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?
Best Answer
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.