Open Mapping Theorem (complex Analysis) - Proof

Proof

Assume f : U → C is a non-constant holomorphic function and U is a domain of the complex plane. We have to show that every point in f(U) is an interior point of f(U), i.e. that every point in f(U) is contained in a disk which is contained in f(U).

Consider an arbitrary w₀ in f(U). Then there exists a point z₀ in U such that w₀ = f(z₀). Since U is open, we can find d > 0 such that the closed disk B around z₀ with radius d is fully contained in U. Consider the function g(z) = f(z)−w₀. Note that z₀ is a root of the function.

We know that g(z) is not constant and holomorphic. The roots of g are isolated and by further decreasing the radius of the image disk d, we can assure that g(z) has only a single root in B (although this single root may have multiplicity greater than 1).

The boundary of B is a circle and hence a compact set, on which |g(z)| is a positive continuous function, so the extreme value theorem guarantees the existence of a positive minimum e, that is, e is the minimum of |g(z)| for z on the boundary of B and e > 0.

Denote by D the open disk around w₀ with radius e. By Rouché's theorem, the function g(z) = f(z)−w₀ will have the same number of roots (counted with multiplicity) in B as h(z):=f(z)−w₁ for any w₁ in D. This is because h(z) = g(z) + (w₀ - w₁), and for z on the boundary of B, |g(z)| ≥ e > |w₀ - w₁|. Thus, for every w₁ in D, there exists at least one z₁ in B such that f(z₁) = w₁. This means that the disk D is contained in f(B).

The image of the ball B, f(B) is a subset of the image of U, f(U). Thus w₀ is an interior point of f(U). Since w₀ was arbitrary in f(U) we know that f(U) is open. Since U was arbitrary, the function f is open.