TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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The image is the least subobject through which a morphism factors

Statement

Let $f : A \to B$ be a morphism in an abelian category, and let $i_{f} : im (f) \to B$ be the image inclusion. Then $f$ factors through $i_{f}$ , and if $f = n \circ g$ with $n : N \to B$ monic, then

$[im (f)] \leq [n]$

in the subobject order of $B$ .

Facts & Assumptions

Given: An abelian category, a morphism $f : A \to B$ , and a factorization $f = n \circ g$ through a monomorphism $n : N \to B$ .

[L1]

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

[L2]

A subobject is a mutual-factorization class of monomorphisms, ordered by factorization (Subobject and quotient object as mutual-factorisation classes of monomorphisms and epimorphisms, Subobjects and quotient objects form oppositely oriented partially ordered collections).

[L3]

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

1.1L1L2

By [L1], $f$ admits an epic-monic factorization $f = i_{f} \circ e_{f}$ , so it factors through its image.

2.1L1L3step 1.1

Because $f = n g$ , the composite $coker (n) \circ f$ is zero. Using step 1.1, this becomes $coker (n) i_{f} e_{f} = 0$ . Since $e_{f}$ is epic, $coker (n) i_{f} = 0$ . Now [L3] says that $n$ is a kernel of $coker (n)$ , so $i_{f}$ factors uniquely through $n$ .

3.1L2step 2.1∎

The factorization in step 2.1 is exactly the order relation $[im (f)] \leq [n]$ from [L2]. Hence the image is the least subobject of $B$ through which $f$ factors.

Depends on

Used by

The axioms AB5 and AB5* Definition
Lifting a morphism from an exact complex into an injective resolution Lemma
AB5 is equivalent to exactness of filtered colimits Theorem
Direct and inverse image of subobjects form a Galois connection Theorem
Epimorphy is detected by members Theorem
Members modulo equivalence correspond to subobjects Theorem
The join of two subobjects is their least upper bound Theorem
The subobject inequalities underlying exactness Theorem

Dependency tree · two levels

18 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, VIII.3 (standard reference, not scraped)