TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30‡ 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.

Four lemma in an abelian category

Statement

Consider a commutative diagram in an abelian category with exact rows

Then:

if $α$ and $γ$ are epic and $δ$ is monic, then $β$ is epic;
if $β$ and $δ$ are monic and $α$ is epic, then $γ$ is monic.

Facts & Assumptions

Given: The commutative exact-row diagram in the statement.

[L1]

Monicity is equivalent to cancellation on members (Monicity by member cancellation).

[L2]

Epicity is equivalent to the member-lifting property (Epimorphy is detected by members).

[L3]

Exactness at a node is equivalent to the member-lifting condition (Exactness is detected by members).

[L4]

The common-refinement construction for member equivalence puts finitely many witness equalities on one epic domain, where hom-set subtraction is defined (Equivalence of members, Member equivalence is transitive, Abelian category).

Proof

technique · direct

1.1

Write the top row as $W \overset{i}{\to} X \overset{j}{\to} Y \overset{p}{\to} Z$ and the bottom row as $W^{'} \overset{i^{'}}{\to} X^{'} \overset{j^{'}}{\to} Y^{'} \overset{p^{'}}{\to} Z^{'}$ . Assume that $α$ and $γ$ are epic and that $δ$ is monic. To prove that $β$ is epic, let $x^{'}$ be a member of $X^{'}$ . Since $γ$ is epic, [L2] gives a member $y$ of $Y$ with $γ y \equiv j^{'} x^{'}$ . Then $δ p y = p^{'} γ y \equiv p^{'} j^{'} x^{'} \equiv 0.$ Because $δ$ is monic, [L1] gives $p y \equiv 0$ . Exactness of the top row at $Y$ now gives a member $x_{0}$ of $X$ with $j x_{0} \equiv y$ by [L3].

L1L2L3assume-hypconstruct

1.2

Assume instead that $β$ and $δ$ are monic and that $α$ is epic. To prove that $γ$ is monic, let $y$ and $y^{'}$ be members of $Y$ with $γ y \equiv γ y^{'}$ . Then $δ p y = p^{'} γ y \equiv p^{'} γ y^{'} = δ p y^{'},$ so [L1] gives $p y \equiv p y^{'}$ . By [L4], replace $y$ and $y^{'}$ by representatives on one common epic refinement of the witnesses for both equalities and define $t : = y - y^{'}$ . Then $γ t = p t = 0$ and $y = y^{'} + t$ on that domain. Exactness of the top row at $Y$ gives a member $x$ of $X$ with $j x \equiv t$ by [L3].

L1L3L4assume-hypchooseconstructalgebra

2.1

From step 1.1 we get $j^{'} β x_{0} = γ j x_{0} \equiv γ y \equiv j^{'} x^{'} .$ By [L4], replace $β x_{0}$ and $x^{'}$ by representatives on a common epic domain and define $z^{'} : = x^{'} - β x_{0}$ . Then $j^{'} z^{'} = 0$ and $x^{'} = β x_{0} + z^{'}$ on that domain. Exactness of the bottom row at $X^{'}$ gives a member $w^{'}$ of $W^{'}$ with $i^{'} w^{'} \equiv z^{'}$ by [L3], and epicity of $α$ gives a member $w$ of $W$ with $α w \equiv w^{'}$ by [L2]. Therefore $β (x_{0} + i w) \equiv β x_{0} + i^{'} α w \equiv β x_{0} + i^{'} w^{'} \equiv β x_{0} + z^{'} \equiv x^{'} .$ So every member of $X^{'}$ lifts along $β$ , and [L2] makes $β$ epic.

L2L3L4step 1.1chooseconstructalgebra

2.2

From step 1.2 we get $j^{'} β x = γ j x \equiv γ t \equiv 0.$ Exactness of the bottom row at $X^{'}$ therefore gives a member $w^{'}$ of $W^{'}$ with $i^{'} w^{'} \equiv β x$ by [L3]. Because $α$ is epic, [L2] gives a member $w$ of $W$ with $α w \equiv w^{'}$ . Then $β (x - i w) \equiv β x - i^{'} α w \equiv β x - i^{'} w^{'} \equiv 0.$ Since $β$ is monic, [L1] yields $x \equiv i w$ . Therefore $t \equiv j x \equiv j i w \equiv 0,$ because the top row is a complex. So $y \equiv y^{'}$ , and [L1] makes $γ$ monic.

L1L2L3step 1.2algebra

3.1

Therefore the four lemma holds in both the epic and the monic form stated above.

step 1.2step 2.2∎

Depends on

Used by

The published module four lemma as an instance Example
The two halves of the four lemma are mutually dual Remark
Why the five lemma asks for isomorphisms in the middle Remark
An exact functor transports every diagram lemma Theorem
Sharp five lemma in an abelian category Theorem
The diagram lemmas hold in the opposite category Theorem

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 12.5.19 (standard reference, not scraped)