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Two riemann surfaces $S$ and $R$ are said to be conformally equivalent if there exist a holomorphic map $f:S\rightarrow R$ which is one-one and onto, and inverse is also holomorphic.

I have to show No two of them are conformaly equivalent: $a$) $\hat{\mathbb{C}}$, $b$) $\mathbb{C}$, $c$) $D$(Open Unit Disc)

Ok let $f:a\rightarrow b$ be a bijective holomorphic map and $f^{-1}: b\rightarrow a$ is also holomorphic, well,the contracdiction is $f(\hat{\mathbb{C}})=\mathbb{C}$ as $\mathbb{C}$ is non-compact? for $b\rightarrow c$ riemann mapping theorem can be applied to show the contradiction? and $a\rightarrow c$ is same argument as $a\rightarrow b$?

please help me.

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It depends on how you've stated RMT. Note that a map $\mathbb C \to D$ is bounded. Also, note that the open unit disk is not compact. – Dylan Moreland Jul 23 '12 at 19:33
well, every simply connected domain which is not equal to $\mathbb{C}$ is biholomorphic to $D$, This is the statement I know for RMT, and I understand that the map will be bounded and hence by Lioville's Theroem it will be constant. – Un Chien Andalou Jul 23 '12 at 19:39
up vote 3 down vote accepted

The sphere is the only compact one, so neither of the other two can be a continuous image of it, so cannot be conformally equivalent to it.

$\mathbb{D}$ cannot be the holomorphic image of $\mathbb{C}$ by Liouville's Theorem. (If it were we would have a bounded non constant entire analytic function).

And we're done!

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