LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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 extension from a closed neighbourhood

Statement

Let $C$ be a closed subset of a smooth manifold $M$ , let $U \subseteq M$ be open with $C \subseteq U$ , and let $f : U \to R$ be smooth. Then there exists a smooth function $F : M \to R$ such that $F = f$ on an open neighbourhood of $C$ and $supp (F) \subseteq U$ .

Facts & Assumptions

Given: A closed set $C \subseteq M$ , an open set $U \subseteq M$ containing $C$ , and a smooth function $f : U \to R$ .

[L1]

There is a smooth cutoff $χ : M \to [0, 1]$ equal to $1$ on a neighbourhood of $C$ and supported in $U$ (A smooth Urysohn lemma for a closed set in an open set).

[L2]

Smooth maps paste over an open cover (Smooth maps paste over an open cover).

[A1]

Products of smooth real-valued functions on the same open set are smooth.

Proof

technique · direct

1.1

Let $χ$ be as in [L1]; then $χ f$ is smooth on $U$ by [A1] and equals $f$ on an open neighbourhood of $C$ because $χ = 1$ there.

L1A1given

2.1

Since $supp (χ) \subseteq U$ , the function $χ$ vanishes on an open neighbourhood of $M ∖ U$ , so the local formula $χ f$ on $U$ agrees with the constant zero map on an open neighbourhood of every boundary point of $U$ .

L1step 1.1

3.1

Pasting these two local formulas by [L2] yields a smooth function $F : M \to R$ with $F = f$ near $C$ and $supp (F) \subseteq U$ .

L2step 1.1step 2.1∎

Depends on

Used by

Extension by zero without support away from the boundary is not smooth Counterexample
Naive extension by zero from an open set need not be smooth False statement

Dependency tree · two levels

7 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 (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)