PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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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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The Kunneth cross product is graded commutative under the twist map

Statement

Let $R$ be commutative and let $C, D$ be chain complexes of $R$ -modules. Under the signed chain isomorphism $τ : C \otimes_{R} D \to D \otimes_{R} C$ , $x \otimes y \mapsto (- 1)^{p q} y \otimes x$ , the cross product of degrees $p, q$ is graded commutative.

Proof

Given: homogeneous cycles $x \in C_{p}$ , $y \in D_{q}$ and the signed twist.

1.1given

The twist sends $x \otimes y$ to $(- 1)^{p q} y \otimes x$ and intertwines the signed tensor differentials.

2.1step 1.1algebra∎

Passing to homology gives $τ_{*} ([x] \times [y]) = (- 1)^{p q} [y] \times [x]$ , which is the asserted graded commutativity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Weibel, An Introduction to Homological Algebra (standard reference, not scraped)