TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Every morphism factors as an epimorphism followed by a monomorphism, uniquely up to unique isomorphism

Statement

Every morphism $f : A \to B$ in an abelian category admits a factorization

$A \overset{e_{f}}{⟶} I_{f} \overset{m_{f}}{⟶} B$

with $e_{f}$ epic and $m_{f}$ monic. If also $f = m^{'} e^{'}$ with $e^{'}$ epic and $m^{'}$ monic, then there is a unique isomorphism $u : I_{f} \to I^{'}$ such that

$m^{'} = m_{f} u^{- 1}, u \circ e_{f} = e^{'} .$

Facts & Assumptions

Given: An abelian category and a morphism $f : A \to B$ .

[L1]

The canonical coimage-to-image map exists (The canonical morphism from the coimage to the image exists and is unique).

[L2]

The coimage projection is epic and the image inclusion is monic (The coimage projection is epic and the image inclusion is monic).

[L3]

In an abelian category the canonical coimage-to-image map is an isomorphism (Abelian category).

[L4]

Abelian categories are balanced (An abelian category is balanced).

[L5]

Every monomorphism is the kernel of its cokernel (Every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel).

Proof

technique · direct

1.1

Let $q_{f} : A \to coim (f)$ and $i_{f} : im (f) \to B$ be the defining maps. By [L1], $f$ factors as $f = i_{f} \overline{f} q_{f}$ , and [L2] makes $q_{f}$ epic and $i_{f}$ monic. Since $\overline{f}$ is an isomorphism by [L3], the composite $e_{f} : = \overline{f} q_{f} : A \to im (f)$ is epic, so $f = i_{f} e_{f}$ is an epic-monic factorization.

L1L2L3

2.1

Suppose also $f = m^{'} e^{'}$ with $e^{'}$ epic and $m^{'}$ monic. Then $coker (m^{'}) f = 0$ , so $coker (m^{'}) i_{f} e_{f} = 0$ , and the epicity of $e_{f}$ gives $coker (m^{'}) i_{f} = 0$ . Since $m^{'}$ is a kernel of its cokernel by [L5], $i_{f}$ factors uniquely through $m^{'}$ as $i_{f} = m^{'} u$ . Reversing the roles of $i_{f}$ and $m^{'}$ gives $m^{'} = i_{f} v$ for a unique $v$ .

L2L5step 1.1

3.1

From step 2.1 one gets $m^{'} = m^{'} u v$ and $i_{f} = i_{f} v u$ , so monicity gives $u v = 1$ and $v u = 1$ . Thus $u$ is an isomorphism. Finally, $m^{'} u e_{f} = i_{f} e_{f} = f = m^{'} e^{'}$ , and monicity of $m^{'}$ gives $u e_{f} = e^{'}$ . Any other comparison map with the same property equals $u$ by the uniqueness in step 2.1.

L4step 2.1∎

Depends on

Used by

Dependency tree · two levels

13 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

Saunders Mac Lane, Categories for the Working Mathematician, Proposition VIII.3.1 (standard reference, not scraped)