False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: an everywhere-invertible derivative gives a global inverse

Statement

False claim: a $C^{1}$ map between open subsets of $R^{n}$ whose derivative is invertible everywhere must have a global inverse.

Facts & Assumptions

Given: On the punctured plane $U = R^{2} ∖ {(0, 0)}$ define $S (x, y) = (x^{2} - y^{2}, 2 x y)$ . The domain is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and scalar derivative algebra is supplied by Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ and Continuously differentiable maps, local inverses, and local diffeomorphisms.

[L1]

If $f$ is $C^{1}$ on an open Euclidean domain and $D f (a)$ is invertible, the inverse function theorem supplies open neighbourhoods on which $f$ has a $C^{1}$ inverse (The Euclidean inverse function theorem).

[L2]

A Euclidean linear map is invertible when it has a two-sided linear inverse (Invertible Euclidean linear maps).

[L3]

Continuous partial derivatives give total differentiability, with total derivative represented by the Jacobian matrix (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

Refutation

technique · direct

1.1

By [L3], $D S (x, y) = (\begin{array}{cc} 2 x & - 2 y \\ 2 y & 2 x \end{array}) .$ At $(x, y) \neq (0, 0)$ , the matrix $\frac{1}{2 (x^{2} + y^{2})} (\begin{array}{cc} x & y \\ - y & x \end{array})$ is its two-sided inverse, so [L2] and [L1] make $S$ locally invertible at every point of $U$ .

L1L2L3givenalgebra

2.1

Nevertheless $S (1, 0) = (1, 0) = S (- 1, 0)$ . Thus $S$ is not injective and has no global inverse, despite its everywhere-invertible derivative.

step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

34 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, Basic Analysis II, Example 8.5.4 (standard reference, not scraped)