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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)
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A universal-coefficient splitting cannot in general be chosen naturally

Statement

The noncanonical splitting in the homological UCT cannot in general be chosen naturally in the complex and coefficient group.

Counterexample

Given: the free complex C1=Ze⊕Zf, C0=Zg, with d(e)=2g and d(f)=0, together with G=Z/2.

1.1givenalgebra

Since H1(C)=Zf, H0(C)=Z/2, and d⊗G=0, direct kernel and cokernel calculations give the exact degree-one sequence 0⟶(Z/2)f⟶(Z/2)e⊕(Z/2)f→qZ/2⟶0, with q(ae+bf)=a. This is the UCT sequence in this finite example: its left map tensors the cycle f, and its right map records the coefficient of e modulo 2.

2.1step 1.1algebra

The chain automorphism u1(e)=e+f, u1(f)=f, u0(g)=g induces the identity on both integral homology groups and hence on both outer UCT terms, while its action on the middle term is (a,b)↦(a,a+b).

3.1step 2.1contradiction∎

A section must send 1 to (1,t) for some t∈Z/2. Naturality with respect to u would force (1,t+1)=(1,t), which is impossible. Therefore a UCT splitting cannot be chosen naturally for all complexes and coefficient groups.

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