7.5 The open mapping theorem and the inverse function theorem  (Page 11/2)

We turn next to a question about functions of a complex variable that is related to [link] , the Inverse Function Theorem.That result asserts, subject to a couple of hypotheses, that the inverse of a one-to-one differentiable function of a real variable is also differentiable.Since a function is only differentiable at points in the interior of its domain, it is necessary to verify that the point $f\left(c\right)$ is in the interior of the domain $f\left(S\right)$ of the inverse function $f^{-1}$ before the question of differentiability at that point can be addressed. And, the peculiar thing is that it is this point about $f\left(c\right)$ being in the interior of $f\left(S\right)$ that is the subtle part. The fact that the inverse function is differentiable there, andhas the prescribed form, is then only a careful $ϵ-\delta$ argument. For continuous real-valued functions of real variables, the fact that $f\left(c\right)$ belongs to the interior of $f\left(S\right)$ boils down to the fact that intervals get mapped onto intervals by continuous functions,which is basically a consequence of the Intermediate Value Theorem. However, for complex-valued functions of complex variables, thesituation is much deeper. For instance, the continuous image of a disk is just not always another disk, and it may not even be an open set.Well, all is not lost; we just have to work a little harder.

Open mapping theorem

Let $S$ be a piecewise smooth geometric set, and write $U$ for the (open) interior $S^0$ of $S.$ Suppose $f$ is a nonconstant differentiable, complex-valued function on the set $U.$ Then the range $f\left(U\right)$ of $f$ is an open subset of $C.$

Let $c$ be in $U.$ Because $f$ is not a constant function, there must exist an $r>0$ such that $f\left(c\right)\ne f\left(z\right)$ for all $z$ on the boundary $C_r$ of the disk $B_r\left(c\right).$ See part (b) of [link] . Let $z_0$ be a point in the compact set $C_r$ at which the continuous real-valued function $\left|f\right(z)-f(c\left)\right|$ attains its minimum value $s.$ Since $f\left(z\right)\ne f\left(c\right)$ for any $z\in C_r,$ we must have that $s>0.$ We claim that the disk $B_{s/2}\left(f\left(c\right)\right)$ belongs to the range $f\left(U\right)$ of $f.$ This will show that the point $f\left(c\right)$ belongs to the ihnterior of the set $f\left(U\right),$ and that will finish the proof.