LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-05 (gpt-6.1-sol)

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.

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Basic forms descend along surjective submersions

Statement

Let $π : Y \to M$ be a surjective smooth submersion that admits local smooth sections. If $α \in Ω^{k} (Y)$ satisfies ${pr}_{1}^{*} α = {pr}_{2}^{*} α$ on the fiber product $Y \times_{M} Y$ , then there is a unique $β \in Ω^{k} (M)$ with $π^{*} β = α$ .

Statement references

A smooth differential $k$ -form

Facts & Assumptions

Given: A surjective smooth submersion $π : Y \to M$ , local smooth sections, and a form $α$ with equal pullbacks on $Y \times_{M} Y$ .

[F1]

A smooth differential $k$ -form: a smooth $k$ -form is a smooth section of the exterior cotangent bundle, and smooth pullback is defined pointwise by the differential.

Proof

1.1F1givenconstruct

Choose an open cover $(U_{i})$ of $M$ and smooth sections $s_{i} : U_{i} \to Y$ with $π s_{i} = {id}_{U_{i}}$ . Define $β_{i} = s_{i}^{*} α$ . If $x \in U_{i} \cap U_{j}$ , the map $(s_{i}, s_{j}) : U_{i} \cap U_{j} \to Y \times_{M} Y$ is smooth and its two projections are $s_{i}$ and $s_{j}$ . Pulling the assumed identity back along this map gives $β_{i} = β_{j}$ on $U_{i} \cap U_{j}$ .

2.1F1step 1.1given∎

The forms $β_{i}$ therefore glue to a unique smooth form $β$ on $M$ . Pullback along $s_{i}$ gives $s_{i}^{*} π^{*} β = β_{i} = s_{i}^{*} α$ ; for any $y \in π^{- 1} (U_{i})$ , the pair $(y, s_{i} (π (y)))$ lies in $Y \times_{M} Y$ , so the assumed identity implies $α_{y} = (s_{i} \circ π)^{*} α_{y} = π^{*} s_{i}^{*} α_{y} = π^{*} β_{y}$ . Thus $π^{*} β = α$ . If $β^{'}$ also pulls back to $α$ , then $s_{i}^{*} α = s_{i}^{*} π^{*} β^{'} = β^{'}$ on each $U_{i}$ , proving uniqueness.

Basic forms descend along surjective submersions

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources