CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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An injective holomorphic map has no critical point and is biholomorphic onto its image

Statement

An injective holomorphic map on a complex domain has nowhere-zero derivative and is biholomorphic onto its open image.

Precisely, if $f : Ω \to C$ is holomorphic and injective on a complex domain, then $f^{'} (a) \neq 0$ for every $a \in Ω$ , the set $f [Ω]$ is a complex domain, and $f : Ω \to f [Ω]$ is biholomorphic (Biholomorphic maps between complex domains).

Facts & Assumptions

Given: A holomorphic injective map $f : Ω \to C$ on a complex domain. Injectivity and bijectivity have their set-theoretic meanings (Injection, surjection, bijection).

[L1]

If $f$ is nonconstant and holomorphic on a complex domain $Ω$ and $a \in Ω$ , then $f^{'} (a) \neq 0$ , $\deg_{a} f = 1$ , local injectivity at $a$ , and biholomorphy between neighbourhoods of $a$ and $f (a)$ are equivalent (Holomorphic inverse function theorem and local-degree criterion).

[L2]

Every nonconstant holomorphic function on a complex domain is an open map (Open mapping theorem for holomorphic functions).

[L3]

A complex differentiable function is continuous at every point of complex differentiability (Complex differentiability at a point implies continuity there).

[L4]

A continuous image of a connected space is connected (A continuous image of a connected space is connected, and connectedness is a topological property, claim 1).

Proof

technique · direct

1.1

The map $f$ is nonconstant because an open complex domain has distinct points and $f$ is injective. It is locally injective at every $a \in Ω$ , so [L1] gives $f^{'} (a) \neq 0$ and a holomorphic local inverse near $f (a)$ .

L1given

1.2

By [L2], the image $f [Ω]$ is open. By [L3], $f$ is continuous, so [L4] makes its image connected; it is therefore a complex domain.

L2L3L4given

2.1

The global set-theoretic inverse $f^{- 1} : f [Ω] \to Ω$ agrees near every image point with the holomorphic local inverse from step 1.1. Hence $f^{- 1}$ is holomorphic throughout the image.

step 1.1step 1.2

3.1

The map $f$ is bijective onto its image, and step 2.1 makes its inverse holomorphic; therefore it is biholomorphic onto the open image.

step 2.1∎

Depends on

Used by

FALSE: every injective real-differentiable planar map has nonzero Jacobian False statement

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Theorem 5.6.3 (standard reference, not scraped)
B. V. Shabat, Introduction to Complex Analysis, §1.2 (standard reference, not scraped)