LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 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 map is well defined and natural

Statement

Let $R$ be commutative and let $C, D$ be chain complexes of $R$ -modules. Then $[x] \otimes [y] \mapsto [x \otimes y]$ defines a natural map $⨁_{p + q = n} H_{p} C \otimes_{R} H_{q} D \to H_{n} (C \otimes_{R} D)$ .

Proof

Given: cycles $x \in Z_{p} C$ , $y \in Z_{q} D$ .

1.1

The signed tensor differential gives $d (x \otimes y) = d x \otimes y + (- 1)^{p} x \otimes d y = 0$ .

given

2.1

Replacing $x$ or $y$ by a boundary changes $x \otimes y$ by a signed boundary; chain maps commute with the formula, proving well-definedness and naturality.

step 1.1algebra∎

Depends on

The homology cross product for tensor complexes

Used by

The Kunneth cross product is graded commutative under the twist map Proposition
The Kunneth theorem for free complexes over a PID Theorem

Dependency tree · two levels

3 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)