Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

One-step extension of a partial projective resolution

Statement

Let PnPn1P0εA0 be an augmented chain complex that is exact at every displayed term except possibly at Pn. Let Kn be the kernel of the previous displayed map, so K0=ker(ε) and Kn=ker(PnPn1) for n1.

If q:Pn+1Kn is an epimorphism from a projective object Pn+1, then composing q with the kernel inclusion KnPn extends the complex by one term and makes it exact at Pn.

Facts & Assumptions

Given: The displayed partial augmented complex and a chosen epimorphism q:Pn+1Kn with Pn+1 projective.

[L1]

Exactness at a degree means that the image of the incoming differential is the kernel subobject of the outgoing differential (Exactness of a complex at a degree and acyclic complexes).

[L2]

An augmented chain complex records the extra map to the resolved object (Augmented chain complexes over an object).

[L3]

Projective objects are the allowable terms in a projective resolution (Projective object).

Proof

technique · direct
1.1

Let in:KnPn be the kernel inclusion, and define the new differential by dn+1:=inq:Pn+1Pn. Because in lands in the kernel of the previous displayed map, the composite of the new differential with that previous map is zero, so the extended row is again an augmented chain complex in the sense of [L2].

givenL2construct
2.1

The image of dn+1 is the image of inq, which is exactly Kn because q is epic. By [L1], this is precisely the exactness condition at Pn. The new term is projective by the given hypothesis and [L3].

L1L3step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources