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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Free-presentation total homology and its degree-one edges

Statement

For the free-presentation bicomplex, Hn(Tot⁡D)=0 for n>1, H1(Tot⁡D)=Fab, E012=R/[F,R], and Ep02=Hp(G;Z). The degree-one maps are inclusion and quotient on abelianizations. This does not assert vanishing of every positive-degree E2 term. Retain DC and supplied-resolution conventions.

Facts & Assumptions

Given: The free presentation and bicomplex of the Definition, with the inherited homology conventions.

[F1]

Total homology is H*(F), E2 is H_p(G;H_q(R)), and the degree-one edges are identified (The free-presentation Lyndon bar bicomplex).

[F2]

A free group has a length-one free resolution (Generator differences form a basis of the free-group augmentation ideal).

[F3]

H1(R) conjugation coinvariants equal R/[F,R] (First integral homology and conjugation coinvariants).

Proof

1.1F1F2given

The total homology equals H*(F) by F1. The free resolution of F2 has no terms above degree one, so after tensoring it has zero homology there; in degree one it gives the free abelian group on the free generators, which is Fab. Hence the asserted total vanishing holds even for an infinite free generating set.

2.1F1F3step 1.1algebra∎

The zero homology of R with trivial integers is Z: every vertex is identified, and each degree-one boundary is a difference of vertices. Conjugation fixes this generator. Consequently Ep02=Hp(G;Z). Also E012=H0(G;H1(R;Z))=R/[F,R] by F3. The edge calculations of F1 send r to its class in Fab and f to its image in Gab, with positive signs.

Depends on

Used by

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Sources