LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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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.

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Smooth chart compatibility is symmetric and reflexive

Statement

Let $M$ be a topological manifold and let $(U, φ)$ and $(V, ψ)$ be charts on $M$ . Then:

Every chart is smoothly compatible with itself.
If $(U, φ)$ and $(V, ψ)$ are smoothly compatible, then $(V, ψ)$ and $(U, φ)$ are smoothly compatible.

Facts & Assumptions

Given: Charts $(U, φ)$ and $(V, ψ)$ on a topological manifold $M$ .

[F1]

Two charts are smoothly compatible when their domains are disjoint, or when both transition maps are smooth; in dimension zero overlapping charts are declared compatible (Smoothly compatible charts and the smoothness of Euclidean transition maps).

[A1]

A constant real-valued function on an open subset of $R^{m}$ is continuous (its preimage of any open set is either the empty set or the whole domain, and both are open).

Proof

technique · direct

1.1

The transition maps of $(U, φ)$ with itself are both the identity ${i d}_{φ (U)}$ . Each component $x \mapsto x_{j}$ of the identity has first partials the constant functions $1$ and $0$ and all higher partials equal to the zero function; every one of these is constant, hence continuous by [A1]. Therefore every component is of class $C^{k}$ for every $k$ , and the identity is smooth. By [F1], this makes $(U, φ)$ smoothly compatible with itself, which is claim 1.

F1A1given

2.1

The definition in [F1] requires both transition maps to be smooth whenever the overlap is nonempty, so interchanging the two charts interchanges the same two smoothness requirements; the hypothesis that $(U, φ)$ and $(V, ψ)$ are compatible therefore makes $(V, ψ)$ and $(U, φ)$ compatible, which is claim 2.

F1given∎

Depends on

Smoothly compatible charts and the smoothness of Euclidean transition maps

Used by

All charts compatible with a smooth atlas form a smooth atlas Lemma

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

Nigel Hitchin, Differentiable Manifolds, §2.2 (standard reference, not scraped)