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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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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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