TITLE: The inverse of a bijective holomorphic function is also holomorphic
QUESTION [23 upvotes]: I'm confused about the following proposition
Proposition. Let $U,V$ are open sets in $\mathbf{C}$. If $f:U\to V$ is holomorphic and bijective, then  the inverse  $f^{-1}:V\to U$ is also holomorphic.
The proof of the proposition think that the  continuity of $f^{-1}$ is obvious, but I find it is really difficult to prove  using $\epsilon-\delta$ definition. Can anyone give some hints?

REPLY [2 votes]: Here is an approach using differential forms. Once you establish $f'$ has no roots and is therefore a local diffeomorphism, and hence a diffeomorphism $U\to V$, put $w = f(z)$ and $g(w) = f^{-1}(w)$. Then if $\tilde\gamma = f\circ \gamma$ is a closed, piecewise $C^1$ curve,
\begin{align*}
\oint_{\tilde\gamma} g(w)\,dw &= \oint_\gamma f^*(g\,dw)\\
&= \oint_\gamma z\,d(fz) \\
&= \oint_\gamma z\,\big(\partial_z f\,dz + \underbrace{\partial_{\bar z}f}_{=\  0\ \text{because $f$ is holomorphic}}\,d\bar z\big) \\
&= \oint_\gamma z\,\partial_zf\,dz = 0,
\end{align*}
where the last integral is $0$ because $f$ is holomorphic. Hence $g$ is holomorphic by Morera's theorem.