LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Reindexing a chain complex cohomologically

PDF p. 4: Property (4): cycle-complex convention

Statement

Let C be a chain complex of abelian groups, with differential d from $C_{n}$ to $C_{n - 1}$ . Define the cochain complex $C^{\lor}$ by $(C^{\lor})^{q}$ = $C_{- q}$ , with differential d in each degree. Then $H^{q} (C^{\lor})$ is canonically $H_{- q} (C)$ .

A chain complex has consecutive differentials whose composite is zero; its homology is the kernel modulo the image. A cochain complex is defined analogously, with differential increasing degree, and its cohomology is the corresponding kernel modulo image.

Facts & Assumptions

Given: A chain complex C and an integer q.

Goal: Identify the cohomology of its reindexing with its homology.

Proof

1.1givenconstruct

The reindexed differential goes from $C_{- q}$ to $C_{- q - 1}$ , and its square is zero because d squared is zero in C. Thus $C^{\lor}$ is a cochain complex.

2.1givenstep 1.1∎

Its degree-q kernel and incoming image are exactly the degree-(−q) cycles and boundaries of C. Their quotients are therefore identical, giving the claimed canonical isomorphism.

Used by · 1

Cycle-complex degree translation