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✓ 7 results · all verified · 5 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Delta Functors and Universality — Examples

1 · Prerequisites

2 · Summary

These examples keep the abstract delta-functor axioms tied to the concrete derived-functor constructions already on disk. They show how long exact sequences, dimension shifting, and universality are used in practice, and they also isolate the main failure mode: exactness alone is not enough without naturality of the connecting maps.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Homology as a homological delta functor

Example

Fix an abelian category A. For chain complexes in A that vanish in negative degrees, the family Hn:Ch⁡≥0(A)→A,n≥0, together with the usual connecting morphisms of homology, is a homological delta functor.

Facts & Assumptions

Given: An abelian category A and its abelian category of nonnegatively graded chain complexes.

[L1]

This page's concrete homology family is the candidate homological delta functor (The homological delta-functor carried by homology of complexes).

[L2]

Homology of complexes satisfies the exactness and naturality axioms (Homology of complexes satisfies the delta-functor naturality and exactness laws).

[L3]

A homological delta functor is exactly such a family of additive functors with natural connecting maps (Homological delta functor).

Verification

technique · direct
1.1L1given

Restrict the integer-indexed homology family of [L1] to complexes that vanish in negative degrees. This full subcategory is closed under kernels and cokernels degreewise, hence is abelian, and its short exact sequences have H−1=0.

2.1L2L3step 1.1∎

The exactness and naturality axioms required in [L3] are supplied by [L2]; step 1.1 makes the integer-indexed long sequence terminate as H0(C)→0. Therefore the restricted family is a homological delta functor indexed by n≥0.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The trivial delta functor of an exact functor

Example

Assume the Axiom of Dependent Choice.

If A and B are abelian categories and F:A→B is exact, then the delta functor concentrated in degree 0, T0=F,Tn=0 for n>0, with zero connecting maps, is a universal homological delta functor; likewise the cohomological family T0=F,Tn=0 for n>0 with zero connecting maps is universal.

Facts & Assumptions

Given: An exact functor F between abelian categories.

[L1]

An exact base functor has precisely these trivial universal delta functors (An exact base functor has the trivial universal delta functor).

Verification

technique · direct
1.1L1given

The displayed homological and cohomological families are exactly the two families identified in [L1].

2.1L1step 1.1∎

Therefore both are universal delta functors.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

One dimension shift along a projective presentation

Example

Assume the Axiom of Dependent Choice.

Let P be supplied projective resolution data on a class D of objects of A, let F:A→B be additive and right exact, and let 0→K→Q→A→0 be a short exact sequence in D with Q projective. Then for every n>1 the connecting map of the derived long exact sequence gives an isomorphism LnPF(A)→ ∼ Ln−1PF(K).

Facts & Assumptions

Given: A short exact sequence 0→K→Q→A→0 in D with Q projective and an integer n>1.

[L1]

The left derived functors form a homological delta functor (Left derived functors form a homological delta functor).

[L2]

Positive left derived functors vanish on projective objects (Positive left derived functors vanish on projective objects).

[L3]

When the outer maps in the exact segment vanish, the connecting map is an isomorphism (Dimension shift for a homological delta functor effaced in the middle).

Verification

technique · direct
1.1L1givenconstruct

By [L1], the given short exact sequence yields an exact segment LnPF(Q)→LnPF(A)→∂nLn−1PF(K)→Ln−1PF(Q).

2.1L2L3step 1.1∎

Since Q∈D is projective and n>1, [L2] gives LnPF(Q)=Ln−1PF(Q)=0. Therefore [L3] turns the connecting map in step 1.1 into the displayed isomorphism.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

One dimension shift along an injective copresentation

Example

Assume the Axiom of Dependent Choice.

Let I be supplied injective resolution data on a class D of objects of A, let F:A→B be additive and left exact, and let 0→A→J→C→0 be a short exact sequence in D with J injective. Then for every n>1 the connecting map gives an isomorphism RIn−1F(C)→ ∼ RInF(A).

Facts & Assumptions

Given: A short exact sequence 0→A→J→C→0 in D with J injective and an integer n>1.

[L1]

The right derived functors form a cohomological delta functor (Right derived functors form a cohomological delta functor).

[L2]

Positive right derived functors vanish on injective objects (Positive right derived functors vanish on injective objects).

[L3]

Vanishing of the adjacent injective terms makes the connecting map an isomorphism (Dimension shift for a cohomological delta functor effaced in the middle).

Verification

technique · direct
1.1L1givenconstruct

By [L1], the given short exact sequence yields an exact segment RIn−1F(J)→RIn−1F(C)→∂n−1RInF(A)→RInF(J).

2.1L2L3step 1.1∎

Since J∈D is injective and n>1, [L2] gives RIn−1F(J)=RInF(J)=0. Hence [L3] makes the connecting map in step 1.1 an isomorphism.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-05Open item page →

Extending a degree-zero natural transformation

Example

Let S=(Sn,∂S) be a homological delta functor, let T=(Tn,∂T) be an effaceable homological delta functor, and let u0:S0⇒T0 be a natural transformation. For an object A, choose an effacement 0→K→P→pA→0 with T1(p)=0. Then the degree-one component of the universal extension is the unique map u1(A):S1(A)→T1(A) satisfying ∂1T u1(A)=u0(K) ∂1S, and this map is independent of the chosen effacement and compatible with the connecting morphisms.

Facts & Assumptions

Given: A degree-zero natural transformation u0:S0⇒T0 and a chosen effacement of one object A.

[L1]

One dimension shift defines the next-degree component from the chosen effacement (A partial morphism of delta functors extends through one dimension shift).

[L2]

The resulting component is independent of the effacing morphism and commutes with connecting maps (The effacement extension is independent of the effacing morphism, The effacement extension commutes with connecting morphisms).

[L3]

Derived-functor universality is built from exactly this extension mechanism (Derived functors are universal delta functors).

Verification

technique · direct
1.1L1given

The defining equation for u1(A) is exactly the homological case of [L1] with n=1.

2.1L2L3step 1.1∎

Item [L2] removes dependence on the chosen effacement and supplies the required compatibility with connecting morphisms, so the map from step 1.1 is the correct first higher component of the universal extension. This is the degree-one pattern used abstractly in [L3].

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12Open item page →

A nonnatural choice of connecting maps does not form a delta functor

Statement refuted

A family of additive functors together with arbitrary exact connecting maps is automatically a homological delta functor.

Facts & Assumptions

Given: The ordinary homology delta functor on complexes and the canonical cone sequence of the identity on the stalk complex Z[0].

[L1]

A homological delta functor requires both exactness and naturality of the connecting maps (Homological delta functor).

[L2]

Homology of complexes is a genuine homological delta functor (Homology of complexes satisfies the delta-functor naturality and exactness laws).

[L3]

The canonical mapping-cone sequence is degreewise split short exact makes the cone sequence of 1Z[0] short exact, while The cone connecting map agrees with the shifted identity up to the declared sign identifies its map H1(Z[1])→H0(Z[0]) with ±1:Z→Z, hence nonzero.

Counterexample

1.1L2L3givenconstruct

Start with the homology delta functor from [L2]. Keep every functor Hn unchanged and keep every connecting map unchanged except on one chosen short exact sequence with nonzero connector from [L3], where replace ∂ by −∂. Each individual long exact sequence remains exact.

2.1L1L2L3step 1.1algebra∎

Choose a distinct isomorphic copy of the cone sequence in [L3], and alter the connector on only the original sequence. The chosen isomorphism of short exact sequences induces isomorphisms on the two homology groups. Before the alteration, naturality identifies the two routes around the connecting square with the same map ±1:Z→Z. After changing exactly one connector to its negative, the two routes are opposite maps ±1 and ∓1, which are unequal over Z. Hence this naturality square fails, and [L1] shows that the altered family is not a homological delta functor.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-05Open item page →

Two universal delta functors and their unique isomorphism

Example

Assume the Axiom of Dependent Choice.

Let A and B be abelian categories, suppose A has enough projectives, and let F:A→B be additive and right exact. If P and Q are two supplied projective resolution data on all objects of A, then the two universal homological delta functors (LnPF)n≥0and(LnQF)n≥0 are uniquely isomorphic. After choosing the standard natural identifications L0PF≅F≅L0QF, the isomorphism is the unique one whose degree-zero component corresponds to id⁡F.

Facts & Assumptions

Given: Two supplied projective resolution data P and Q on all objects of A for the same right exact functor F.

[L1]

Each of the two derived constructions is a universal homological delta functor (Derived functors are universal delta functors).

[L2]

Two universal delta functors with the same degree-zero term are uniquely isomorphic (Universal delta functors extending the same degree-zero functor are uniquely isomorphic).

[L3]

For every supplied projective resolution datum, the zeroth left derived functor is naturally isomorphic to the original right exact functor (Left derived functors form a homological delta functor).

Verification

technique · direct
1.1L1L3given

By [L1], both (LnPF) and (LnQF) are universal homological delta functors, and [L3] supplies their natural degree-zero identifications with F.

2.1L2step 1.1∎

Choose the natural degree-zero identifications with F from step 1.1. Applying [L2] yields the unique isomorphism of delta functors whose degree-zero part corresponds under those identifications to id⁡F; its higher components are then forced.

Sources