TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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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.

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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

1.1L1given

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 $k_{f} u = α k_{β}$ .

2.1step 1.1L1∎

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.

Depends on

Used by

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)