24 - Inverse Function Theorem

Inverse Function Theorem

Suppose $f:U\to \mathbb{C}$ is injective and holomorphic and $f^{\prime }(z)\ne 0$ for all $z\in U$

If $g:f(U)\to U$ is the inverse of $f$ then $g$ is holomorphic with

$$g^{\prime }(w)=\frac{1}{f^{\prime }(g(w))}$$

Proof

By the Open Mapping Theorem| then $g$ is continuous and
If $V$ is open in $f(U)$ then $g^{-1}(V)=f(V)$ is open by the theorem

Fix $w_0\in f(U)$
Let $z_0=g(w_0)$ then since $f$ and $g$ are continuous then

$$w-w_0⟹z=g(w)\to z_0$$

Thus

$$\underset{w\to w_0}{\lim }\frac{g(w)-g(w_0)}{w-w_0}=\underset{z\to z_0}{\lim }\frac{z-z_0}{f(z)-f(z_0)}=\frac{1}{f^{\prime }(z_0)}$$