TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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 arrow-theoretic criterion for exactness

Statement

Let $A \overset{f}{\to} B \overset{g}{\to} C$ be composable morphisms in an abelian category, let $K \overset{k}{\to} B$ be a kernel of $g$ , and let $B \overset{q}{\to} Q$ be a cokernel of $f$ .

Then the pair is exact at $B$ if and only if both $g f = 0 and q k = 0.$

Facts & Assumptions

Given: The composable pair $A \overset{f}{\to} B \overset{g}{\to} C$ , a kernel $k : K \to B$ of $g$ , and a cokernel $q : B \to Q$ of $f$ .

[L1]

Exactness at $B$ means $[im (f)] = [\ker (g)]$ , equivalently $[coker (f)] = [coim (g)]$ (Exactness at a node, Image and coimage in a category with kernels and cokernels).

[L2]

For the image factorization $f = m e$ , one has $[⁣ m ⁣] \leq [⁣ k ⁣]$ if and only if $g f = 0$ , and $[⁣ k ⁣] \leq [⁣ m ⁣]$ if and only if every morphism killed by $g$ factors through $m$ (The subobject inequalities underlying exactness).

[L3]

Kernels and cokernels are characterized by the usual vanishing and universal factorization properties (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1

Assume the pair is exact at $B$ . Then $[im (f)] = [\ker (g)]$ , so [L2] gives $g f = 0$ .

L1L2

1.2

Write $f = m e$ for an image factorization. Exactness gives $k = m u$ for some $u : K \to im (f)$ , and $q f = q m e = 0$ implies $q m = 0$ because $e$ is epic. Hence $q k = q m u = 0$ .

L1L2L3algebra

1.3

Assume now that $g f = 0$ and $q k = 0$ . Writing again $f = m e$ , the equality $g f = 0$ gives $[⁣ m ⁣] \leq [⁣ k ⁣]$ by [L2].

2.1

Since $q f = q m e = 0$ and $e$ is epic, one has $q m = 0$ . Together with the hypothesis $q k = 0$ , the cokernel property in [L3] yields $u : K \to im (f)$ with $m u = k$ , hence $[⁣ k ⁣] \leq [⁣ m ⁣]$ .

L3step 1.3algebra

3.1

Steps 1.3 and 2.1 give $[im (f)] = [\ker (g)]$ , so the pair is exact at $B$ by [L1]. With steps 1.1 and 1.2, this proves the equivalence.

L1step 1.1step 1.2step 1.3step 2.1∎

Depends on

Used by

A short exact sequence is a kernel-cokernel pair Theorem

Dependency tree · two levels

10 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

Peter Freyd, Abelian Categories, Theorem 2.21 (standard reference, not scraped)
David Mehrle, Category Theory, Part III, Definition 7.20 (standard reference, not scraped)