False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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The Kunneth short exact sequence has no generally canonical splitting

Statement

The assertion that the Kunneth short exact sequence has a canonical splitting is false.

Refutation

Given: let $C_{1} = Z e \oplus Z f$ , $C_{0} = Z g$ with $d (e) = 2 g$ , $d (f) = 0$ , and let $D_{1} = Z a$ , $D_{0} = Z b$ with $d (a) = 2 b$ .

1.1givenalgebra

We have $H_{1} (C) = Z f$ , $H_{0} (C) = H_{0} (D) = Z / 2$ , and $H_{1} (D) = 0$ . In total degree one the Kunneth sequence is therefore $0 ⟶ Z / 2 ⟶ H_{1} (C \otimes D) ⟶ Z / 2 ⟶ 0$ . Writing $v = e \otimes b - g \otimes a$ , a direct kernel/modulo-boundary calculation gives $H_{1} (C \otimes D) = (Z / 2) v \oplus (Z / 2) (f \otimes b)$ ; the left Kunneth term is generated by $f \otimes b$ and the quotient by the class of $v$ .

2.1step 1.1algebra

The chain automorphism $u (e) = e + f$ , $u (f) = f$ , $u (g) = g$ of $C$ induces the identity on $H_{*} (C)$ , hence on both outer Kunneth terms, but sends $v$ to $v + f \otimes b$ in the middle term.

3.1step 2.1contradiction∎

Every section of the quotient sends its generator to $v + t (f \otimes b)$ for some $t \in Z / 2$ . Naturality under $u \otimes 1_{D}$ would require this lift to be fixed, but $u$ changes $t$ to $t + 1$ . Thus no splitting of all Kunneth sequences can be canonical or natural.

Depends on

The Kunneth sequence splits nonnaturally

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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)