LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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The union of two compatible smooth atlases is a smooth atlas

Statement

Let $A$ and $B$ be compatible smooth atlases on a topological manifold $M$ . Then the family $A \cup B$ of all charts belonging to $A$ or to $B$ is again a smooth atlas on $M$ .

Facts & Assumptions

Given: Compatible smooth atlases $A$ and $B$ on $M$ .

[F1]

A smooth atlas is a family of charts whose domains cover $M$ and whose members are pairwise smoothly compatible, and compatible atlases have every chart of one smoothly compatible with every chart of the other (Smooth atlases).

Proof

technique · direct

1.1

The union of the two domain covers is a cover of $M$ , so the family [F1, given] $A \cup B$ satisfies the covering condition.

F1given

1.2

Two charts both from $A$ are compatible by the pairwise condition [F1, given] inside $A$ , and likewise two charts both from $B$ ; a chart of $A$ and a chart of $B$ are compatible because the two atlases are compatible.

F1given

2.1

Therefore every pair of members of $A \cup B$ is smoothly [F1, step 1.1, step 1.2] compatible, and step 1.1 gives the covering condition. Hence $A \cup B$ is a smooth atlas.

F1step 1.1step 1.2∎

Depends on

Smooth atlases

Used by

Dependency tree · two levels

3 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

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