I am having trouble trying to prove the following:

Given f an holomorphic function from an open $U$ to $\mathbb{C}$, we choose $a\in U$. Then:

1) If $f'(a) \neq 0$ then it exists a nighborhood $U_1$ of $a$ suche that $f|_{U_1}$ is injective;

2)If $f$ is injective and $f'(z) \neq 0 \forall z \in U$, then $f(U)$ is open and the inverse $f^{-1}$ is also holomorphic.

I have been told that this should follow quite simply from the implicit function theorem in $\mathbb{R^2}$ and the Cauchy-Riemann condition. I tried to look for a proof in the Cartan, but it looks like there's not. Thanks for the help.


2 Answers 2


Let $f$ be holomorphic on $U \subseteq \mathbb{C}$. We define, $F(z,w)$, $\forall \ z, w \in U $, as follows :

\begin{align} F(z,w) &= \frac{f(z)-f(w)}{z-w} & & z \neq w \\ &= f'(z) & & z = w \\ \end{align}

It's easy to see that $F$ is a continuous function on $U\times U$.

Next, notice that $|F(z,w)|> 0 $, for some neighborhood $ W\subseteq U$ of $a$, since $|f'(a)| \ne 0$.

This is because $F(a,a)= f'(a)$ and $F$ is continuous therefore, $\exists \ W \subseteq U$ an open neighborhood of $a$ such that $|F(z,w) - f'(a)|< \frac{1}{2}|f'(a)|,\ \forall \ z,w \in W$.

And therefore,

\begin{align} |F(z,w)| &> \frac{1}{2}|f'(a)| && (\Delta \ \text{inequality})\\ \implies |f(z)-f(w)| &> \frac{1}{2}|f'(a)||z-w|> 0 & & (\ \forall \ z,w \in W, z \ne w)\\ \end{align}

Thus we have proved that $f$ is one-to-one in a neighborhood of $a$.

To see that $f$ is an open map, we must prove that for every $x\in W$, $\exists \ \epsilon >0 $ such that $D(f(x), \epsilon) \subseteq f(W)$. Suppose not, then for every $\epsilon > 0$, $\exists \ y \in D(f(x), \epsilon)$ such that $y\ne f(z)$ for any $z\in W$.

From what we have already observed, for every $x\in W$, $|f(x) - f(z)| > \frac{1}{2}|f'(a)||x-z|$, in particular if $ \overline{D(x. \delta)} \subseteq W$, then $|f(x+\delta e^{i\theta}) - f(x)|> \frac{1}{2} \delta|f'(a)|$.

Pick $\epsilon < \frac{1}{4} \delta|f'(a)|$.

Then $\exists \ y$ as before, i.e $|y-f(x)|< \epsilon$ but $\nexists \ z \in W$ such that $y= f(z)$. Therefore, the function $g(z) = \frac{1}{y-f(z)}$ is holomorphic. We claim $|g(z)|$ attains its maximum at an interior point on $\overline{D(x,\delta )}$.

Since, $|y-f(x+ \delta e^{i\theta})| \ge |f(x+\delta e^{i\theta}) - f(x)| - |y-f(x)| > 2\epsilon -\epsilon = \epsilon > |y- f(x)|$.

Thus $|g(x)| > |g(x+\delta e^{i\theta})|$, $\forall \ \theta \in [0, 2\pi]$, contradicting the maximum modulus principle for the holomorphic function $g(z)$.

Having established that $f$ is a local homeomorphism, differentiability is actually straightforward. If h is a local inverse of f, i.e $h: f(W)\rightarrow W $ is the inverse to the map $f|_W$, then

\begin{align} \frac{h(z)-h(z_0)}{z-z_0} = \frac{z-z_0}{f(z)-f(z_0)} \ \ & (\ \forall \ z,z_0 \in f(W), z \ne z_0)\\ \end{align} Therefore,

\begin{align} h'(z_0) = \lim_{z\to z_0}\frac{h(z)-h(z_0)}{z-z_0} = \lim_{z\to z_0}\frac{z-z_0}{f(z)-f(z_0)} = \frac{1}{f'(z_0)} \end{align}

$\textbf{Ps}$: The proof is sourced from Walter Rudin's $\textbf{Real and Complex Analysis}$. In Rudin's sequence Maximum modulus is proved using Parseval's formula. It should be possible to reverse these arguments to obtain a proof of Maximum Modulus using Open mapping, while deriving the Open mapping theorem from Inverse Function theorem for $\mathbb{R}^2$, as observed by the previous commentator.

  • $\begingroup$ Thanks for the answer, I also have Rudin's book you mentioned and I'll surely look there if I have some other doubts $\endgroup$
    – tommy1996q
    Commented May 22, 2017 at 15:27

We're using the Inverse Function Theorem, not the Implicit Function Theorem! And the crucial thing you may have missed is that the Inverse Function Theorem not only provides the existence of a local inverse function, but also provides a very useful formula for the derivative of this inverse function.

So, we started with a holomorphic function $$ f \ : \ x + iy \mapsto f(x+iy) = u(x,y) + iv(x,y),$$ Since $f'(x+iy)$ is non-zero, and since $f$ obeys the Cauchy-Riemann equations $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \ \ \ \ \ \ \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \ \ \ \ \ \ (\ast) $$ we can easily verify that the Jacobian matrix $$ \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} $$ has non-zero determinant.

So we may apply the Inverse Function Theorem. This gives an inverse function $$ f^{-1} \ : \ u + iv \mapsto f^{-1}(u+iv) = x(u,v) + i y(u,v),$$ mapping from $f(U_1)$ to $U_1$, where $U_1$ is a sufficiently small open neighbourhood of $a$ inside $U$. This already addresses point (1) in the claim. Please forgive me for my abuse of notation: I have used the symbols $x,y,u,v$ both as coordinates and as functions, though I do find this notation quite helpful.

Our ultimate goal is to verify the Cauchy-Riemann equations for $f^{-1}$. These state that: $$ \frac{\partial x}{\partial u} = \frac{\partial y}{\partial v}, \ \ \ \ \ \ \frac{\partial x}{\partial v} = - \frac{\partial y}{\partial u} \ \ \ \ \ \ (\ast \ast) $$ To do this, we must somehow make use of the Cauchy-Riemann equations for the original holomorphic function $f$, given in $(\ast)$.

But the Inverse Function Theorem gives a relationship between our two sets of partial derivatives! It says that the Jacobian matrix for $(x(u,v), y(u,v))$ is the inverse of the Jacobian matrix for $(u(x,y), v(x,y))$: $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}\Bigg\vert_{(u,v)} = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix}^{-1}\Bigg\vert_{(x(u,v),y(u,v))} = \frac{1}{\frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}} \begin{bmatrix} \frac{\partial v}{\partial y} & -\frac{\partial u}{\partial y} \\ -\frac{\partial v}{\partial x} & \frac{\partial u}{\partial x} \end{bmatrix}\Bigg\vert_{(x(u,v),y(u,v))} $$

In view of the Cauchy-Riemann equations for $f$, equation $(\ast)$, this reduces to $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}\Bigg\vert_{(u,v)} =\frac{1}{\left(\frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y}\right)^2} \begin{bmatrix} \frac{\partial u}{\partial x} & -\frac{\partial u}{\partial y} \\ \frac{\partial u}{\partial y} & \frac{\partial u}{\partial x} \end{bmatrix}\Bigg\vert_{(x(u,v),y(u,v))} $$ From this, one can easily see that the Cauchy-Riemann equations for $f^{-1}$, equation $(\ast \ast)$, are satisfied too. This proves part (2) of the claim.

  • $\begingroup$ thanks! I think I've understood almost everything now! In fact I didn't understand how we could use the implicit function theorem in such a situation, I guess I simply got it wrong while writing. So (correct me if I am wrong) you look at $\mathbb{C}$ just like $\mathbb{R}^2$ with Cauchy-Riemann equations for differentiable functions in $\mathbb{C}$. By using them, you can see that the jacobian of $f$ is non-zero, then I have an inverse which is differentiable in $\mathbb{R}^2$ for the inverse function theorem, but we still have to verify it satisfies Cauchy-Riemann. How ca I do that? $\endgroup$
    – tommy1996q
    Commented May 20, 2017 at 19:57
  • $\begingroup$ @tommy1996 You're absolutely right about thinking of $\mathbb C$ as $\mathbb R^2$ etc. That is exactly the idea. I tried to explain the derivation of Cauchy-Riemann for $f^{-1}$ in the second half of my answer. I have now edited my answer to include more steps, so hopefully you can follow it now. $\endgroup$
    – Kenny Wong
    Commented May 20, 2017 at 20:13
  • $\begingroup$ ok I see it now, thank you very much! One last thing, though, just to be absolutely sure. When you use the notation $|_{(u,v)}$ for instance at the right of a matrix, you mean that you consider $(u,v)$ to be the variables right? 'Cause I only use that when evaluating things (integrals most of the times), but seeing what you wrote I am quite sure you mean that they are to be considered variables. Am I right? $\endgroup$
    – tommy1996q
    Commented May 21, 2017 at 9:33
  • $\begingroup$ @tommy1996q Oh I see, that can be confusing. I mean that if the matrix on the left-hand side is evaluated at some $(u,v)$, then the matrix on the right-hand side should be evaluated at $(x,y) = (x(u,v), y(u,v))$. $\endgroup$
    – Kenny Wong
    Commented May 21, 2017 at 11:01

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