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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Changing the section changes the factor set by a coboundary

Statement

If s and s are normalized sections of the same abelian-kernel extension, then there is a normalized one-cochain u:GM such that

fs=fs+δu.

Facts & Assumptions

Given: Two normalized sections s,s:GE of the same extension.

[F1]

Two-coboundaries have the form (δu)(g,h)=gu(h)u(gh)+u(g) (Normalized two-cocycle and two-coboundary).

[F2]

Factor sets are defined by the section formula (Normalized set-theoretic section and factor set).

[L1]

Each factor set is a normalized two-cocycle (The factor set of a section is a normalized two-cocycle).

Proof

technique · direct
1.1

Since π(s(g)s(g)1)=1, each quotient s(g)s(g)1 lies in the kernel. So there is a unique u(g)M with s(g)=i(u(g))s(g). The normalization s(1)=s(1)=1 gives u(1)=0.

F2givenchoosealgebra
2.1

Substitute s(g)=i(u(g))s(g) into the factor-set formula of [F2]. After moving kernel terms past lifts by the prescribed action, the result is fs(g,h)=u(g)+gu(h)u(gh)+fs(g,h)=fs(g,h)+(δu)(g,h).

F1F2step 1.1algebra
3.1

Thus the two factor sets differ by a coboundary; [L1] shows that both are indeed cocycles.

L1step 2.1

Depends on

Used by

Dependency tree · two levels

4 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