The Open Mapping Theorem — Mathematics Knowledge System

Overview

The open mapping theorem states that a non-constant holomorphic function on a connected open domain sends open sets to open sets: if f is holomorphic and non-constant on a domain Ω, then for every open U ⊆ Ω, f(U) is open in ℂ. This is sharply different from smooth real functions, where a non-constant smooth map (like f(x) = x², sending an open interval around 0 to a half-open interval) can easily fail to be open. The theorem follows from a local counting argument: near any point z₀, a non-constant holomorphic f behaves like f(z) = f(z₀) + c(z − z₀)^k + higher order terms for some k ≥ 1 (its local degree), and this k-to-1 behavior near z₀ means f actually covers a whole neighborhood of f(z₀), not just the single point or a boundary sliver. The open mapping theorem is the direct route to a one-line proof of the maximum modulus principle (an open image can't have |f| maximized at an interior point, since nearby image points would have to include ones of strictly larger modulus), and it underlies the inverse function theorem for holomorphic maps at points where f' ≠ 0.

Intuition

Zoom in near a point z₀ where a non-constant holomorphic function takes the value w₀ = f(z₀). Subtracting w₀, the function f(z) − w₀ has an isolated zero at z₀ of some finite order k ≥ 1 (finite because f is non-constant, so its Taylor series around z₀ isn't identically zero). Near z₀, f(z) − w₀ behaves essentially like c(z−z₀)^k, and raising to the k-th power is itself an open map — z ↦ z^k sends every small disk around 0 onto a full small disk around 0 (traversed k times), not just an arc or a single ray. So f, being locally like this k-th power map (up to a change of coordinates given by the holomorphic k-th root, which exists locally), also sends small disks around z₀ onto full small disks around w₀. Since this happens at every point, the image of any open set is a union of open disks, hence open. This is the precise sense in which holomorphic maps 'spread out' rather than 'flatten' near a point — the total absence of any flattening (any k=0 behavior) is guaranteed by f being non-constant.

Formal Definition

Definition

Let f be holomorphic and non-constant on a connected open domain Ω ⊆ ℂ:

U \subseteq \Omega \text{ open} \;\Longrightarrow\; f(U) \text{ is open in } \mathbb{C}

Open mapping theorem

f(z) = f(z_0) + c_k(z-z_0)^k + c_{k+1}(z-z_0)^{k+1} + \cdots, \qquad c_k \ne 0,\ k\ge 1

Local normal form near z_0 (k = order of the zero of f - f(z_0))

f'(z_0) \ne 0 \;\Longrightarrow\; f \text{ is a local biholomorphism (conformal bijection) near } z_0

Local invertibility when the local degree k=1

Worked Examples

Near z₀=0, f(z)-f(0) = z² has a zero of order k=2 at 0 (c_2=1, no lower-order terms).
$f(z) - f(0) = z^2, \qquad k=2$
The map z ↦ z² sends the disk |z|<r onto the disk |w|<r², since every w with |w|<r² has a square root of modulus √|w|<r.
$\{|z|<r\} \xrightarrow{z^2} \{|w|<r^2\}$
The image {|w|<r²} is indeed an open disk, confirming the open mapping theorem's conclusion in this example, even though z↦z² is 2-to-1 (not injective) on the punctured disk.
$\{|w| < r^2\} \text{ is open in } \mathbb{C}$

Answer: Yes: f(z)=z² sends the open disk |z|<r onto the open disk |w|<r², confirming the theorem even though the map is 2-to-1 near 0 (local degree k=2).

Practice Problems

Difficulty 5/10

What is the local degree k of f(z) = z³ + 5 at z₀ = 0 (i.e., the order of the zero of f(z) - f(0) at z=0)?

Difficulty 6/10

For f(z) = e^z, what is the local degree k at any point z₀ (i.e., the order of the zero of f(z)-f(z₀) at z=z₀), and what does the open mapping theorem's proof say about local invertibility there?

Difficulty 8/10

A student claims that g(x) = x² (as a map ℝ→ℝ) is a counterexample to 'holomorphic-like' open mapping behavior, since g sends the open interval (-1,1) to the half-open-looking image [0,1). Explain why this does not contradict the open mapping theorem, and identify precisely why real smooth maps can fail to be open while non-constant holomorphic maps cannot.

Quiz

Summary

The open mapping theorem: a non-constant holomorphic function on a domain sends open sets to open sets.
Locally near z₀, f behaves like f(z₀) + c(z-z₀)^k + higher order terms, and this k-to-1 covering behavior (like z↦z^k sending disks to disks) is the mechanism behind the theorem.
When the local degree k=1 (equivalently f'(z₀)≠0), f is a local biholomorphism — a conformal bijection near that point.
The open mapping theorem gives a one-line proof of the maximum modulus principle: an interior local max of |f| would force the open image to contain points of strictly larger modulus, a contradiction.

References

WebsiteWikipedia — Open mapping theorem (complex analysis)