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I have read this statement in several places: "Two surfaces are homeomorphic iff are diffeomorphic". I think the nontrivial implication follows in this manner: First, we triangulate the surface and then we smooth the resulting piecewise linear manifold, but I haven't found and well-written proof. Could you possibly give me a reference for the proof? Thanks in advance.

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What about a square and a sphere? Or you assume that both surfaces are smooth? –  Ilya Jan 31 '13 at 19:04
Could you give a reference where you've read it? This page has some references for something similar to your question. –  anon271828 Jan 31 '13 at 19:11
See this MathOverflow question, which gives a list of references. –  David Moews Jan 31 '13 at 19:23
Ilya: Sorry, you're right, I assume smooth surfaces :) –  math_failure Jan 31 '13 at 23:49
anon271828, I just read that in lecture notes on the net and google books without proof, because of that I need an strong reference –  math_failure Jan 31 '13 at 23:51

1 Answer 1

This is clearly false. Consider the two-dimensional case. For each positive integer $k$ we have a cusp given by $\gamma : (I,0) \to (\mathbb{R}^2,0)$ where $t \mapsto (t^2,t^{2k+1})$. Each one of these cusps is distinct up to diffeomorphism, i.e. there is no local diffeomorphim $\psi : (\mathbb{R}^2,0) \to (\mathbb{R}^2,0)$, which carries the image of one onto another. However, all of the images are homeomorphic to an open interval.

The space of diffeomorphisms is a proper subset of the space of homeomorphisms.

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Presumably when OP wrote "surface", s/he meant "smooth surface". It is true (but not obvious) that two smooth surfaces are homeomorphic if and only if they are diffeomorphic. –  Paul Siegel Jan 31 '13 at 19:59
@PaulSiegel Perhaps. But I can only answer the question before me... –  Fly by Night Jan 31 '13 at 20:12
@PaulSiegel indeed, thank you for clarifying my point :D –  math_failure Jan 31 '13 at 23:54
@FlybyNight, sorry, I made a wrong statement, but thanks anyway... –  math_failure Jan 31 '13 at 23:54
I don't understand: your cusps are not surfaces and your argument about diffeomorphisms not carrying one cusp to another doesn't seem to address the question, which is whether homeomorphic surfaces are diffeomorphic. What are your homeomorphic but not diffeomorphic surfaces? –  Georges Elencwajg Sep 30 '13 at 8:47

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