LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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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 $n \in Z$ , the boundary inclusion $B_{n} (C) ↪ C_{n}$ factors uniquely through the cycle inclusion $Z_{n} (C) ↪ C_{n}$ .

Facts & Assumptions

Given: A chain complex $C_{∙}$ and an integer $n$ .

[L1]

In a chain complex, $d_{n} d_{n + 1} = 0$ (Chain complex in an abelian category).

[L2]

A kernel $k : Z_{n} (C) \to C_{n}$ is characterized by $d_{n} k = 0$ and universal factorization among arrows annihilated by $d_{n}$ (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

[L3]

Every morphism admits an epic-monic factorization; in particular $d_{n + 1}$ factors as $C_{n + 1} \overset{e}{\to} B_{n} (C) \overset{i}{\to} C_{n}$ with $e$ epic and $i$ monic (Every morphism factors as an epimorphism followed by a monomorphism, uniquely up to unique isomorphism).

Proof

technique · direct

1.1L1L3givenalgebra

By [L3], write $d_{n + 1} = i e$ with $i : B_{n} (C) \to C_{n}$ monic and $e$ epic. Then $d_{n} i e = d_{n} d_{n + 1} = 0$ by [L1], so epicity of $e$ gives $d_{n} i = 0$ .

2.1L2step 1.1constructdischarge-construct∎

Since $d_{n} i = 0$ , the kernel property in [L2] gives a unique morphism $β_{n} : B_{n} (C) \to Z_{n} (C)$ with $k β_{n} = i$ , where $k : Z_{n} (C) \to C_{n}$ is the cycle inclusion. That is exactly the required factorization.

The boundary subobject factors through the cycle subobject

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources