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

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The map $(z_{1}, z_{2}) \mapsto (e^{z_{1}}, z_{2})$ has invertible complex Jacobian everywhere and is not injective

Statement refuted

Refuted claim: a holomorphic map with everywhere-invertible complex Jacobian must be injective.

Facts & Assumptions

Given: The map $F (z_{1}, z_{2}) = (e^{z_{1}}, z_{2})$ .

[L1]

The complex Jacobian is computed from the complex differential (Holomorphic maps $C^{m} \to C^{n}$ and the complex Jacobian matrix).

[L2]

The complex exponential satisfies $e^{z + w} = e^{z} e^{w}$ , and $e^{i π} = - 1$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

Counterexample

technique · direct

1.1

The complex Jacobian of $F$ is $J_{C} F (z_{1}, z_{2}) = (\begin{array}{cc} e^{z_{1}} & 0 \\ 0 & 1 \end{array}),$ so $\det J_{C} F (z_{1}, z_{2}) = e^{z_{1}} \neq 0$ for every $(z_{1}, z_{2})$ .

givenL1algebra

2.1

By [L2], $e^{2 π i} = e^{i π} e^{i π} = (- 1) (- 1) = 1$ , so $F (z_{1} + 2 π i, z_{2}) = (e^{z_{1} + 2 π i}, z_{2}) = (e^{z_{1}}, z_{2}) = F (z_{1}, z_{2}) .$ Thus distinct points have the same image, and $F$ is not injective despite step 1.1.

step 1.1L2algebra∎

Depends on

Used by

FALSE: an everywhere-invertible complex Jacobian forces global injectivity False statement
FALSE: the holomorphic inverse function theorem is global False statement

Dependency tree · two levels

33 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

Jiří Lebl, Tasty Bits of Several Complex Variables, Section 1.6 (standard reference, not scraped)
Jiří Lebl, Guide to Cultivating Complex Analysis, Section 4.6 (standard reference, not scraped)

The map (z1,z2)↦(ez1,z2)(z1​,z2​)↦(ez1​,z2​) has invertible complex Jacobian everywhere and is not injective