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Can we completely classify the simply-connected surfaces (with or without boundary) in $\mathbb R^3$ up to homeomorphism?

Summer
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    Relevant (and almost a duplicate): http://math.stackexchange.com/questions/5588/classification-theorem-for-non-compact-2-manifolds-2-manifolds-with-boundary – Jesse Madnick Nov 26 '11 at 01:35
  • It looks like this question was never properly answered. Part of the problem is that there is a continuum of such surfaces (up to a homeomorphism). However, if you take a classification of closed subsets of the unit circle for granted, then there is a rather clear answer: These surfaces are homeomorphic to the closed disk with a compact subset $C$ of the boundary circle removed, where $C$ is either the full circle or a subset not containing any nondegenerate intervals. – Moishe Kohan Jan 09 '23 at 16:29

2 Answers2

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The classification of surfaces is well-known. Every surface is homeomorphic to either $mT+nD$ or $mP+nD$ where $T$ is a torus, $P$ is a projective plane, $D$ is a disk, $m,n$ are non-negative integers, and $+$ is connected sum.

Now you want the surfaces to be in ${\bf R}^3$, so that rules out $mP$ for $m\ge1$.

You also want them simply-connected. That rules out $mT+nD$ for $m\ge1$ and $mP+nD$ for $m\ge1$ and $nD$ for $n\ge2$, leaving only two surfaces: the sphere, and the disk.

Gerry Myerson
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    This is only for compact surfaces. What about noncompact ones? – Damian Sobota Nov 25 '11 at 23:32
  • Ah, I wondered if that was what OP was getting at. Those are beyond me. See Ian Richards, On the classification of noncompact surfaces, Trans. Amer. Math. Soc. 106 (1963) 259-269. There is some discussion of the Richards result at http://mathoverflow.net/questions/4155/classification-problem-for-non-compact-manifolds – Gerry Myerson Nov 26 '11 at 00:08
  • The OP might also want surfaces with boundary. – Jesse Madnick Nov 26 '11 at 01:34
  • @JesseMadnick, compact surfaces with boundary are already taken care of - the $D$ terms in the connected sums are the boundaries. Noncompact surfaces with boundary I leave to those better informed than myself. – Gerry Myerson Nov 26 '11 at 01:59
  • @GerryMyerson Ah, right, sorry, it didn't occur to me that $D$ is a closed disk. – Jesse Madnick Nov 26 '11 at 03:06
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For noncompact surfaces --- with additional assumptions that the boundary is empty and the surface is connected --- the answer is the similar to the one given for compact surfaces in the answer of @GerryMyerson, namely the only possibility is the open disc also known as $\mathbb{R}^2$.

To see why one needs a bit of the classification theory for noncompact surfaces. The surface may contain neither a Möbius band nor a punctured torus, for then it would not be simply connected. It follows from the classification theory that the surface embeds in $\mathbb{R}^2$. Nor may it have more than one end (in the sense of Freudenthal), for then there would be an embedded circle separating the two ends from each other, and again it would not be simply connected. It remains to know that every planar, one-ended surface is homeomorphic to $\mathbb{R}^2$.

Lee Mosher
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