\times S^1$ in $S^1 \times S^1$ . I came to know the difference between meridians and parallels, i.e . which is which, by remembering Greenwich as the $0$ meridian, dividing 'East and West' . May 18, 2014 at 0:16