# Open mapping theorem

Today we’ll prove the open mapping theorem:

Theorem 1 (Open mapping Theorem) Let ${f\in{\mathcal H}(U)}$, for some open set ${U}$ in ${{\mathbb C}}$. Then ${f(U)}$, the set of all possible images of ${f}$ is either constant, or is open in ${{\mathbb C}}$.

Let ${z_0\in f(U)}$, and let ${w_0}$ be a preimage. That is, ${f(w_0)=z_0}$. Since ${U}$ is open, there must be a small closed ball around ${w_0}$ completely contained in ${U}$. Let ${r}$ be it’s radius. Then ${\overline{B_r(w_0)}\subseteq U}$. Let ${g(z)=f(z)-z_0}$. What do we know about ${g}$?

We know that the roots of ${g}$ are isolated points. After all, if not, we would have a convergent sequence of roots, contradiction one of the theorems we proved here. So we can choose ${r}$ to be even smaller so that the only root of ${g}$ in ${\overline{B_r(w_0)}}$ is ${w_0}$. Since the boundary of ${\overline{B_r(w_0)}}$ is a circle, and hence compact, and ${|g(z)|}$ is continuous it attains its minimum when restricted to the boundary. Let ${m}$ be the minimum value of ${|g(z)|}$ for ${z}$ on the boundary of ${\overline{B_r(w_0)}}$.

Now, by Rouché’s theorem, ${g}$ will have the same number of roots in in ${B_r(w_0)}$ as ${f(z)-z_1}$ for any ${z_1}$ in ${B_m(z_0)}$. (Application of Rouché’s theorem is left as an exercise.)

In particular, this means that for every complex number in ${z_1\in B_m(z_0)}$, the function ${f(z)-z_1}$ has at least one root in ${B_r(w_0)}$, and so ${B_m(z_0)}$ is contained in ${f(U)}$, meaning our arbitrarily chosen point ${z_0}$ is in the interior of ${f(U)}$. Thus ${f(U)}$ is open.