Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28
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.

Projective object characterisations

Statement

For an object P of an abelian category, the following are equivalent:

  1. P is projective.
  2. For every short exact sequence 0→K→E→M→0, the induced sequence 0→A(P,K)→A(P,E)→A(P,M)→0 is exact.
  3. Every epimorphism E↠P splits.

Facts & Assumptions

Given: An object P in an abelian category.

[L1]

Projectivity is the lifting property against epimorphisms (Projective object).

[L2]

In an abelian category, the pullback of an epimorphism is an epimorphism (The pullback of an epimorphism is an epimorphism).

[F1]

For every short exact sequence, the functor A(P,−) is left exact; projectivity is exactly the extra surjectivity at the right-hand end.

Proof

technique · direct
1.1L1F1assume-hyp

Assume P is projective. Then [L1] gives a lift of every map P→M across every epimorphism E↠M, so the last map in condition 2 is surjective. Together with the left exactness in [F1], this proves condition 2.

1.2L1assume-hyp

Condition 2 clearly implies condition 1, because surjectivity of A(P,E)→A(P,M) for every short exact sequence is exactly the lifting property [L1].

1.3L1assume-hyp

If P is projective and q:E↠P is epic, apply [L1] to 1P:P→P. A lift s:P→E with qs=1P is a section, so q splits.

1.4L2assume-hypconstruct

Assume condition 3. Given an epimorphism q:E↠M and a map f:P→M, form the pullback of q along f. By [L2], its projection to P is epic, so condition 3 makes it split. Composing such a section with the other pullback leg gives a lift of f across q. Thus P is projective.

2.1step 1.1step 1.2step 1.3step 1.4∎

Steps 1.1 and 1.2 prove 1⇔2, and steps 1.3 and 1.4 prove 1⇔3. Hence all three conditions are equivalent.

Depends on

Used by

Dependency tree · two levels

7 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