Proving the Independence of Connected Sum on Open Discs: A Topological Approach

[Diophantus]

I am trying to show that the connected sum of two topological surfaces does not depend on the open discs removed.

Any hints?

 

[StatusX]

This is obviously only true if the surface is path connected. You want to show that for every such surface, any two discs on the surface, and any homeomorphism between their boundaries, there exists a homeomorphism from the surface to itself which restricts to the given homeomorphism on the boundaries of the discs (do you see why this is what you want to show?). The easiest way I can think to do this is to use a theorem that states every surface is homeomorphic to some polygon with certain edges identified. It'll still be difficult though, as you'll need to account for several different cases, eg, when the disc lies on an edge.

 

[Diophantus]

Yes I should have said path-connected.

You want to show that for every such surface, any two discs on the surface, and any homeomorphism between their boundaries, there exists a homeomorphism from the surface to itself which restricts to the given homeomorphism on the boundaries of the discs (do you see why this is what you want to show?).

I do indeed see why this is equivalent to the statement - we apply this to both surfaces and then apply a 'gluing lemma' to complete the proof.

The easiest way I can think to do this is to use a theorem that states every surface is homeomorphic to some polygon with certain edges identified.

I know a theorem which does this for closed surfaces so I suppose the general one would be the same but would not necessarily involve EVERY edge in our topological polygon to be identified. Is this correct?

It'll still be difficult though, as you'll need to account for several different cases, eg, when the disc lies on an edge.

Counting all cases does indeed sound tricky. I suppose just finding an open set which covers both discs and finding a homeomorphism between the respective cases resticted to this open set would do but I don't know if that makes things any easier.

I certainly have more insight into the problem now anyway, thanks.