TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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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

1.1

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 $i π = t = 1_{B} - s p$ .

L1givenconstructalgebra

1.2

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 $s p = u = 1_{B} - i π$ .

L1givenconstructalgebra

2.1

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 $i π + s p = 1_{B}$ .

L1step 1.1algebra

2.2

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 $i π + s p = 1_{B}$ .

L1step 1.2algebra

3.1

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.

L1L2step 1.1step 2.1

3.2

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.

L1L2step 1.2step 2.2

4.1

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

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

Depends on

Used by

A non-split short exact sequence of abelian groups Counterexample
The splitting lemma instantiated at the published module theorem Example
FALSE: a short exact sequence splits whenever its middle object is isomorphic to the biproduct of the outer two False statement

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)