Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31
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Hom in the homotopy category is zero-degree homology of the Hom complex

Statement

For chain complexes C∙,D∙ in an abelian category, there is a natural isomorphism Hom⁡K(A)(C,D)≅H0(Hom⁡‾(C,D)∙).

Facts & Assumptions

Given: Chain complexes C∙ and D∙.

[L1]

Degree-0 cycles in the Hom complex are exactly chain maps (Zero cocycles in the Hom complex are chain maps).

[L2]

A degree-0 chain map is null-homotopic exactly when it is a boundary in the Hom complex (A null-homotopic chain map).

[L3]

Homotopy classes are chain maps modulo null-homotopic maps (Homotopy classes of chain maps).

[L4]

Morphisms in K(A) are those homotopy classes (The homotopy category of chain complexes).

Proof

technique · direct
1.1L1L2givenalgebra

By [L1], the group of 0-cycles in Hom⁡‾(C,D)∙ is exactly the group of chain maps C∙→D∙. By [L2], its subgroup of 0-boundaries is exactly the null-homotopic chain maps.

2.1L3L4step 1.1algebra∎

Therefore H0(Hom⁡‾(C,D))=Z0/B0 is the quotient of the chain maps by the null-homotopic ones. By [L3] this is [C,D]K, and [L4] identifies that quotient with Hom⁡K(A)(C,D).

Depends on

Used by

Dependency tree · two levels

11 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