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

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Equivalent higher extensions need not have one middle-term isomorphism

Statement refuted

Give a zigzag of endpoint-identity chain maps between two two-fold extensions whose middle-object shapes prevent a single isomorphism, illustrating the generated equivalence relation.

Facts & Assumptions

Given: $M = N = Z$ in abelian groups and the two two-fold extensions below.

Counterexample

technique · direct

1.1givenconstruct

Let $E$ be $0 \to Z \overset{1}{\to} Z \overset{0}{\to} Z \overset{1}{\to} Z \to 0$ , and let $E^{'}$ be $0 \to Z \to Z^{2} \overset{(a, c) \mapsto (0, c)}{\to} Z^{2} \overset{(b, c) \mapsto b}{\to} Z \to 0$ , where the left inclusion is $a \mapsto (a, 0)$ . Both are exact.

2.1step 1.1algebra∎

The maps $(a) \mapsto (a, 0)$ and $(b) \mapsto (b, 0)$ give an endpoint-identity chain map $E \to E^{'}$ , while the coordinate projections give one $E^{'} \to E$ . Thus the two extensions are equivalent in the generated sense. But their first middle terms are $Z$ and $Z^{2}$ , which are not isomorphic, so no single middle-term isomorphism can witness this equivalence.

Depends on

Equivalence of n-fold extensions

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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

Charles A. Weibel, An Introduction to Homological Algebra, Chapters 3–4 (standard reference, not scraped)