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Let $X$ be a Riemann surface with complex structure $\{(U_i,\phi_i)\}$. Is it the case that $\phi_i:U_i\rightarrow V_i$ is a biholomorphic map in the sense of Riemann surfaces?

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up vote 1 down vote accepted

The answer is obviously yes. Indeed we only need check that $\phi_i^{-1}\phi_i:V_i\rightarrow V_i$ is holomorphic in the classical sense, which it trivially is.

This is however a useful observation, since it means that the maps we get in the local representation theorem are genuinely biholomorphic, which is perhaps not obvious when you first see them.

Documenting this has certainly helped my understanding - I hope it might do the same for someone else!

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