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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-12
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The homology Kunneth sequence splits nonnaturally

Statement

Assume AC. For any commutative PID R and spaces X,Y, the homology Künneth short exact sequence admits an R-linear section of its Tor quotient in every degree n≥0. Consequently its middle term is abstractly the direct sum of its tensor and Tor terms. The section is constructed after choices; no natural choice of splitting is asserted.

Facts & Assumptions

[F1]

Topological Kunneth short exact sequence for homology gives the exact sequence 0→K→α′W→γQ→0.

[F2]

The PID Kunneth sequence admits a section after choices supplies, for the tensor complex, maps r:V→K and j:Q→V satisfying rα=1, βj=1, jβ=1−αr, and rj=0.

[F3]

Singular product chain equivalence by simplex models supplies the shuffle homology isomorphism s:V→W with inverse t, where V=Hn(C(X;R)⊗RC(Y;R)). The topological maps of [F1] are α′=sα, γ=βt. Assume The Axiom of Choice.

Proof

Given: R,X,Y,n and the maps in [F1]–[F3], under AC.

1.1F1F2F3given

The singular complexes are nonnegative free PID complexes, so the section theorem [F2] applies. Define j′=sj:Q→W and r′=rt:W→K. Then γj′=βtsj=βj=1, r′α′=rtsα=1, and r′j′=rtsj=0. All maps are R-linear.

2.1F1F2F3step 1.1

The remaining composite satisfies j′γ=sjβt=s(1−αr)t=1−α′r′. Therefore (k,q)↦α′k+j′q and w↦(r′w,γw) are inverse: one composite is (k,q)↦(k,q) using r′j′=0, γα′=0 and the identity composites; the other is w↦(α′r′+j′γ)w=w. This proves the direct-sum assertion with the actual cross product and quotient.

3.1F1F2step 1.1step 2.1

The maps r,j of [F2] use chosen retractions Cp→ZpC and Dq→ZqD, obtained by splitting the surjections onto Bp−1C and Bq−1D. No splitting of the inclusions BpC↪ZpC is used. Since no compatibility of those retractions with all space maps is supplied, steps 1.1 and 2.1 assert existence of a section, without asserting its naturality. The canonical exact sequence itself retains its naturality from [F1]. This observation alone is not a proof that every possible natural section is impossible.

4.1F1F2F3step 1.1step 2.1step 3.1∎

At n=0, Q=0 and j′ is the unique map from zero; α′r′=1 gives the degree-zero cross-product isomorphism. If a factor is empty, K=V=W=Q=0 and every formula remains valid. If K=0, the section is the inverse of γ; if Q=0, no Tor summand is added. In degree zero on a pair of points, α′ sends the tensor of the two unit vertex classes to the unit product vertex class. AC is inherited from [F2] for arbitrary-rank cycle retractions and simultaneous degreewise choices; composing the maps adds no choices.

Depends on

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Sources