False statementConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Two atlases on the same topological manifold need not have a union atlas

Statement

False claim: any two smooth atlases on the same topological manifold have a union that is again a smooth atlas.

Facts & Assumptions

Given: The real line $R$ with the two singleton atlases $A = {(R, id)}$ and $B = {(R, ψ)}$ , where $ψ (x) = x^{3}$ .

[F1]

A smooth atlas is a covering family of pairwise smoothly compatible charts (Smooth atlases).

[F2]

Two charts are smoothly compatible only when both transition maps on the overlap are smooth (Smoothly compatible charts and the smoothness of Euclidean transition maps).

Refutation

technique · direct

1.1

Both $A$ and $B$ are atlases on $R$ : each consists of one global chart and therefore covers the space.

given

Their only cross-transition maps are

$ψ \circ {id}^{- 1} (x) = x^{3}, id \circ ψ^{- 1} (x) = x^{1 / 3} .$

The first is smooth on $R$ , while the second is not $C^{1}$ at $0$ . [given]

2.1

Since one of the two required transition maps fails to be smooth, [F2] shows that the chart in $A$ is not smoothly compatible with the chart in $B$ . Hence $A \cup B$ is not pairwise compatible and therefore is not a smooth atlas by [F1].

F1F2step 1.1

3.1

Thus two atlases on the same topological manifold need not have a union atlas.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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, §2.2 (standard reference, not scraped)
Rob van der Vorst, Introduction to differentiable manifolds, §2 (standard reference, not scraped)