CorollaryStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-01

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.

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The cone criterion from the general long exact sequence

Statement

For a chain map $f : C_{∙} \to D_{∙}$ , the cone $Cone (f)$ is acyclic if and only if $f$ is a quasi-isomorphism. This is exactly the same criterion already recorded on the mapping-cone page.

Facts & Assumptions

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

[L1]

The cone long exact sequence has the form $\dots \to H_{n} (C) \overset{H_{n} (f)}{\to} H_{n} (D) \to H_{n} (Cone (f)) \to H_{n - 1} (C) \to \dots$ (The cone long exact sequence).

[L2]

Acyclic means vanishing homology in every degree (Exactness of a complex at a degree and acyclic complexes).

[L3]

The mapping-cone page already proves the same equivalence (A chain map is a quasi-isomorphism exactly when its cone is acyclic).

Proof

technique · direct

1.1L1L2givenalgebra

If $Cone (f)$ is acyclic, then the middle terms in [L1] vanish by [L2]. Exactness therefore makes every $H_{n} (f)$ both monic and epic, so $f$ is a quasi-isomorphism.

2.1L1L2L3givenalgebra∎

If $f$ is a quasi-isomorphism, then every map $H_{n} (f)$ in [L1] is an isomorphism. Exactness forces each $H_{n} (Cone (f))$ to vanish, so the cone is acyclic by [L2]. This is the same criterion already established in [L3].

Depends on

Used by

Dependency tree · two levels

19 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

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)
The Stacks Project, Section 12.13: Complexes (standard reference, not scraped)