TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The covering criterion for exactness

Statement

For a composable pair $X \overset{f}{\to} Y \overset{g}{\to} Z$ in an abelian category, the following are equivalent:

the pair is exact at $Y$ ;
$g f = 0$ , and for every morphism $h : W \to Y$ with $g h = 0$ , there exist an object $V$ , an epimorphism $k : V \to W$ , and a morphism $l : V \to X$ such that $h k = f l .$

Facts & Assumptions

Given: The composable pair $X \overset{f}{\to} Y \overset{g}{\to} Z$ .

[L1]

Exactness at $Y$ is equivalent to the member criterion $g y \equiv 0 \Rightarrow \exists x, f x \equiv y$ (Exactness is detected by members).

[L2]

Member equivalence means equality after precomposition by one common pair of epimorphisms (Equivalence of members).

Proof

technique · direct

1.1

Assume the pair is exact. Then [L1] gives $g f = 0$ . Now let $h : W \to Y$ satisfy $g h = 0$ . Then $g h \equiv 0$ , so [L1] gives a member $x : U \to X$ with $f x \equiv h$ . By [L2], there exist an object $V$ and epimorphisms $a : V \to U$ and $k : V \to W$ such that $f x a = h k$ . Putting $l : = x a$ proves the covering condition.

L1L2assume-hypchooseconstruct

1.2

Assume $g f = 0$ and the covering condition. Let $y : W \to Y$ be a member with $g y \equiv 0$ . Choose an epic $u : W^{'} \to W$ with $g y u = 0$ , and apply the covering condition to $h : = y u$ . This gives an epic $k : V \to W^{'}$ and a map $l : V \to X$ with $y u k = f l$ . Since $u k$ is epic, [L2] says exactly that $f l \equiv y$ . Therefore [L1] gives exactness at $Y$ .

L1L2assume-hypconstruct

2.1

Thus the covering condition is equivalent to exactness.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

20 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

The Stacks Project, Section 12.5, Lemma 12.5.15 (standard reference, not scraped)
David Mehrle, Category Theory, Part III, Chapter 7 (standard reference, not scraped)