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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. –  Berci Nov 28 '12 at 16:42
    
I don't understand your second sentence. –  dcs24 Nov 28 '12 at 16:43
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@Berci Sorry, the second sentence didn't parse. Now it makes sense. –  dcs24 Nov 28 '12 at 16:55
    
@WillieWong Thanks. So if I take $(0,1)^2$ then I'm good? (Set of measure zero is nothing to me) –  dcs24 Nov 28 '12 at 16:56
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up vote 5 down vote accepted

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 $\mathbb R^2$. –  JSchlather Nov 28 '12 at 17:01
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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? –  Willie Wong Nov 28 '12 at 17:07
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I had the Riemannian mapping theorem in mind. –  JSchlather Nov 28 '12 at 18:56
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