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.

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