TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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Homology is an additive functor

Statement

For each $n \in Z$ , homology defines an additive functor $H_{n} : Ch (A) \to A .$

Facts & Assumptions

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

[L1]

A chain map induces a map on homology (A chain map induces a well-defined map on homology).

[L2]

Those induced maps respect identities and composition (Homology respects identities and composition).

[L3]

An additive functor is a functor that is additive on each hom-group (Additive functor).

[L4]

An abelian category is additive, so $Ch (A)$ is also additive and sums of chain maps are defined degreewise (Abelian category, The category of complexes in an additive category is additive).

[L5]

Kernels are universal among arrows annihilated by the displayed map (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1

By [L1] and [L2], the assignment $C \mapsto H_{n} (C)$ and $f \mapsto H_{n} (f)$ is already a functor.

L1L2

1.2

Let $f, g : C \to D$ be chain maps. By [L4], their sum $f + g$ is the chain map with components $f_{n} + g_{n}$ . Let $k_{C} : Z_{n} (C) \to C_{n}, k_{D} : Z_{n} (D) \to D_{n}$ be the cycle inclusions and $q_{C} : Z_{n} (C) \to H_{n} (C), q_{D} : Z_{n} (D) \to H_{n} (D)$ the homology quotients. Then $k_{D} (Z_{n} (f) + Z_{n} (g)) = f_{n} k_{C} + g_{n} k_{C} = (f_{n} + g_{n}) k_{C} = k_{D} Z_{n} (f + g) .$ Since both maps on cycles are killed by $d_{n}^{D}$ , the uniqueness in the kernel property [L5] for $k_{D}$ gives $Z_{n} (f + g) = Z_{n} (f) + Z_{n} (g) .$ Therefore $(H_{n} (f) + H_{n} (g)) q_{C} = H_{n} (f) q_{C} + H_{n} (g) q_{C} = q_{D} Z_{n} (f) + q_{D} Z_{n} (g) = q_{D} Z_{n} (f + g) .$ By the uniqueness clause in [L1], this forces $H_{n} (f + g) = H_{n} (f) + H_{n} (g) .$

L1L4L5algebra

2.1

Steps 1.1 and 1.2 are exactly the functoriality and additivity demanded by [L3]. Hence $H_{n}$ is an additive functor.

L3step 1.1step 1.2∎

Depends on

Used by

Finite biproducts of complexes are computed degreewise Proposition

Dependency tree · two levels

17 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 12.13: Complexes (standard reference, not scraped)
Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)