False statementConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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A bijective smooth map need not be a diffeomorphism

Statement

False claim: every bijective smooth map is a diffeomorphism.

Facts & Assumptions

Given: The map $F : R \to R$ , $F (x) = x^{3}$ .

[F1]

A smooth map is a continuous map whose coordinate representatives are smooth ( $C^{r}$ and smooth maps between smooth manifolds).

[F2]

A diffeomorphism is a bijective smooth map whose inverse is also smooth (Diffeomorphisms and local diffeomorphisms of manifolds).

Refutation

technique · direct

1.1

The map $F (x) = x^{3}$ is smooth on $R$ and bijective, with inverse [F1] $F^{- 1} (y) = y^{1 / 3}$ .

2.1

The inverse is not differentiable at $0$ , because

step 1.1

$\frac{d}{d y} y^{1 / 3} = \frac{1}{3} y^{- 2 / 3}$

for $y \neq 0$ , and these derivatives are unbounded near $0$ . So $F^{- 1}$ is not smooth. [step 1.1]

3.1

By [F2], steps 1.1 and 2.1 show that $F$ is a bijective smooth map that is [F2, step 1.1, step 2.1] not a diffeomorphism.

F2step 1.1step 2.1∎

Depends on

Used by

A bijective smooth map with nonsmooth inverse Counterexample

Dependency tree · two levels

7 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

Nigel Hitchin, Differentiable Manifolds, §2.4 (standard reference, not scraped)
Rob van der Vorst, Introduction to differentiable manifolds, §2 (standard reference, not scraped)