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I'm taking a course in topology and we just defined the connected sum of connected surfaces (2-manifolds) as follows:

Given $S_1,S_2$ surfaces and $D_i\subseteq S_i$ open disks with $\varphi : \partial D_1 \rightarrow \partial D_2$ a homeomorphism the connected sum of $S_1,S_2$ is $S_1\#S_2=(S_1-D_1\cup S_2-D_2)/\varphi$

This definition is said to be independent of the choice of $D_1,D_2,\varphi$ (up to homeomorphism) but I can't find an explanation besides the ones that tackle this in the general case (n-varieties). I'm quite sure there exist one since in every one of the explanations I found they were remarking that the complications arise for the case $n\geq 3$

Do you have a source or an explanation?

Orazio Cherubini
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  • Intuitively, if a surface is connected, it doesn't matter where you make a hole in it, because you can move the hole anywhere you want by stretching things. – Karl Feb 27 '24 at 17:04
  • "Independent" here means "independent up to homeomorphism." It's not true that different choices of disks give literally the same connected sum, but they give homeomorphic connected sums. – Andreas Blass Feb 27 '24 at 17:30
  • @AndreasBlass Yeah of course, I mean up to homeomorphism. Thank you, I'll edit – Orazio Cherubini Feb 27 '24 at 17:34
  • @Karl Ok but I can't find something more rigorous anywhere – Orazio Cherubini Feb 27 '24 at 17:35
  • It can be seen as a consequence of the "Annulus theorem" (see https://en.wikipedia.org/wiki/Annulus_theorem). Although it is very easy to state, its proof, especially in higher dimensions, is non-trivial. In dimensions 2 and 3 it follows basically from the fact that manifolds of this dimension are triangulizable (Theorems of Rado and Moise-Bing). For higher dimensions, it was only proven by Kirby in 1969 (dimension >5) and Quinn in 1982 (dimension 4). – G. Blaickner Feb 27 '24 at 17:42
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    See also the answers to the question here. – G. Blaickner Feb 27 '24 at 17:48

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