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How does the classification of closed (compact, boundaryless) surfaces imply the classification of all orientable not-necessarily-compact surfaces with boundary? It seems to be that they are all obtained by deleting points and the interiors of closed disks from closed surfaces. This gives three indices: (genus, number of disks deleted, number of points deleted). Why is this exhaustive? Why is there only one such triple for every surface? Clearly the second index is unique (it determines the number of boundary components), but how do you know you can't change both the first and third and wind up with the same space?

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  • $\begingroup$ Since you don't require compactness, you could start with a closed surface and remove a Cantor set. $\endgroup$ – Andreas Blass Aug 4 '13 at 4:08
  • $\begingroup$ Very relevant: math.stackexchange.com/questions/5588/… $\endgroup$ – Jesse Madnick Aug 4 '13 at 4:28
  • $\begingroup$ @AndreasBlass : excellent point! $\endgroup$ – Nate Aug 4 '13 at 6:37
  • $\begingroup$ @Jesse Madnick : Yes, sorry for the duplicate. Don't know why it didnt show up in the "related posts" thing $\endgroup$ – Nate Aug 4 '13 at 6:38
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It is a long way from classification of closed 2-dimensional manifolds to the classification of all (connected) 2-dimensional manifolds (possibly with boundary). This was accomplished by E. Brown and R. Messer, "The classification of two-dimensional manifolds", Trans. Amer. Math. Soc., vol. 255 (1979), 377–402, about 100 years after closed surfaces were classified.

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