PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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.

A chain map is a quasi-isomorphism exactly when its cochain reindexing is

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map, and read both complexes as cochain complexes by the reindexing convention $(C^{♯})^{- n} = C_{n}$ , $(D^{♯})^{- n} = D_{n}$ . Then $f$ is a quasi-isomorphism if and only if the induced cochain map $f^{♯} : C^{♯} \to D^{♯}$ is a quasi-isomorphism in cohomology.

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ .

[L1]

A cochain complex may be read as a reindexed chain complex (Cochain complex in an abelian category).

[L2]

Cohomology is the cokernel of the coboundary subobject inside the cocycle subobject (Cohomology object of a cochain complex).

[L3]

A quasi-isomorphism is a chain map inducing isomorphisms on every homology object (Quasi-isomorphism).

Proof

technique · direct

1.1

Under the reindexing convention [L1], the cocycles and coboundaries of $C^{♯}$ in degree $- n$ are exactly $Z_{n} (C)$ and $B_{n} (C)$ . Therefore [L2] identifies $H^{- n} (C^{♯}) ≅ H_{n} (C), H^{- n} (D^{♯}) ≅ H_{n} (D) .$

L1L2given

2.1

The map induced by $f^{♯}$ on $H^{- n}$ is therefore the same morphism as $H_{n} (f)$ under the identifications of step 1.1. Hence $H_{n} (f)$ is an isomorphism for every $n$ if and only if $H^{- n} (f^{♯})$ is an isomorphism for every $n$ . By [L3], this is exactly the claimed equivalence.

L3step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Romyar Sharifi, Homological Algebra, Remark 2.7.11 (standard reference, not scraped)
Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)