TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The subobject inequalities underlying exactness

Statement

Let $A \overset{f}{\to} B \overset{g}{\to} C$ be composable morphisms in an abelian category. Choose an epi-mono factorization $A \overset{e}{\to} im (f) \overset{m}{\to} B$ of $f$ , and choose a kernel $K \overset{k}{\to} B$ of $g$ .

Then:

$[⁣ m ⁣] \leq [⁣ k ⁣]$ if and only if $g f = 0$ .
$[⁣ k ⁣] \leq [⁣ m ⁣]$ if and only if every morphism $u : U \to B$ with $g u = 0$ factors through $m$ .

Facts & Assumptions

Given: The composable pair $A \overset{f}{\to} B \overset{g}{\to} C$ , the factorization $f = m e$ with $e$ epic and $m$ monic, and the kernel $k : K \to B$ of $g$ .

[L1]

Every morphism in an abelian category admits an epimorphism-monomorphism factorization, unique up to unique isomorphism (Every morphism factors as an epimorphism followed by a monomorphism, uniquely up to unique isomorphism).

[L2]

Subobjects are ordered by factorization of monomorphisms, and the image is the least subobject through which the morphism factors (Subobject and quotient object as mutual-factorisation classes of monomorphisms and epimorphisms, The image is the least subobject through which a morphism factors).

[L3]

The kernel $k$ satisfies $g k = 0$ , and every morphism killed by $g$ factors uniquely through $k$ (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1L2L3algebra

If $[⁣ m ⁣] \leq [⁣ k ⁣]$ , then $m = k t$ for some $t : im (f) \to K$ by [L2], so $g f = g m e = g k t e = 0$ by [L3].

1.2L1L2L3algebra

If $g f = 0$ , then $g m e = 0$ , and the epicity of $e$ from [L1] gives $g m = 0$ . So [L3] gives $t : im (f) \to K$ with $k t = m$ , hence $[⁣ m ⁣] \leq [⁣ k ⁣]$ by [L2].

1.3L2L3

If $[⁣ k ⁣] \leq [⁣ m ⁣]$ , then $k = m s$ for some $s : K \to im (f)$ by [L2]. For any $u : U \to B$ with $g u = 0$ , [L3] gives $v : U \to K$ with $u = k v = m s v$ , so $u$ factors through $m$ .

1.4L2L3

Conversely, if every $u : U \to B$ with $g u = 0$ factors through $m$ , then in particular the kernel arrow $k$ does, because $g k = 0$ by [L3]. Thus $k = m s$ for some $s$ , so $[⁣ k ⁣] \leq [⁣ m ⁣]$ by [L2].

2.1step 1.1step 1.2

Steps 1.1 and 1.2 prove the first biconditional.

3.1step 1.3step 1.4∎

Steps 1.3 and 1.4 prove the second biconditional.

Depends on

Used by

Exactness is detected by members Theorem
The arrow-theoretic criterion for exactness Theorem

Cited to discharge well-definedness by Exactness at a node.

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.4 (standard reference, not scraped)
David Mehrle, Category Theory, Part III, Chapter 7 (standard reference, not scraped)