# Connected sum is well-defined for surfaces, proof?

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.

• 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. – user1688 Oct 23 '16 at 13:41
• This was answered in all dimensions in mathoverflow.net/questions/121571/… – Lee Mosher Oct 23 '16 at 14:29
• As promised, I requested reconsideration at meta: meta.mathoverflow.net/a/3017/2926 Please verify that it reflects what you wanted. – Todd Trimble Oct 24 '16 at 12:18
• @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