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

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A chain map induces a well-defined map on homology

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map. For every $n \in Z$ there is a unique morphism $H_{n} (f) : H_{n} (C) \to H_{n} (D)$ such that the quotient maps from cycles to homology commute with $Z_{n} (f)$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ and an integer $n$ .

[L1]

The maps $Z_{n} (f) : Z_{n} (C) \to Z_{n} (D)$ and $B_{n} (f) : B_{n} (C) \to B_{n} (D)$ exist and are compatible with the boundary and cycle inclusions (A chain map carries cycles to cycles and boundaries to boundaries).

[L2]

$H_{n} (C)$ and $H_{n} (D)$ are the cokernels of the canonical maps $B_{n} (C) \to Z_{n} (C)$ and $B_{n} (D) \to Z_{n} (D)$ (Homology object of a chain complex).

[L3]

A cokernel is universal among arrows that kill the map being quotiented (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1

Let $β_{C} : B_{n} (C) \to Z_{n} (C)$ and $β_{D} : B_{n} (D) \to Z_{n} (D)$ be the boundary-to-cycle maps. Compatibility in [L1] means $Z_{n} (f) β_{C} = β_{D} B_{n} (f) .$ Therefore the composite $q_{D} Z_{n} (f)$ kills $β_{C}$ , where $q_{D} : Z_{n} (D) \to H_{n} (D)$ is the homology quotient.

L1L2givenalgebra

2.1

Since $q_{D} Z_{n} (f)$ annihilates $β_{C}$ , the cokernel property [L3] for $q_{C} : Z_{n} (C) \to H_{n} (C)$ gives a unique morphism $H_{n} (f) : H_{n} (C) \to H_{n} (D)$ with $H_{n} (f) q_{C} = q_{D} Z_{n} (f) .$ By [L2], this is exactly the induced map on homology.

L2L3step 1.1constructdischarge-construct∎

Depends on

Used by

Quasi-isomorphism Definition
A chain map computed on cycles, boundaries, and homology Example
FALSE: a chain map is determined by its maps on homology False statement
Homology respects identities and composition Proposition
Homology is an additive functor Theorem

Dependency tree · two levels

10 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)
The Stacks Project, Section 12.13: Complexes (standard reference, not scraped)