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

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Relevant (and almost a duplicate): math.stackexchange.com/questions/5588/… –  Jesse Madnick Nov 26 '11 at 1:35
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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.

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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 mathoverflow.net/questions/4155/… –  Gerry Myerson Nov 26 '11 at 0:08
    
The OP might also want surfaces with boundary. –  Jesse Madnick Nov 26 '11 at 1:34
    
@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 1:59
    
@GerryMyerson Ah, right, sorry, it didn't occur to me that $D$ is a closed disk. –  Jesse Madnick Nov 26 '11 at 3:06
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