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

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Singular cohomology is graded commutative

Statement

Let $X$ be a space, $R$ a commutative unital ring, $a \in H^{p} (X; R)$ and $b \in H^{q} (X; R)$ , with $p, q \geq 0$ . Then $a ⌣ b = (- 1)^{p q} b ⌣ a .$ The defining cochain operation need not be graded commutative. No AC is required.

Facts & Assumptions

[F1]

Factor reversal gives the commutativity chain homotopy supplies $H = K Δ_{#}$ with $d H + H \partial = W D_{X} - D_{X}$ , where $W (x \otimes y) = (- 1)^{∣ x ∣ ∣ y ∣} y \otimes x$ .

[F2]

Singular cohomology ring defines multiplication by taking the class of the front/back cup of representative cocycles.

Proof

Given: Cocycles $φ, ψ$ representing $a, b$ . On the tensor complex define $J (x \otimes y) = φ (x) ψ (y)$ in bidegree $(p, q)$ and zero in other bidegrees of total degree $p + q$ .

1.1

The functional $J$ is $R$ -balanced by commutativity. On tensors of bidegree $(p + 1, q)$ , $J d = δ φ (x) ψ (y) = 0$ ; on tensors of bidegree $(p, q + 1)$ , $J d = (- 1)^{p} φ (x) δ ψ (y) = 0$ . All other bidegrees contribute zero. Thus $J d = 0$ . Evaluating [F1]'s identity yields $J W D_{X} - J D_{X} = J H \partial = δ (J H) .$ The right side is an explicit coboundary, with degree- $(p + q - 1)$ primitive.

F1given

2.1

The functional $J D_{X}$ is $φ ⌣ ψ$ . In $J W D_{X}$ , only the original diagonal cut of bidegree $(q, p)$ contributes, and its value is $(- 1)^{p q} φ (σ [q, \dots, p + q]) ψ (σ [0, \dots, q])$ . Commutativity in $R$ identifies this with $(- 1)^{p q} (ψ ⌣ φ) (σ)$ . Step 1.1 and [F2] therefore give the asserted equality of classes.

F2step 1.1given

3.1

If $p = q = 0$ , the primitive in step 1.1 has negative degree and is zero; the two cochains agree by ordinary commutativity in $R$ . If just one degree is zero, the sign is positive and the same homotopy calculation applies. Empty spaces, zero classes and the zero ring give zero products. All formulas include points and degenerate simplices. Only the two given cocycle representatives and the specified homotopy are used, so no choice principle is invoked.

F1F2step 1.1step 2.1∎

Depends on

Used by

Poincaré duality gives a nonsingular cup pairing Corollary
Cochain cup product is not strictly graded commutative Counterexample
Cup length over a coefficient ring Definition
Integral cohomology ring of a closed orientable surface Example
Integral surface cup pairing from the oriented polygon Lemma
Cohomological Kunneth cross product is a ring isomorphism Theorem
The de Rham theorem Theorem

Dependency tree · two levels

6 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

Hatcher Theorem 3.11; Miller Lecture 29 (standard reference, not scraped)