TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Homology factors uniquely through the homotopy category

Statement

Fix $n \in Z$ . Let $Q : Ch (A) \to K (A)$ be the canonical quotient functor for an abelian category $A$ . Then there is a unique additive functor ${\overline{H}}_{n} : K (A) \to A$ such that $H_{n} = {\overline{H}}_{n} \circ Q .$

Facts & Assumptions

Given: An abelian category $A$ and an integer $n$ .

[L1]

Homotopic chain maps induce the same map on homology (Chain-homotopic maps induce the same map on homology).

[L2]

Homology is an additive functor on chain complexes (Homology is an additive functor).

[L3]

The quotient functor $Q$ is additive (The canonical functor from complexes to the homotopy category is additive).

[L4]

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

Proof

technique · direct

1.1

Define ${\overline{H}}_{n} (C) : = H_{n} (C)$ on objects and ${\overline{H}}_{n} ([f]) : = H_{n} (f)$ on morphisms. This is well defined because [L1] shows that homotopic representatives have the same homology map, and [L4] says those are exactly the equal morphisms in $K (A)$ .

L1L4givenconstruct

2.1

Because $H_{n}$ is additive by [L2] and $Q$ is additive by [L3], the definition in step 1.1 gives an additive functor with $H_{n} = {\overline{H}}_{n} Q$ . Uniqueness is immediate from [L4]: every morphism of $K (A)$ is a class $[f]$ , so any factorization must send $[f]$ to $H_{n} (f)$ .

L2L3L4step 1.1discharge-construct∎

Depends on

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Sources

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)
The Stacks Project, Section 13.8: The homotopy category (standard reference, not scraped)