TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

Cup product Leibniz identity

Statement

For cochains $φ \in C^{p} (X; R)$ and $ψ \in C^{q} (X; R)$ , $p, q \geq 0$ , over a commutative unital ring with positive coboundary, $δ (φ ⌣ ψ) = δ φ ⌣ ψ + (- 1)^{p} φ ⌣ δ ψ .$ Thus the product of cocycles is a cocycle, and changing either cocycle representative by a coboundary changes their product by a coboundary.

Facts & Assumptions

[F1]

Singular cup product on cochains defines $φ ⌣ ψ = J (φ, ψ) D_{X}$ , with $D_{X}$ a chain map.

[F2]

The additive singular cohomology cross product is well-defined proves $δ J (φ, ψ) = J (δ φ, ψ) + (- 1)^{p} J (φ, δ ψ)$ on the signed tensor complex.

Proof

Given: $X, R, p, q, φ, ψ$ as stated. Negative cochain degrees are zero, and $δ^{2} = 0$ .

1.1F1F2given

Since $D_{X} \partial = d D_{X}$ , precomposing the tensor-functional identity with $D_{X}$ gives $δ (φ ⌣ ψ) = J (φ, ψ) D_{X} \partial = J (φ, ψ) d D_{X} = (J (δ φ, ψ) + (- 1)^{p} J (φ, δ ψ)) D_{X} .$ By the cup formula this is exactly the asserted identity. If both inputs are closed its right side is zero.

2.1F1step 1.1given

Now assume $δ φ = δ ψ = 0$ . Let $u \in C^{p - 1} (X; R)$ and $v \in C^{q - 1} (X; R)$ . Bilinearity expands the change to $(φ + δ u) ⌣ (ψ + δ v) - φ ⌣ ψ = δ u ⌣ ψ + φ ⌣ δ v + δ u ⌣ δ v$ . Step 1.1, applied to each pair, identifies this as $δ (u ⌣ ψ + (- 1)^{p} φ ⌣ v + u ⌣ δ v) .$ Indeed the respective other Leibniz terms contain $δ ψ$ , $δ φ$ , or $δ^{2} v$ , and vanish. This proves simultaneous descent and, by setting $u = 0$ or $v = 0$ , each separate descent.

3.1F1F2step 1.1step 2.1∎

If $p = 0$ , then $u = 0$ and its two terms in the primitive are absent; if $q = 0$ , then $v = 0$ and its terms are absent. For $p = q = 0$ both representatives are unchanged, while the Leibniz identity itself still holds, with sign $+ 1$ . At zero input or zero coefficient ring the equality is zero by bilinearity. An empty space has zero cochains, and a point or a degenerate simplex satisfies the same chain-map and tensor identities. No representative, basis, or primitive was chosen from an arbitrary family: the primitive is the displayed expression in the given $u, v$ . Thus no AC is used.

Depends on

Dependency tree · 2 levels

Sources

Hatcher Lemma 3.6; Miller Lecture 28