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 cochains as chains

PDF p. 4: Shift [2j] in equation (2.1)

Statement

Let $C$ be a cochain complex of abelian groups, with differentials $d^{q} : C^{q} \to C^{q + 1}$ . Define a chain complex $D$ by $D_{n} = C^{- n}$ and $\partial_{n} = d^{- n}$ . Then $H_{n} (D) = H^{- n} (C)$ naturally in $C$ . Here homology and cohomology mean the kernel of the outgoing differential modulo the image of the incoming differential.

Facts & Assumptions

Given: A cochain complex $C$ as stated.

Goal: Construct $D$ and identify its homology.

Proof

1.1givenconstruct

The composite $\partial_{n - 1} \partial_{n} = d^{- n + 1} d^{- n}$ is zero, so the displayed groups and maps form a chain complex. A cochain map reindexes to a chain map.

2.1givenstep 1.1∎

At degree $n$ , both the kernel and the image in the quotient $H_{n} (D)$ are exactly those defining $H^{- n} (C)$ . The identity on representatives gives the natural isomorphism.

Used by · 2