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

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.

Smooth maps paste over an open cover

Statement

Let $M$ and $N$ be smooth manifolds, let $(U_{i})_{i \in I}$ be an open cover of $M$ , and let $F_{i} : U_{i} \to N$ be smooth maps such that $F_{i} ∣_{U_{i} \cap U_{j}} = F_{j} ∣_{U_{i} \cap U_{j}}$ for all $i, j \in I$ . Then there is a unique map $F : M \to N$ with $F ∣_{U_{i}} = F_{i}$ for every $i \in I$ , and $F$ is smooth.

Facts & Assumptions

Given: An open cover $(U_{i})_{i \in I}$ of $M$ and smooth maps $F_{i} : U_{i} \to N$ agreeing on all overlaps.

[L1]

A family of continuous maps on an open cover that agree on overlaps determines a unique continuous map on the whole space (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L2]

A continuous map into $N$ is smooth exactly when its restriction to every member of an open cover is smooth (Smoothness is local on the source).

Proof

technique · direct

1.1

Define $F : M \to N$ by $F (x) : = F_{i} (x)$ whenever $x \in U_{i}$ . The overlap [given, L1] hypothesis makes this single-valued, and each $F_{i}$ is continuous because smooth maps are continuous, so [L1] pastes the pieces into a unique continuous map $F$ with $F ∣_{U_{i}} = F_{i}$ for every $i$ .

givenL1

2.1

Every restriction $F ∣_{U_{i}} = F_{i}$ is smooth by the hypothesis, and $F$ is [given, L2, step 1.1] continuous by step 1.1, so [L2] makes $F$ smooth.

givenL2step 1.1

3.1

The uniqueness and the defining restriction property come from step 1.1, [step 1.1, step 2.1] and smoothness comes from step 2.1. This proves the claim.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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, Theorem 2.15 (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds, §2.4 (standard reference, not scraped)