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Prove that any two simply connected open set in the plane R^2 are diffeomorphic. I know that in the complex plane any simply connected open set is diffeomorphic to either complex plane or open unit disk.How to adapt the proof here. edited later

Also I need to show that: C considered as a complex-analytic manifold is not isomorphic to the unit disk

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up vote 2 down vote accepted

I will use the fact that any simply connected open set is diffeomorphic to $D$, the complex unit disc. (And clearly $\mathbb{C} \simeq D$).

Given two sets $S$ and $S'$, we have diffeomorphisms $\varphi$ and $\psi'$, from $S$ (resp. $S'$) to $D$. These maps are bijective, and thus invertible. The inverse of a diffeomorphism is again a diffeomorphism, so $\varphi^{-1}$ and $\psi^{-1}$ are diffeomorphisms. We may now take the map $$\varphi \circ \psi^{-1}: S \to D \to S'$$which is again a diffeomorphism, and it is proved.

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is it C≃D? I think it is not.but I think R^2≃D and that is I think one of the necessary things to be shown – Koushik Jan 14 '13 at 16:20
If you're looking at the standard (metric topology), they are the same (homeomorphic). And yes, $\mathbb{C} \simeq D$ by the map $\frac{z}{1-|z|^2} : D \to \mathbb{C}$. – andybenji Jan 14 '13 at 16:26
You may look at $\mathbb{C}$ as $(\mathbb{R}^2, + , *)$ where $(a,b)*(c,d) = (ac-bd, ad + bc)$. – andybenji Jan 14 '13 at 16:28
@K.Ghosh: what does Liouville's theorem tell you about an an analytic map $\mathbb{C} \to \mathbb{D}$? – Martin Jan 14 '13 at 17:04
@K.Ghosh: Yes, such a map must be constant. Be careful: andybenji showed you in the first comment how $\mathbb{C}$ and $D$ are diffeomorphic (in particular homeomorphic). What Liouville tells you is that there can be no analytic isomorphism because every analytic map $\mathbb{C}\to D$ is constant. – Martin Jan 14 '13 at 20:38

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