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Is there always a diffeomorphism between $(0,1)^2$ and any given (not degenerate) triangle?

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    *Diffeo*-morphism?? I would doubt. At maximum between the open *inner* part of the square and the triangle.2012-11-28
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    I don't understand your second sentence.2012-11-28
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    @Berci is referring to the fact that [every convex open subsets of the Euclidean plane is diffeomorphic to the unit open ball](http://mathoverflow.net/questions/4468/what-are-the-open-subsets-of-mathbbrn-that-are-diffeomorphic-to-mathbb/4516#4516).2012-11-28
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    @Berci Sorry, the second sentence didn't parse. Now it makes sense.2012-11-28
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    @WillieWong Thanks. So if I take $(0,1)^2$ then I'm good? (Set of measure zero is nothing to me)2012-11-28
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    Question changed, somebody go for the answer.2012-11-28

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If by "triangle" you mean the open set bounded by three line segments (the boundaries themselves are not included), then yes, every convex open subset of the Euclidean plane is diffeomorphic to $\mathbb{R}^2$.

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    In fact every simply connected open subset of the Euclidean plane is diffeomorphic to $\mathbb R^2$.2012-11-28
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    @Jacob: I see two ways to conclude that, one of which involves the Riemannian mapping theorem. Do you have in mind a easier proof?2012-11-28
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    I had the Riemannian mapping theorem in mind.2012-11-28