The inverse of a bijective holomorphic function is also holomorphic 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?
 A: Hint: Prove that $f^{\prime}(z)\neq 0$ for all $z\in U$. Then use the inverse function theorem for analytic functions.
A: Step 1: $f'$ is never zero.
Indeed, if $f'(a)=0$ for some $a\in U$, then the Taylor expansion at $a$ is of the form $f(z)=f(a) + c_n (z-a)^n+\dots$ with $n\ge 2$, $c_n\ne 0$. This implies that $g(z) = (f(z)-f(a))/(z-a)^n$ is a nonzero holomorphic function near $a$, hence admits an $n$th degree root (a function $h$ such that $h^n=g$). Hence $$f(z) = f(a) +  [(z-a)h(z)]^n$$
Since $z\mapsto (z-a) h(z)  $ is an open map, its image contains a neighborhood of $0$; in particular it contains the points $\epsilon$ and $\epsilon \exp(2\pi i /n)$ for small $\epsilon$. These two points are sent into one, contradicting  the injectivity of $f$. 
Step 2: Inverse is smooth
This is just the Inverse function theorem: writing out $f=u+iv$, one can see that the Jacobian determinant of $(x,y)\mapsto (u,v)$ is $|f'(z)|^2\ne 0$. 
Step 3: Inverse is holomorphic
Also the Inverse function theorem. Writing the derivative of $f$ as a $2\times 2$ real matrix, we get something of the form
$$\begin{pmatrix} a & b \\ -b& a\end{pmatrix}$$
due to the Cauchy-Riemann equations. The inverse of such a matrix is also of this form: hence, $f^{-1}$ satisfies the Cauchy-Riemann equations. 
A: 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.
A: Another proof: 
Without loss of generality we may assume that $U$ is connected. Obviously $f'$ does not vanish identically on $U$, so the set $Z$ of zeros of $f'$ is closed and discrete in $U$. By the open mapping theorem, $f:U\to V$ is a homeomorphism, so $f(Z)$ must also be closed and discrete. It is easy to check that $g=f^{-1}$ is holomorphic on $V\setminus f(Z)$, and the Riemann's theorem on removable singularities now shows that $g$ must be holomorphic on all of $V$.
