Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Left total derived functor is independent up to a unique augmentation-compatible natural isomorphism

Statement

Two supplied projective replacement systems for the same additive F give a natural isomorphism of left total derived functors, unique among natural comparisons commuting with the augmentations. This is not uniqueness of unrestricted natural automorphisms.

Facts & Assumptions

Given: Two supplied projective replacement systems for the same additive F give a natural isomorphism of left total derived functors, unique among natural comparisons commuting with the augmentations. This is not uniqueness of unrestricted natural automorphisms.

[F1]

The replacement construction has augmentation QF(pX) (Left total derived functor on the bounded above derived category).

[F2]

Hom out of a K-projective complex needs no roof (Morphisms from a homotopically projective complex need no roof).

Proof

1.1

Write pX:PXX and pX:PXX. The no-roof bijection gives a unique map cX:PXPX in K such that pXcX=pX in K. The opposite comparison is inverse by the same uniqueness. This works for zero complexes and identity replacements. Applying additive F preserves these homotopy identities.

F1F2
2.1

For a derived arrow u:XY the two paths between the chosen projective models have the same localized image, hence coincide in K by the no-roof bijection. Thus QF(cX) is a natural isomorphism commuting with augmentations.

F2step 1.1
3.1

On any projective complex P, the augmentation of either construction is invertible: its replacement map is a homotopy equivalence by step 1.1 applied to the identity replacement. Therefore augmentation compatibility forces a comparison at P to be (ϵP)1ϵP. Naturality along the isomorphism Q(pX):QPXQX then forces its value at every X. This proves precisely the stated uniqueness.

F1step 1.1step 2.1algebra

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