Transition of complex charts are biholomorphic [Riemann Surfaces, Miranda Exercise]

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I'm reading Rick Miranda, Algebraic Curves and Riemann Surfaces. I'm having problems with this exercise about complex charts.

Let $\phi_i: U_i \to V_i$, $i=1,2$, be complex charts on $X$ with $U_1\cap U_2 \ne \emptyset$. Suppose that $\phi_2 \circ \phi_1^{-1}: \phi_1 \left(U_1 \cap U_2 \right) \to \phi_2\left(U_1 \cap U_2 \right)$ is holomorphic. Show that it is bijective with inverse $\phi_1 \circ \phi_2^{-1}: \phi_2 \left(U_1 \cap U_2 \right) \to \phi_1\left(U_1 \cap U_2 \right)$.

It's easy to prove that the map $\phi_2 \circ \phi_1^{-1}$ is bijective with inverse $\phi_1 \circ \phi_2^{-1}$ (over the sets $\phi_i \left( U_1 \cap U_2 \right)$. But I don't know how to prove that this inverse is also holomorphic. Maybe a holomorphism which is also a homeomorphism it's biholomorphism but I'm not sure.

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Yes, this follows from the holomorphic inverse function theorem (which should not be too hard to prove if you know how to prove the regular inverse function theorem.)