Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

Projective comparison maps exist

Statement

Assume the Axiom of Dependent Choice.

Let u:AB be a morphism, and let PA and QB be projective resolutions. Then there exists an augmentation-preserving chain map f:PQ lifting u.

Facts & Assumptions

Given: A morphism u:AB and projective resolutions PA, QB.

[L1]

Degree zero can be lifted across the target augmentation (Lifting a map through degree zero of a projective resolution).

[L2]

A partial comparison map extends one degree at a time (Extending a partial comparison map by one degree).

[L3]

The required notion is an augmentation-preserving chain map (Augmentation-preserving maps of projective resolutions).

[L4]

Dependent choice licenses the countable successor-by-successor selection of compatible lifts (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain).

Proof

technique · direct
1.1

By [L1], choose f0:P0Q0 with εQf0=uεP.

L1construct
2.1

Starting from the degree-zero lift in step 1.1, every partial comparison map through degree n extends to one through degree n+1 by [L2]. The successive choices depend on the previously chosen partial map, so [L4] produces maps fn in every degree.

L2L4step 1.1choose
3.1

The family (fn) is a chain map by construction, and step 1.1 gives the augmentation identity. Hence [L3] is satisfied, so f is a comparison map lifting u.

L3step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources