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I know that any compact orientable 3-manifold can be obtained from the three sphere $S^3$ by an integer surgery. I am not sure why the surgery operation is completely determined by Where we map meridians of the tori. More specifically,

an integer surgery is determined by specifying a link in $S^3$ and a framing for each component of that link (or equivalently an integer for each component). For the sake of simplicity in this question I will assume that my link has a single component (a knot).

To perform the surgery we cut out a regular neighborhood of the knot (which is a solid torus) and then we glue back a solid torus by a certain homeomorphism between the surfaces of these two solid tori. What is not clear to me is that why the final manifold is determined by declaring where the homeomophism maps meridian? (in the case of the integer surgery we map the meridian to the framing of the knot). In other words, why the map between these two curves on the tori determine the final 3-manifold (and hence the surgery).

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    $\begingroup$ The homeomorphism is not determined by the integer. Homeomorphisms of a torus up to isotopy are determined by an element of $GL_2 \mathbb Z$. But the resulting manifold is determined by the integer -- this is because the resulting gluing manifolds will be homeomorphic. A solid torus has non-trivial homeomorphisms, up to isotopy. In particular if you fix a meridian of a solid torus, and demand the homeomorphism be orientation preserving, all such homeomorphisms are determined by the linking number of the longitude. $\endgroup$
    – Ryan Budney
    Commented Apr 9, 2020 at 3:33
  • $\begingroup$ Thank you but that still does not answer my question though, When specify the surgery, one declares the meridian of the first torus goes to the framing of the knot on the second torus and that is sufficient to completely define this operation ( for me it seem that what is needed is to define the entire map between the two tori which is an element of $GL_2\mathbf{Z}$ as you mentioned ), my question is why just the mapping between these two curves this is sufficient (again given what you said that a homeomorphism of a torus is determined by an element of $GL_2\mathbf{Z}$ )? $\endgroup$
    – Steve
    Commented Apr 9, 2020 at 3:44
  • $\begingroup$ I did answer your question: the homeomorphism is not determined by the integer. But the resulting manifold is determined by the integer, up to an essentially canonical homeomorphism. $\endgroup$
    – Ryan Budney
    Commented Apr 9, 2020 at 5:56
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    $\begingroup$ Perhaps it helps to think about a simpler example. Glue the boundary of one interval $[0,1]$ to the boundary of another interval $[0,1]$. There are two gluing homeomorphisms of the boundary $\partial [0,1] = \{0,1\}$. But the resulting manifold, either way, is homeomorphic to the circle. They are not the same manifolds set-theoretically, but up to an essentially canonical homeomorphism, they are the same. This is because the homeomorphisms of the boundary extend over the interval. $\endgroup$
    – Ryan Budney
    Commented Apr 9, 2020 at 5:58

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You can think about attaching a solid torus as a two-step process:

First, attach $D^2 \times I$ where $D^2$ is the meridional disc of the attaching torus.

Secondly, attach the remaining of the solid torus, which is homeomorphic to a ball $B^3$.

Now if we know where the meridian of the solid torus goes, then we know the result of the first step, $M_1$, up to isotopy. On the other hand, the result of the second step is determined by the map $\phi \colon \partial B^3 \rightarrow \partial M_1$, up to isotopy. By Smale's theorem, any orientation-preserving diffeomorphism of the 2-sphere is smoothly isotopic to the identity map. Hence the second step is essentially done uniquely.

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  • $\begingroup$ Thank you but how do you know that $\partial M_1$ is a sphere? $\endgroup$
    – Steve
    Commented Apr 9, 2020 at 12:19
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    $\begingroup$ Assume for the moment that we are talking about closed orientable 3-manifolds. Then the boundaries of B^3 and M_1 coincide since after gluing them together along their boundary, we obtain a closed manifold. $\endgroup$
    – Mehdi Yazdi
    Commented Apr 9, 2020 at 12:40
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    $\begingroup$ If we want to consider the case of compact, orientable 3-manifolds then the statement should be modified as follows: every compact orientable 3-manifold can be obtained by removing a regular neighborhood of some graph G in S^3 and integral surgery on a link L in S^3 - G. In any case, for the above argument we only need to consider a specific boundary component of M_1 which coincides with the boundary of B^3, and the argument is as before. $\endgroup$
    – Mehdi Yazdi
    Commented Apr 9, 2020 at 12:43
  • $\begingroup$ Just to clarify, as Ryan mentioned, the above argument only shows that the diffeomorphism type of the resulting manifold is uniquely determined by the image of the meridian up to isotopy. $\endgroup$
    – Mehdi Yazdi
    Commented Apr 9, 2020 at 20:17

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