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

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A nonnatural choice of universal-coefficient splitting

Statement

Choosing a complement in a free cycle-boundary decomposition gives a UCT splitting that is changed by a chain automorphism moving that complement; it is therefore not natural.

Counterexample

Given: the free complex $C_{1} = Z e \oplus Z f$ , $C_{0} = Z g$ , with $d (e) = 2 g$ and $d (f) = 0$ , and the coefficient group $G = Z / 2$ .

1.1givenalgebra

Here $H_{1} (C) = Z f$ , $H_{0} (C) = Z / 2$ , and the differential of $C \otimes G$ is zero. Thus the degree-one UCT sequence is $0 ⟶ (Z / 2) f ⟶ (Z / 2) e \oplus (Z / 2) f \overset{q}{\to} Z / 2 ⟶ 0$ , where $q (a e + b f) = a$ .

2.1step 1.1algebra

The maps $u_{1} (e) = e + f$ , $u_{1} (f) = f$ , and $u_{0} (g) = g$ define a chain automorphism inducing the identity on $H_{1} (C)$ and $H_{0} (C)$ , but $u_{1} \otimes 1$ sends $(a, b)$ to $(a, a + b)$ on the middle UCT term.

3.1step 2.1contradiction∎

Any section has $s (1) = (1, t)$ . Naturality with respect to $u$ would require $(u_{1} \otimes 1) s (1) = s (1)$ because $u$ acts identically on the two outer terms, but $(1, t + 1) \neq (1, t)$ . Hence no section for this complex and coefficient group is invariant under all chain automorphisms.

Depends on

A universal-coefficient splitting cannot in general be chosen naturally

Used by

Nothing in the library uses this result yet.

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)