CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-28

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The exponential map is a holomorphic surjection C to C^× that is not an automorphism

Statement refuted

Every holomorphic surjection $C \to C^{\times}$ is a biholomorphic automorphism of $C^{\times}$ .

Facts & Assumptions

Given: The complex exponential map $\exp : C \to C^{\times}$ .

[L1]

For real $x, y$ , one has $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , and the real exponential is onto $(0, \infty)$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The exponential is a continuous bijection from $R$ onto $(0, \infty)$ ).

[L2]

$\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

Counterexample

technique · direct

1.1L1givenchoose

If $w = r (\cos θ + i \sin θ)$ with $r > 0$ , [L1] gives a real $x$ with $e^{x} = r$ , and then $\exp (x + i θ) = w$ . So the exponential map is surjective onto $C^{\times}$ .

2.1L2given∎

Fact [L2] gives $\exp (0) = \exp (2 π i) = 1$ , so the exponential map is not injective. It is therefore a holomorphic surjection onto $C^{\times}$ that is not an automorphism.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 3 §§2.2-3.5 (standard reference, not scraped)
Matthias Weber, Complex Analysis, Ch. 1 §§1.3-1.4 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 8 §§1-2 (standard reference, not scraped)