TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-29

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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.
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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

1.1L1L2

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

1.2L1L2L3algebra

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 $q k = q m u = 0$ .

1.3L2

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].

2.1L3step 1.3algebra

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 $[⁣ k ⁣] \leq [⁣ m ⁣]$ .

3.1L1step 1.1step 1.2step 1.3step 2.1∎

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.

Depends on

Used by

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)