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.