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CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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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.

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Universal delta functors extending the same degree-zero functor are uniquely isomorphic

Statement

Let F be a fixed degree-zero functor.

If S and T are universal homological delta functors equipped with chosen natural isomorphisms σ:S0F,τ:T0F, then there is a unique isomorphism of homological delta functors ST whose degree-zero part is τ1σ:S0T0.

If S and T are universal cohomological delta functors equipped with chosen natural isomorphisms σ:S0F,τ:T0F, then there is a unique isomorphism of cohomological delta functors ST whose degree-zero part is τ1σ:S0T0.

Facts & Assumptions

Given: Two universal delta functors together with chosen degree-zero identifications to F.

[L1]

Universality for homological and cohomological delta functors is the unique extension property from degree zero (Universal delta functor).

[L2]

Morphisms of delta functors are the degreewise maps compatible with the connecting morphisms (Morphism of homological delta functors, Morphism of cohomological delta functors).

Proof

technique · direct
1.1

In the homological case, set t0=τ1σ:S0T0. Applying [L1] to t0 gives a morphism ϕ:ST, and applying [L1] to t01 gives a morphism ψ:TS. Their composites extend idS0 and idT0 respectively, so uniqueness in [L1] forces ψϕ=idS and ϕψ=idT. Hence ϕ is the unique isomorphism extending τ1σ.

L1L2givenalgebra
2.1

The cohomological case is identical, with the maps oriented out of the universal functors as required by [L1].

L1L2step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources