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It is a standard excercise in differential geometry to prove that a connected sum $M\#N$ of two smooth manifolds $M,N$ of the same dimension is uniquely defined (under some assumptions regarding orientation). In the proof one needs to show that $M\#N$ does not depend on:

  1. the choice of balls (chart maps) around fixed points $(m,n)\in M\times N$ (assuming that maps in comparison induce the same local orientation near $m$ or $n$),
  2. the choice of points.

I can prove the first for smooth manifolds (for two neighbourhoods of $m\in M$, I choose such a small neighbourhood that it looks like a ball in both charts, and I observe that the claim is trivial for balls). The second follows from homogeneity of both: topological and smooth manifolds.

My question is: Prove or disprove the first point for topological manifolds. If necessary, one can assume orientability, but the smooth case gives hope that the claim may be true also for non-orientable manifolds.

Technical clarifications: For the sake of this question "the" definition of $M\#N$ is as follows. Take $(m,n)\in M\times N$ and chart maps establishing homeo/diffeo of neighbourhoods of $m,n$ with open balls. Then canonically identify the punctured balls with cylinders and glue the cylinders reversing the vertical coordinate: $$B\setminus \{m\} \simeq S^{n-1}\times (0,1)\owns(s,t) \mapsto (s,1-t) \in S^{n-1}\times (0,1) \simeq B'\setminus \{n\}.$$ To avoid problems we can assume that the chart maps can be extended to bigger balls.

P.S. Some related doubts Update: My earlier doubts resolved thanks to discussion in the comments:

  • why wikipedia needs (note that the definition of connected sum in wiki is slightly different than mine):

    • oriented manifolds (in both cases if I understand correctly). Resolved: Thanks to George Lowther's comment I already know that the connected sum depends on the orientation of the spheres/cylinders glued (I missed that in my proof previously).
    • and why it mentions some "canonical glueing" as necessary for the uniqueness (in the smooth case)? Resolved: as studiosus states in the comments, the thing is that wikipedia allows ugly diffeomorphisms of spheres that can not be extended to the whole disk, which is not the case in this question,
  • wikipedia claims the answer is affirmative "crucially" by the disc theorem, however the article cited handles the differential case. Resolved: The disc theorem is false for the topological case, because a sum of Alexander horned sphere and the bounded component of its complement is homeo with the disk, while the unbounded component is not homeo with complement of the unit disk in $\mathbb R^3$.

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Connected sum is not uniquely defined in the smooth category if you allow arbitrary diffeomorphisms of boundary spheres. – studiosus Sep 10 '13 at 15:46
@saviki: Yes. Gluing the boundaries of the balls together could give a different result than reflecting one of them before gluing. In general, even if it does give the same result up to homeomorphism, it would require a global homeomorphism. Although it seems like the orientation should matter, I can't think of a specific example where it definitely does. Also, I can post a proof using the Schoenflies theorem, but its getting a bit late, so I have favourited this question and can come back tomorrow and post an answer if it hasn't already been answered by then. – George Lowther Sep 11 '13 at 23:20
The connected sum of two complex projective planes depends on the orientation of the gluing maps:… – George Lowther Sep 12 '13 at 12:26
You can prove it from the annulus theorem (, which looks like it could be difficult to prove. According to the history, it was proven in different dimensions decades apart. See also…. – George Lowther Sep 12 '13 at 21:43
Yes, in the non-orientable case you don't need to worry about aligning orientations at the join. Also, thinking about this myself a bit, you can consider a special class of manifolds - ones given by charts whose transition maps are stable homeomorphisms (in a local sense). You can prove that you get unique connected sums for these types of manifolds (as long as you match orientations at the joins). The annulus theorem then says that all homeomorphisms of $R^n$ are stable, so these manifolds are the same thing as plain old topological manifolds. – George Lowther Sep 12 '13 at 23:06
up vote 1 down vote accepted

I'll write a self-answer to summarise findings from the comments (mainly by George Lowther).

An assumption previously missing in the question is the one regarding orientation. For non-orientable manifolds it is not important (due to that question or a characterization of non-orientable manifolds as containing $n$-dimensional equivalent of the Möbius strip), but if both manifolds are orientable and do not admit an orientation-inversing automorphism, then depending on using chart maps consistent or inconsistent with the orientation we can obtain two different topological spaces.

The claim is true due to the annulus theorem (rather nontrivial) - which enables a proof similar as in the smooth case. The theorem highly relies on the innocent assumption:

"To avoid problems we can assume that the chart maps can be extended to bigger balls."

A very similar question was asked previously on MO.

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