TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28‡ 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.

In a pullback square, the induced map on the kernels of the two parallel arrows is an isomorphism

Statement

In a pullback square in an abelian category

the induced morphism from $\ker (β)$ to $\ker (f)$ is an isomorphism.

Facts & Assumptions

Given: The displayed pullback square in an abelian category.

[L1]

Abelian categories have pullbacks, and the square above is one (Abelian category, A pullback is the kernel of the difference of the two legs, and dually for pushouts).

Proof

technique · direct

1.1

Let $k_{β} : K_{β} \to P$ be a kernel of $β$ and let $k_{f} : K_{f} \to A$ be a kernel of $f$ . Since $g β k_{β} = f α k_{β} = 0$ , the kernel property of $k_{f}$ gives a unique map $u : K_{β} \to K_{f}$ with $k_{f} u = α k_{β}$ .

L1given

2.1

Because $f k_{f} = 0 = g 0$ , the pullback universal property gives a unique map $s : K_{f} \to P$ with $α s = k_{f}$ and $β s = 0$ . Since $β s = 0$ , the kernel property of $k_{β}$ gives a unique map $v : K_{f} \to K_{β}$ with $k_{β} v = s$ . Then $k_{f} u v = α k_{β} v = α s = k_{f}$ , so $u v = 1$ by monicity of $k_{f}$ ; similarly $k_{β} v u = s = k_{β}$ , so $v u = 1$ by monicity of $k_{β}$ . Thus $u$ is an isomorphism.

step 1.1L1∎

Depends on

Used by

A square with monic legs is a pullback exactly when it identifies the source with the intersection subobject Theorem

Dependency tree · two levels

11 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 12.5.12 (standard reference, not scraped)