Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The boundary subobject factors through the cycle subobject

Statement

Let C be a chain complex in an abelian category. For every nZ, the boundary inclusion Bn(C)Cn factors uniquely through the cycle inclusion Zn(C)Cn.

Facts & Assumptions

Given: A chain complex C and an integer n.

[L1]

In a chain complex, dndn+1=0 (Chain complex in an abelian category).

[L2]

A kernel k:Zn(C)Cn is characterized by dnk=0 and universal factorization among arrows annihilated by dn (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

[L3]

Every morphism admits an epic-monic factorization; in particular dn+1 factors as Cn+1eBn(C)iCn with e epic and i monic (Every morphism factors as an epimorphism followed by a monomorphism, uniquely up to unique isomorphism).

Proof

technique · direct
1.1

By [L3], write dn+1=ie with i:Bn(C)Cn monic and e epic. Then dnie=dndn+1=0 by [L1], so epicity of e gives dni=0.

L1L3givenalgebra
2.1

Since dni=0, the kernel property in [L2] gives a unique morphism βn:Bn(C)Zn(C) with kβn=i, where k:Zn(C)Cn is the cycle inclusion. That is exactly the required factorization.

L2step 1.1constructdischarge-construct

Depends on

Used by

Dependency tree · two levels

12 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