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

Dependency tree · 2 levels

Sources

Saunders Mac Lane, Categories for the Working Mathematician, VIII.3