Can we completely classify the simply-connected surfaces (with or without boundary) in $\mathbb R^3$ up to homeomorphism?
The classification of surfaces
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0Relevant (and almost a duplicate): http://math.stackexchange.com/questions/5588/classification-theorem-for-non-compact-2-manifolds-2-manifolds-with-boundary – 2011-11-26
2 Answers
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.
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0This is only for compact surfaces. What about noncompact ones? – 2011-11-25
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0Ah, 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 – 2011-11-26
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0The OP might also want surfaces _with_ boundary. – 2011-11-26
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0@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. – 2011-11-26
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0@GerryMyerson Ah, right, sorry, it didn't occur to me that $D$ is a _closed_ disk. – 2011-11-26
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 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 $\mathbb{R}^2$.