Inverse of holomorphic map on Riemann surface is holomorphic?

Question

Let $X,Y$ be Riemann surfaces and $f:X→Y$ is holomorphic map between $X$ and $Y$ and $f$ is bijective.

Then, could you tell me the proof of the inverse of $f$ is holomorphic map?

I proved this in special cases but I'm not sure where this holds in general or not.

Does this answer your question? The inverse of a bijective holomorphic function is also holomorphic (The linked post concerns only when $X, Y$ are open sets, but that's exactly what you need, as holomorphicity is a local property) — Arctic Char, Jul 01 '21 at 16:49

score 1 · Accepted Answer · answered Jul 01 '21 at 16:47

Pick $x \in X,$ and say $y = f(x).$ Let $(\phi, U)$ be a chart at $x$ in $X$ and $(\psi, V)$ a chart at $y$ in $Y.$ Then, since $f$ is holomorphic, we know that $\psi \circ f \circ \phi^{-1}$ holomorphic in the typical sense, since it's domain and codomain are both open subsets of $\mathbb{C}.$ Hence, it's inverse $\phi \circ f^{-1} \circ \psi^{-1}$ is holomorphic, by standard complex analysis in the plane (since it is a composition of bijections, it is still invertible). But then, by definition of holomorphic functions on Riemann surfaces, $f^{-1}$ will be holomorphic at $y.$ Repeat for every $x \in X$ and you find $f^{-1}$ is holomorphic on all of $Y.$

Inverse of holomorphic map on Riemann surface is holomorphic?

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