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I once read (I believe in Ravi Vakil's notes on Algebraic Geometry) that the connected sum of a pair of surfaces can be defined in terms of a universal property. This gives a slick proof that the connected sum is unique up to homeomorphism. Unfortunately, I am unable to find where exactly I read this or remember what exactly universal property was; if anyone could help me out in either regard it would be much appreciated.

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A quick search of Foundations of Algebraic Geometry reveals nothing. Certainly one can define the wedge sum by a universal property: it is the coproduct in the category of pointed topological spaces, or equivalently the pushout of $X \leftarrow \{ * \} \rightarrow Y$ in the category of topological spaces. Perhaps something similar works for connected sums? – Zhen Lin Jul 7 '11 at 3:44
Well, that does not quite give what you're looking for, but by definition the connected sum is the push-out of $(M\smallsetminus D) \gets \partial D \to (N \smallsetminus D)$ where $D$ is an embedded open disk in $M$ and $N$. – t.b. Jul 7 '11 at 4:36
I am afraid that the complex structure or algebraic structure will be totally ruined in general cases when you do a surgery like that in the category of topological spaces or smooth manifolds. However, a definition can always be made via the universal property, regardless of the existence of such an object. – Yang Zhou Oct 1 '12 at 11:20
Even if one found a universal property that characterized the direct sum, you'd have to prove that whatever disc you removed satisfied that property at the end of the day. For this I can think of no way to avoid the use of the disc theorem: which is hard. But, if you've gotten this far, then you've already proven that the connected sum is well-defined up to homeomorphism. So I don't know if the end result would be "slick." – Dylan Wilson Nov 29 '12 at 14:06

1 Answer 1

As already noted in the comments, there is an obvious universal property (since the connected sum is a special pushout) once the embeddings of the discs have been chosen. For different embeddings, there exists some homeomorphism. There are lots of them, but even abstract nonsense cannot replace the nontrivial proof of existence. But since there is no canonical homeomorphism, I strongly doubt that there is a ny universal property which does not depend on the embeddings of the discs.

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