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EDIT: So my question is distinct from the question asked here because I am asking an easier question. Why should we have to invoke something as powerful as the "Annulus Theorem" to show that the connected sum is well-defined here?

Define the connected sum of two surfaces $\Sigma_1$, $\Sigma_2$ to be the surface we get by taking away a small disk from each of $\Sigma_1$, $\Sigma_2$ and sewing the two boundary components together. The surface we get is denoted by $\Sigma_1 \# \Sigma_2$.

Is it possible someone could provide a concise but complete proof that for orientable $\Sigma_1$, $\Sigma_2$ this operation is well-defined, i.e. the resulting surface doesn't depend on which disks are removed from $\Sigma_1$, $\Sigma_2$ or how the boundaries are glued?

I had searched a bit, but out of the two sources I looked at, one simply asserted that the operation was well-defined, which does not seem to be obvious at all, and the other one simply gave some vague "intuition" as to why the statement should be true.

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  • If the surfaces are compact, you know that they are classified by the genus, which is the number of handles on a ball. In a connected sum, whichever way you do it, you simple add the genera. –  Oct 23 '16 at 13:41
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    This was answered in all dimensions in http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds/121635#121635 – Lee Mosher Oct 23 '16 at 14:29
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    As promised, I requested reconsideration at meta: http://meta.mathoverflow.net/a/3017/2926 Please verify that it reflects what you wanted. – Todd Trimble Oct 24 '16 at 12:18
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    @ToddTrimble Yes that is what I wanted, thanks so much! – user380206 Oct 25 '16 at 18:27
  • The surfaces should be assumed to be connected (otherwise it's not true). But maybe this is already included in the meaning of "surface"? I'm not a topologist. – HeinrichD Oct 25 '16 at 19:15
  • It should be an easy reduction from surfaces with more than one connected component to those with one, assuming anomalies like the sphere cross the rationals are excluded. Gerhard "Likes Surfaces With High Ceilings" Paseman, 2016.10.25. – Gerhard Paseman Oct 25 '16 at 19:42
  • I am not sure there is a concise proof. There some issues to be dealt with here. You can read the proof (with some details left as exercises) in Section 2.5 of Terry Lawson's "Topology: a geometric approach", see http://users.metu.edu.tr/serge/courses/422-2014/supplementary/TGeometric.pdf. – Igor Belegradek Oct 25 '16 at 21:00
  • http://mathoverflow.net/questions/87705/connected-sum-of-surfaces – Igor Belegradek Oct 26 '16 at 00:09

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