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 ${Hom}_{K (A)} (C, D) ≅ H_{0} (\underset{}{Hom} (C, D)_{∙}) .$

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

1.1

By [L1], the group of $0$ -cycles in $\underset{}{Hom} (C, D)_{∙}$ is exactly the group of chain maps $C_{∙} \to D_{∙}$ . By [L2], its subgroup of $0$ -boundaries is exactly the null-homotopic chain maps.

L1L2givenalgebra

2.1

Therefore $H_{0} (\underset{}{Hom} (C, D)) = Z_{0} / B_{0}$ 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 ${Hom}_{K (A)} (C, D)$ .

L3L4step 1.1algebra∎

Depends on

Used by

Homotopy classes as H-zero of a Hom complex Example

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

The Stacks Project, Section 12.14: Homotopy and the shift functor (standard reference, not scraped)
The Stacks Project, Section 12.16: Graded objects (standard reference, not scraped)