TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29

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.

Splitting lemma in an abelian category

Statement

Let $0 \to A \overset{i}{\to} B \overset{p}{\to} C \to 0$ be a short exact sequence in an abelian category.

If $s : C \to B$ satisfies $p s = 1_{C}$ , then there is a unique morphism $π : B \to A$ such that $π i = 1_{A}, i π + s p = 1_{B} .$
If $π : B \to A$ satisfies $π i = 1_{A}$ , then there is a unique morphism $s : C \to B$ such that $p s = 1_{C}, i π + s p = 1_{B} .$

In either case the sequence is split.

Facts & Assumptions

Given: The short exact sequence in the statement.

[L1]

In a short exact sequence, $i$ is a kernel of $p$ and $p$ is a cokernel of $i$ (A short exact sequence is a kernel-cokernel pair).

[L2]

A split short exact sequence is exactly one equipped with maps $s, π$ satisfying $p s = 1_{C}$ , $π i = 1_{A}$ , and $i π + s p = 1_{B}$ (Split short exact sequence in an abelian category).

Proof

technique · constructive

1.1L1givenconstructalgebra

Assume $p s = 1_{C}$ and put $t : = 1_{B} - s p$ . Then $p t = 0$ , so because $i$ is a kernel of $p$ by [L1], there is a unique $π : B \to A$ with $i π = t = 1_{B} - s p$ .

1.2L1givenconstructalgebra

Conversely assume $π i = 1_{A}$ and put $u : = 1_{B} - i π$ . Then $u i = 0$ , so because $p$ is a cokernel of $i$ by [L1], there is a unique $s : C \to B$ with $s p = u = 1_{B} - i π$ .

2.1L1step 1.1algebra

Composing $i π = 1_{B} - s p$ with $i$ gives $i π i = i - s p i = i$ , because $p i = 0$ by [L1]. Since $i$ is monic, $π i = 1_{A}$ , and the defining equation also yields $i π + s p = 1_{B}$ .

2.2L1step 1.2algebra

Composing $s p = 1_{B} - i π$ with $p$ gives $p s p = p$ , and epicity of $p$ forces $p s = 1_{C}$ . The same equation already gives $i π + s p = 1_{B}$ .

3.1L1L2step 1.1step 2.1

If $π^{'}$ satisfies the same two identities, then $i π^{'} = 1_{B} - s p = i π$ , so monicity of $i$ gives $π^{'} = π$ . Hence the map is unique, and [L2] says the sequence is split.

3.2L1L2step 1.2step 2.2

If $s^{'}$ satisfies the same two identities, then $s^{'} p = 1_{B} - i π = s p$ , so epicity of $p$ gives $s^{'} = s$ . Hence the map is unique, and [L2] again says the sequence is split.

4.1step 1.1step 2.1step 3.1step 1.2step 2.2step 3.2discharge-construct∎

Steps 1.1 to 3.1 prove claim 1, and steps 1.2 to 3.2 prove claim 2.

Depends on

Used by

Dependency tree · two levels

15 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 12.5.10 (standard reference, not scraped)
Saunders Mac Lane, Categories for the Working Mathematician, VIII.4 (standard reference, not scraped)