False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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: every smooth vector field can be pushed forward by every smooth map

Statement

False claim: every smooth vector field $X$ on $M$ has a canonically defined pushforward by every smooth map $F : M \to N$ .

Facts & Assumptions

Given: The projection $F : R^{2} \to R$ , $F (x, y) = x$ , and the vector field $X = y \partial / \partial x$ on $R^{2}$ .

[L1]

For a general smooth map, the correct comparison notion is $F$ -relatedness; an actual pushforward is defined here only for diffeomorphisms (Pushforwards and pullbacks of vector fields by a diffeomorphism).

Refutation

technique · direct

1.1

At a point $(x, y)$ , the differential of $F$ sends $X_{(x, y)} = y \partial / \partial x$ to the tangent vector $y d / d x ∣_{x}$ in $T_{x} R$ .

given

2.1

If a pushforward vector field $F_{*} X$ on $R$ existed, its value at $x$ would have to equal $y d / d x ∣_{x}$ for every point $(x, y)$ in the fibre $F^{- 1} (x)$ . That is impossible because different values of $y$ in the same fibre give different target vectors.

step 1.1given

3.1

Therefore a smooth map need not push a vector field forward to a well-defined vector field on the target, in agreement with [L1].

L1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)