PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31

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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.
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A chain isomorphism is a chain homotopy equivalence

Statement

If a chain map $f : C_{∙} \to D_{∙}$ admits a chain map $g : D_{∙} \to C_{∙}$ with $g \circ f = 1_{C}, f \circ g = 1_{D},$ then $f$ is a chain homotopy equivalence.

Facts & Assumptions

Given: Chain maps $f : C_{∙} \to D_{∙}$ and $g : D_{∙} \to C_{∙}$ with $g f = 1_{C}$ and $f g = 1_{D}$ .

[L1]

A homotopy equivalence is a chain map with a homotopy inverse up to homotopy (A chain homotopy equivalence).

[L2]

Identities and composites of chain maps are chain maps (Identities and composites of chain maps are chain maps).

Proof

technique · direct

1.1

By [L2], the composites $g f$ and $f g$ are chain maps, and by hypothesis they are exactly the identity chain maps.

L2given

2.1

Exact equality implies chain homotopy, via the zero homotopies. Therefore $g$ is a homotopy inverse of $f$ , and [L1] shows that $f$ is a chain homotopy equivalence.

L1step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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)