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30 results · all verified · 22 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 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Ext and Balanced Resolutions

1 · Prerequisites

2 · Summary

This draft compares the projective and injective resolution constructions of Ext. The comparison uses a first-quadrant Hom double complex with direct-sum totalisation on finite diagonals.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Ext via an injective resolution of the second variable

Definition

Let A be abelian and let I be supplied injective-resolution data. For M and N in its domain, set CIq(M,N)=HomA(M,Iq(N)),dIq(f)=dI(N)qf. Thus CI(M,N) is a cochain complex. Define ExtIn(M,N):=Hn(CI(M,N))(n0). The subscript I remains part of the notation: this is a construction relative to the supplied data, not yet an intrinsic Ext bifunctor.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Ext via a projective resolution of the first variable

Definition

Let A be abelian and let P be supplied projective-resolution data, written homologically with dP:Pq+1(M)Pq(M). For M,N in its domain, set CPq(M,N)=HomA(Pq(M),N),dPq(f)=fdPq+1. Then dPq+1dPq=0 because consecutive differentials of P(M) compose to zero. Define ExtPn(M,N):=Hn(CP(M,N))(n0). The subscript P records the supplied choice; no equality with the injective construction is being made here.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The degree-zero injective construction of Ext is Hom

Statement

Assume the Axiom of Dependent Choice, and let I be supplied injective resolution data on the objects under consideration. For 0NιI0d0I1, fιf gives a natural isomorphism Hom(M,N)ExtI0(M,N).

Facts & Assumptions

Given: The displayed injective resolution and an object M.

Proof

technique · direct
1.1

Exactness gives kerd0=imι; thus ιf is a degree-zero cocycle, and fιf is injective because ι is monic.

givenconstruct
2.1

If d0g=0, then g factors uniquely through ι. Since there are no negative-degree coboundaries, this is the required H0 identification. Precomposition makes it natural in M. For b:NN, choose a comparison extension between the supplied resolutions; its degree-zero square with the coaugmentations commutes, so the identification is natural in N, and the independence lemma makes this map independent of the chosen extension.

step 1.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The degree-zero projective construction of Ext is Hom

Statement

For P1d1P0εM0, precomposition with ε gives a natural isomorphism Hom(M,N)ExtP0(M,N).

Facts & Assumptions

Given: The displayed projective resolution and an object N.

Proof

technique · direct
1.1

A map h:P0N is a zero-cocycle exactly if hd1=0, equivalently if it vanishes on kerε=imd1.

givenconstruct
2.1

Thus h factors uniquely as fε; there are no negative-degree coboundaries. This is the asserted isomorphism and is natural under pre- and postcomposition.

step 1.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06Open item page →

Injective-resolution Ext has the stated bifunctor variance

Statement

Assume the Axiom of Dependent Choice. Let I be supplied injective resolution data on a class D in an abelian category. For every q, injective-resolution Ext is contravariant in M and covariant in N: a:MM and b:NN induce ExtIq(M,N)ExtIq(M,N) and ExtIq(M,N)ExtIq(M,N).

Facts & Assumptions

Given: Objects N,ND, objects M,M, and maps a,b as stated.

Proof

technique · direct
1.1

Precomposition by a is a cochain map Hom(M,I)Hom(M,I). A comparison extension of b is a cochain map II, hence postcomposition gives the second cochain map.

givenconstruct
2.1

Homotopic comparison extensions induce the same cohomology map, so the second map is choice-independent. Identity and composition of the comparison maps give the functor laws, with a reversing arrows.

step 1.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06Open item page →

Projective-resolution Ext has the stated bifunctor variance

Statement

Assume the Axiom of Dependent Choice. Let A be an abelian category and let P() be supplied projective resolution data on every object of A. For every n0, ExtPn(,):Aop×AAb is contravariant in its first variable and covariant in its second variable.

Facts & Assumptions

Given: Morphisms u:MM and v:NN in A.

Proof

technique · direct
1.1

Postcomposition with v is a cochain map Hom(P(M),N)Hom(P(M),N). A comparison lift u~:P(M)P(M) from A morphism has a comparison lift between the supplied projective resolutions gives precomposition u~ in the opposite direction.

givenconstruct
2.1

These maps commute because pre- and postcomposition commute. Two lifts of u are chain-homotopic by Projective comparison maps are unique up to chain homotopy. Precomposing with the homotopy gives a cochain homotopy between the two induced maps on Hom(,N), so they induce the same map on Hn. The comparison identity and composition laws hold up to such homotopy, while postcomposition is strictly functorial. Hence the maps on Hn define the asserted bifunctor.

step 1.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06Open item page →

Positive injective-resolution Ext vanishes on an injective second variable

Statement

Assume the Axiom of Dependent Choice. Let A be an abelian category and let I be a supplied injective-resolution datum on a class D of its objects. If JD is injective, then ExtIn(M,J)=0 for every MD and every integer n>0.

Facts & Assumptions

Given: The supplied datum I on D, an object MD, an injective object JD, and an integer n>0.

Proof

technique · direct
1.1

Apply Positive right derived functors vanish on injective objects to the supplied injective-resolution datum and the additive left exact functor Hom(M,). It compares the supplied resolution of J with the length-zero injective resolution and gives vanishing in every positive degree.

givenconstruct
2.1

By Ext via an injective resolution of the second variable, that right-derived group is precisely ExtIn(M,J), so it is zero for n>0.

step 1.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06Open item page →

Positive projective-resolution Ext vanishes on a projective first variable

Statement

Assume the Axiom of Dependent Choice. Let A be an abelian category with the supplied projective-resolution construction. If P is projective, then for every object N and every q>0, ExtPq(P,N)=0.

Facts & Assumptions

Given: A projective object P and an object N.

[L1]

Projective-resolution Ext is the cohomology of Hom(P,) (Ext via a projective resolution of the first variable).

[L2]

Positive right derived functors computed from arbitrary supplied injective-resolution data vanish on injective objects, assuming Dependent Choice (Positive right derived functors vanish on injective objects).

Proof

technique · direct
1.1

For the additive functor F=HomA(,N) on Aop, the supplied projective resolution of P in A is an injective resolution in Aop, and P is injective there. Thus the cohomology in [L1] is the corresponding right-derived construction, and [L2] compares it with the length-zero resolution of P.

L1L2givenconstruct
2.1

The Hom cochain complex of that length-zero resolution is concentrated in degree zero, so its cohomology is zero for q>0. The comparison in step 1.1 therefore gives ExtPq(P,N)=0.

step 1.1algebra
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The Hom double complex of projective and injective resolutions

Definition

For supplied resolutions P(M)M and NI(N), define the first-quadrant bigraded object Kp,q=HomA(Pp(M),Iq(N))(p,q0). Its horizontal and vertical maps are hp,q(f)=fdP,p+1,vp,q(f)=dIqf. They have bidegrees (1,0) and (0,1) respectively. The first-quadrant restriction is part of the definition used below.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The two Hom double-complex differentials commute before signing

Statement

For the Hom double complex Kp,q=HomA(Pp,Iq), the unsigned horizontal and vertical maps commute:

hp,q+1vp,q=vp+1,qhp,q:Kp,qKp+1,q+1.

Moreover, h2=0=v2. Consequently the signed total differential DKp,q=h+(1)pv satisfies D2=0.

Facts & Assumptions

Given: Projective and injective resolutions PM and NI, with the maps h(f)=fdP and v(f)=dIf.

Proof

technique · direct
1.1

For fKp,q, both mixed composites equal dIqfdP,p+1, so hv(f)=vh(f). The resolution identities also give h2(f)=fdP,p+1dP,p+2=0 and v2(f)=dIq+1dIqf=0.

givenalgebra
2.1

Since the horizontal degree increases from p to p+1 after applying h, D2f=h2f+(1)p+1vhf+(1)phvf+v2f=0.

step 1.1algebra
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The direct-sum total complex on finite diagonals

Definition

For the first-quadrant Hom bicomplex K, put TotnK=p+q=nKp,q,DKp,q=hp,q+(1)pvp,q. Each diagonal has precisely n+1 possible bidegrees, so this direct sum is finite. Consequently there is no choice here between direct-sum and product totalisations.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Acyclic assembly by exact columns

Statement

Let Kp,q be a first-quadrant double cochain complex whose signed total complex uses finite direct sums on every diagonal. Suppose a cochain complex C maps to the bottom edge so that, for every p, the augmented column 0CpKp,0vKp,1vKp,2 is exact and the augmentations commute with the horizontal maps. Then the induced cochain map CTotK is a quasi-isomorphism.

Facts & Assumptions

Given: The first-quadrant double complex, compatible column augmentations, and exact augmented columns stated above.

Proof

technique · direct
1.1

Adjoin Cp in vertical degree 1. Compatibility makes this an augmented double complex, and the cone of CTotK is its signed total complex up to shift. In total degree n, only the finitely many columns 0pn+1 occur; filtering by the largest horizontal degree has successive quotients equal to shifts of the exact augmented columns.

givenalgebra
2.1

Starting with one column and adjoining the others, the short exact sequences of successive filtered complexes show inductively that every finite truncation is acyclic. These truncations stabilize degreewise, so the full cone is acyclic. Hence CTotK is a quasi-isomorphism.

step 1.1algebra
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Acyclic assembly by exact rows

Statement

Let Kp,q be a first-quadrant double cochain complex whose signed total complex uses finite direct sums on every diagonal. Suppose a cochain complex E maps to the left edge so that, for every q, the augmented row 0EqK0,qhK1,qhK2,q is exact and the augmentations commute with the vertical maps. Then the induced cochain map ETotK is a quasi-isomorphism.

Facts & Assumptions

Given: The first-quadrant double complex, compatible row augmentations, and exact augmented rows stated above.

Proof

technique · direct
1.1

Interchange the two indices of K. Multiplying the component in bidegree (p,q) by (1)pq identifies its signed total complex with the total complex after the interchange; the augmented rows become exact augmented columns with compatible edge maps.

givenalgebra
2.1

Apply the exact-column assembly lemma to the interchanged double complex. Transporting its quasi-isomorphism back through the sign identification gives ETotK.

step 1.1algebra
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Hom from a projective makes injective-resolution columns exact

Statement

Let PM be a projective resolution and 0NI0I1 an injective resolution. For every p0, the augmented column 0Hom(Pp,N)Hom(Pp,I0)Hom(Pp,I1) is exact. These augmentations commute with precomposition by dP, so their edge complex is Hom(P,N) and maps naturally to the total Hom complex.

Facts & Assumptions

Given: The two resolutions in the statement and the Hom double complex Kp,q=Hom(Pp,Iq).

Proof

technique · direct
1.1

Each Pp is projective, so Hom(Pp,) is exact. Applying it to the augmented injective resolution gives the displayed exact column. Naturality of Hom shows that these augmentations commute with precomposition by dP.

givenalgebra
2.1

The objects in vertical degree 1 are Hom(Pp,N), with horizontal differential given by precomposition by dP. They therefore form Hom(P,N) and give the asserted natural edge map to the total complex.

step 1.1algebra
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Hom into an injective makes projective-resolution rows exact

Statement

Let P1P0M0 be a projective resolution and NI an injective resolution. For every q0, the augmented row 0Hom(M,Iq)Hom(P0,Iq)Hom(P1,Iq) is exact. These augmentations commute with postcomposition by dI, so their edge complex is Hom(M,I) and maps naturally to the total Hom complex.

Facts & Assumptions

Given: The two resolutions in the statement and the Hom double complex Kp,q=Hom(Pp,Iq).

Proof

technique · direct
1.1

Each Iq is injective, so Hom(,Iq) is exact. Applying it to the augmented projective resolution gives the displayed exact row. Naturality of Hom shows that these augmentations commute with postcomposition by dI.

givenalgebra
2.1

The objects in horizontal degree 1 are Hom(M,Iq), with vertical differential given by postcomposition by dI. They therefore form Hom(M,I) and give the asserted natural edge map to the total complex.

step 1.1algebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Projective and injective constructions of Ext agree for supplied resolutions

Statement

For supplied resolutions of M,N, the canonical maps from Hom(P,N) and Hom(M,I) to TotHom(P,I) are quasi-isomorphisms. Consequently ExtPq(M,N)ExtIq(M,N) for all q0.

Facts & Assumptions

Given: A projective resolution PM and an injective resolution NI.

Proof

technique · direct
1.1

The augmented columns are exact after applying Hom(Pp,), and the augmented rows are exact after applying Hom(,Iq). Both augmentations commute with the other differential.

given
2.1

Finite-diagonal acyclic assembly applied first to columns and then to rows makes both edge-to-total maps quasi-isomorphisms. Taking cohomology yields the displayed isomorphism; the maps themselves are retained for the later naturality proof.

step 1.1
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The Ext balance isomorphism is independent of resolution comparison data

Statement

Assume the Axiom of Dependent Choice. Under the enough-projectives and enough-injectives hypotheses used for the two supplied derived constructions, the balance isomorphism ExtPn(M,N)ExtIn(M,N) is independent of the comparison lifts used after changing either supplied resolution.

Facts & Assumptions

Given: The supplied projective and injective resolution constructions of Ext.

Proof

technique · direct
1.1

For fixed resolutions, the two edge maps defining the balance zigzag are induced by the augmentations PM and NI, so they involve no comparison lift. After replacing a projective or injective resolution, choose a comparison map over or under the resolved object. These maps give a morphism between the two Hom double complexes and commute with both edge augmentations.

givenconstruct
2.1

Any two projective comparison maps are chain-homotopic, and any two injective comparison maps are cochain-homotopic, by the two comparison uniqueness theorems. Applying Hom turns either homotopy into a homotopy of the corresponding total-complex maps. Hence the induced maps on all three cohomologies in the edge-to-total zigzag are independent of the chosen lifts, and the balance isomorphism is independent of those choices.

step 1.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The Ext balance isomorphism is natural in both variables

Statement

Assume the Axiom of Dependent Choice and the supplied resolution hypotheses of the balance theorem. For every n0, the balance maps βM,Nn:ExtPn(M,N)ExtIn(M,N) are natural in both M and N.

Facts & Assumptions

Given: Morphisms u:MM and v:NN in the resolved class.

Proof

technique · direct
1.1

Choose the projective and injective comparison maps for u and v. Projective-resolution Ext has the stated bifunctor variance and Injective-resolution Ext has the stated bifunctor variance give the two routes around the naturality square.

givenconstruct
2.1

The chosen comparison maps induce a morphism of Hom double complexes. Because its squares with the projective augmentation PM and the injective coaugmentation NI commute, both edge-to-total quasi-isomorphisms commute with it. Taking cohomology makes the balance square commute. The independence lemma removes the chosen comparison maps, proving naturality in both variables.

step 1.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Ext balance isomorphisms satisfy change-of-resolution cocycle laws

Statement

If βP,I denotes the balance map attached to supplied projective data P and injective data I, then βP,I is the identity when the comparison data are unchanged, and comparison through any intermediate resolution gives the same map as direct comparison.

Facts & Assumptions

Given: Two choices of resolution data and, where needed, a third intermediate choice.

Proof

technique · direct
1.1

The direct comparison and the composite through the intermediate data are morphisms of the same Ext delta functors and both restrict to the identity on degree-zero Hom.

givenconstruct
2.1

The Ext balance isomorphism is independent of resolution comparison data identifies these morphisms. The same argument with identical data gives the identity law, and The Ext balance isomorphism is natural in both variables makes the laws natural.

step 1.1algebra
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The balanced Ext bifunctor

Definition

Assume the Axiom of Dependent Choice. Let A be an abelian category with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all objects of A. For each n0, define ExtAn(M,N) to mean either ExtPn(M,N) or ExtIn(M,N), identified by the natural comparison isomorphism already proved. This notation is justified by the comparison theorem, its independence of comparison data, its two-variable naturality, and its change-of-resolution cocycle law; it is not a definition by equality of the two complexes.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The long exact Ext sequence in the second variable

Statement

Assume the Axiom of Dependent Choice. Let A be abelian with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all its objects. For 0NNN0 and every M, there is a natural exact sequence 0Hom(M,N)Hom(M,N)Hom(M,N)δ0Ext1(M,N)Ext1(M,N), where δq:Extq(M,N)Extq+1(M,N); it is natural in the short exact sequence and contravariantly natural in M.

Facts & Assumptions

Given: A short exact sequence 0NNN0 and an object M.

Proof

technique · direct
1.1

Apply Right derived functors form a cohomological delta functor to the left exact functor Hom(M,); it supplies the displayed long exact sequence for the injective construction.

givenconstruct
2.1

Replace its terms by balanced Ext using The balanced Ext bifunctor. Delta-functor naturality gives naturality in the short exact sequence, while precomposition in M gives the stated contravariant naturality.

step 1.1algebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The long exact Ext sequence in the first variable

Statement

Assume the Axiom of Dependent Choice. Let A be abelian with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all its objects. For 0MMM0 and every N, there is a natural exact sequence 0Hom(M,N)Hom(M,N)Hom(M,N)0Ext1(M,N)Ext1(M,N), where q:Extq(M,N)Extq+1(M,N); it is natural contravariantly in the short exact sequence and covariantly in N.

Facts & Assumptions

Given: A short exact sequence 0MMM0 and an object N.

Proof

technique · direct
1.1

Regard HomA(,N) as a left exact functor AopAb. A projective resolution in A is an injective resolution in Aop, so Right derived functors form a cohomological delta functor on the opposite category gives the displayed order and connecting maps.

givenconstruct
2.1

Translating the short exact sequence to the opposite category gives the three Hom terms in the displayed order and the maps q. The balanced Ext bifunctor identifies the right-derived groups with Ext, and delta-functor naturality gives contravariant naturality in the short exact sequence.

step 1.1algebra
3.1

To check covariance in N, fix a projective horseshoe 0PPP0 for the given short exact sequence, as supplied by The horseshoe lemma for projective resolutions. Its degreewise splitting makes 0Hom(P,N)Hom(P,N)Hom(P,N)0 a short exact sequence of cochain complexes. Postcomposition with v:NN gives a morphism from this sequence to the analogous one with coefficients N. By Naturality of the cohomology connecting morphism, all connecting squares commute. These are the horseshoe connecting maps used by the right-derived theorem in step 1.1. The comparison isomorphisms transporting the middle horseshoe resolution to the supplied one are induced by precomposition, which commutes with postcomposition by v. Thus the transported connecting maps, and hence the balanced Ext sequence of step 2.1, are covariantly natural in N.

step 2.1construct
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The two Ext long exact sequences agree under balance

Statement

Assume Dependent Choice. Let A be abelian with enough projectives and injectives and supplied projective and injective resolution data on all objects. The balance maps βn:ExtPnExtIn commute with the connecting maps in either variable, for every n0. Here the connecting maps are those obtained from the short exact Hom complexes and the horseshoe constructions, transported to the supplied resolutions by comparison maps. Thus they identify the two long exact Ext sequences, with the degree-zero identification to Hom.

Facts & Assumptions

Given: The stated resolution data and a short exact sequence in either variable.

[F1]

The two edge maps into T(P,I)=TotHom(P,I) are quasi-isomorphisms, and their cohomology ratio is balance (Projective and injective constructions of Ext agree for supplied resolutions); these maps commute with resolution comparisons (The Ext balance isomorphism is natural in both variables).

[F2]

Under DC, projective horseshoes are degreewise split short exact sequences of resolutions; dualizing gives the same assertion for injective horseshoes (The horseshoe lemma for projective resolutions, The horseshoe lemma for injective resolutions).

[F3]

Connecting maps commute with maps of short exact sequences of cochain complexes (Naturality of the cohomology connecting morphism). The derived connecting maps are formed with horseshoes and transported to the supplied data (Right derived functors form a cohomological delta functor, The long exact Ext sequence in the second variable, The long exact Ext sequence in the first variable).

Proof

technique · direct
1.1

For 0NNN0, fix PM and choose an injective horseshoe 0III0. There are three short exact sequences of cochain complexes: Hom(P,N), Hom(M,I), and T(P,I), where denotes the three terms of the short exact sequence, not cochain degree. The first is exact by projectivity of each Pp; the second and third are exact because the horseshoe is split in each degree, and total diagonals are finite. The coaugmentations and augmentation give two morphisms of short exact sequences from the edge sequences to the total sequence.

F1F2givenconstruct
1.2

For 0MMM0, fix NI and choose a projective horseshoe 0PPP0. The three short exact sequences are Hom(P,N), Hom(M,I), and T(P,I), all ordered with double-prime first and prime last. The first and third are exact by the degreewise splitting; the second is exact by injectivity of each Iq. Again the augmentation and coaugmentation give morphisms from both edge sequences to the total sequence.

F1F2givenconstruct
2.1

In each case let a be the projective-edge map to the total and b the injective-edge map. By [F3], H(a) and H(b) commute with the connecting maps of their respective sequences and the total sequence. They are isomorphisms by [F1]. Hence β=H(b)1H(a) also commutes with connecting maps. These are the actual balance maps, not merely some degreewise natural isomorphism. The total differential is h+(1)pv; both edge maps are cochain maps with the page's unsigned Hom differentials, so these are commuting squares with no additional sign.

F1F3step 1.1step 1.2algebra
3.1

The horseshoe middle resolutions may differ from the fixed ones. Transport their cohomology to the supplied data by comparison isomorphisms. By [F3] this is exactly how the derived connecting maps are defined, and [F1] makes balance commute with these comparisons. Therefore the squares proved in step 2.1 hold for the supplied resolutions as well. In degree zero both augmentations identify the common cocycles with Hom, giving its identity identification.

F1F3step 2.1algebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Ext dimension shifting in the first variable

Statement

Assume the Axiom of Dependent Choice. Let A be abelian with enough projectives and enough injectives, fix supplied projective and injective resolution data on all its objects, and let 0ΩMiP0εM0 be the first stage of a projective resolution. For every object N there is an exact sequence 0Hom(M,N)Hom(P0,N)Hom(ΩM,N)Ext1(M,N)0, and for every q1 there is a natural isomorphism Extq(ΩM,N)Extq+1(M,N).

Facts & Assumptions

Given: The displayed first stage and the remaining projective resolution P2P1P0M0.

[L1]

A short exact sequence in the first variable gives the long exact Ext sequence (The long exact Ext sequence in the first variable).

Proof

technique · direct
1.1

Apply [L1] to 0ΩMP0M0. Its relevant terms are Extq(P0,N)Extq(ΩM,N)Extq+1(M,N)Extq+1(P0,N).

L1givenconstruct
2.1

Since P0 is projective, the outer groups vanish for q1 by Positive projective-resolution Ext vanishes on a projective first variable, giving the displayed natural isomorphism. The degree-zero end of the same long exact sequence is exactly the displayed five-term sequence.

step 1.1algebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Ext dimension shifting in the second variable

Statement

Assume the Axiom of Dependent Choice. Let A be abelian with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all its objects. If 0NIΣN0 is an injective copresentation, then for q1 there are natural isomorphisms Extq+1(M,N)Extq(M,ΣN); its low-degree part is 0Hom(M,N)Hom(M,I)Hom(M,ΣN)Ext1(M,N)0.

Facts & Assumptions

Given: The displayed short exact sequence with I injective.

Proof

technique · direct
1.1

The long exact sequence in the second variable contains Extq(M,I)Extq(M,ΣN)Extq+1(M,N)Extq+1(M,I).

given
2.1

The outer groups vanish for q1 because I is injective, so the middle arrow is an isomorphism. At q=0 the same sequence gives exactly the printed low-degree segment.

step 1.1algebra
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Ext can be computed from any projective resolution of the first variable

Statement

Assume the Axiom of Dependent Choice and the hypotheses of The balanced Ext bifunctor. If Q is any other supplied projective resolution datum on the same class of objects, then for every M,N and n0, ExtAn(M,N)HnHom(Q,N), naturally in M and N. In particular, the formula computes Ext from any individual projective resolution Q(M)M; the resulting objectwise isomorphism is canonical on cohomology.

Facts & Assumptions

Given: Dependent Choice, the balanced Ext hypotheses, supplied data P,Q, objects M,N, and n0.

[F1]

Projective comparison maps lifting any object morphism exist: Projective comparison maps exist.

[F2]

Two comparison maps lifting the same morphism are chain-homotopic: Projective comparison maps are unique up to chain homotopy.

[F3]

Resolutions of the same object are homotopy equivalent: Projective resolutions of the same object are homotopy equivalent over that object.

Proof

technique · direct
1.1

Choose aM:Q(M)P(M) lifting 1M. A reverse comparison is its homotopy inverse since both composites lift the identity and [F2] compares them with identity chain maps.

F1F2F3choose
2.1

Precomposition gives aM:Hom(P(M),N)Hom(Q(M),N). A homotopy ab=dh+hd induces the cochain homotopy sn(f)=fhn1, with s0=0. Thus Hn(aM) is a choice-independent isomorphism. Its source is ExtPn(M,N) by Ext via a projective resolution of the first variable, hence balanced Ext by The balanced Ext bifunctor.

step 1.1F2algebra
3.1

For u:MM, choose lifts P(u) and Q(u) by [F1]. Precomposition defines their cohomology actions independently of the lifts by [F2]. Identities and composition follow because composites lift the object composites. The maps P(u)aM and aMQ(u) both lift u, so [F2] makes them homotopic. Precomposition and cohomology therefore give the required contravariant naturality square in M.

F1F2step 2.1construct
4.1

Postcomposition by v:NN commutes exactly with precomposition by aM. This gives naturality in N and hence both variables. For a single supplied resolution Q(M), steps 1.1–2.1 already give the canonical objectwise isomorphism. No simultaneous class-wide choice of comparison maps is required.

step 2.1step 3.1algebra
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Ext can be computed from any injective resolution of the second variable

Statement

Assume the Axiom of Dependent Choice and the hypotheses of The balanced Ext bifunctor. If J is any other supplied injective resolution datum on the same class of objects, then for every M,N and n0, ExtAn(M,N)HnHom(M,J), naturally in M and N. In particular, the formula computes Ext from any individual injective resolution NJ(N); the resulting objectwise isomorphism is canonical on cohomology.

Facts & Assumptions

Given: Dependent Choice, the balanced Ext hypotheses, supplied data I,J, objects M,N, and n0.

[F1]

Comparison maps extending any object morphism exist under Dependent Choice: Injective comparison maps exist.

[F2]

Two such maps extending the same morphism are homotopic: Injective comparison maps are unique up to cochain homotopy.

[F3]

The two resolutions of N are homotopy equivalent under N: Injective resolutions of the same object are homotopy equivalent under that object.

Proof

technique · direct
1.1

Choose aN:I(N)J(N) extending 1N. Its reverse comparison is a homotopy inverse: both composites extend the identity, so [F2] compares them to the identity cochain maps.

F1F2F3choose
2.1

Applying Hom(M,) carries a homotopy ab=dh+hd to the homotopy fhf. Hence HnHom(M,aN) is an isomorphism independent of aN. The definition Ext via an injective resolution of the second variable identifies its source with ExtIn(M,N), which The balanced Ext bifunctor identifies with balanced Ext.

step 1.1F2algebra
3.1

For u:NN, choose comparison maps I(u) and J(u) extending u. Define their actions on cohomology by postcomposition. Independence follows from [F2]; identity and composition laws follow because comparison composites extend the corresponding object composites. Moreover J(u)aN and aNI(u) both extend u, so [F2] makes them homotopic. Applying Hom and cohomology gives precisely the naturality square in N.

F1F2step 2.1construct
4.1

For v:MM, precomposition by v commutes exactly with postcomposition by aN. This proves contravariant naturality in M and therefore naturality in both variables. The same construction at a single N uses only the individual resolution J(N), proving the final assertion without a global choice of comparison maps.

step 2.1step 3.1algebra
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Exact functors compatible with Hom transport Ext under stated adjunction hypotheses

Statement

Let F:AB be exact and have a right adjoint G. If F sends projectives to projectives, then the adjunction isomorphisms induce ExtBn(FM,X)ExtAn(M,GX) for every n0, provided the displayed projective resolutions exist. The dual assertion holds for an exact G that sends injectives to injectives.

Facts & Assumptions

Given: The stated exactness, adjunction, preservation, and resolution hypotheses.

Proof

technique · direct
1.1

Apply F to a projective resolution of M. Exactness preserves its augmentation exactness and the preservation hypothesis makes it a projective resolution of FM. The adjunction Under local smallness, transposition gives the natural hom-set bijection, and conversely identifies its Hom cochain complex into X with the original Hom cochain complex into GX.

givenconstruct
2.1

Taking cohomology and using The balanced Ext bifunctor gives the claimed isomorphism. The dual argument applies the stated injective preservation to the adjoint construction; no assertion is made without these hypotheses.

step 1.1algebra
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Ext one of Z modulo n by Z is Z modulo n

Statement

Assume the Axiom of Dependent Choice and fix the supplied resolution data used for balanced Ext in Ab. For n>1, ExtZ1(Z/n,Z)Z/n; in particular it is nonzero.

Facts & Assumptions

Given: The projective resolution 0ZnZZ/n0.

Proof

technique · direct
1.1

Applying HomZ(,Z) gives 0ZnZ0, in cohomological degrees 0,1.

givenconstruct
2.1

Its first cohomology is coker(n:ZZ)=Z/n. Ext can be computed from any projective resolution of the first variable identifies this cohomology with ExtZ1(Z/n,Z). Since n>1, the quotient is nonzero.

step 1.1algebra

5 · Examples, counterexamples and false statements

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

FALSE: Ext is defined before choosing or supplying resolutions

Statement

FALSE: Ext is defined before choosing or supplying resolutions

Facts & Assumptions

Given: An abelian category, objects M,N, and the two resolution-based Ext constructions of this page.

Refutation

technique · direct
1.1

The injective construction is the cohomology of Hom(M,I(N)) and requires a supplied injective resolution NI(N); the projective construction similarly requires a supplied resolution P(M)M. Neither complex is specified before that datum is supplied.

givenalgebra
2.1

Thus the raw resolution constructions are not definitions of a choice-free object merely from M and N. Choice-independence is a later comparison result, so it cannot make the claim “defined before supplying resolutions” true.

step 1.1algebra
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

FALSE: projective and injective Ext are equal by definition

Statement

FALSE: projective and injective Ext are equal by definition

Facts & Assumptions

Given: Objects M,N with a projective resolution PM and an injective resolution NI.

Refutation

technique · direct
1.1

By definition the two groups are HqHom(P,N) and HqHom(M,I). They are cohomologies of different complexes, with no equality map included in either definition.

givenalgebra
2.1

The finite-diagonal double-complex argument supplies a natural isomorphism between these groups. Because that isomorphism is the conclusion of the balance theorem rather than a definitional identity, the asserted statement is false.

step 1.1algebra
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

FALSE: Ext is covariant in both variables

Statement

FALSE: Ext is covariant in both variables

Facts & Assumptions

Given: A morphism a:MM, a morphism b:NN, and balanced Ext.

Refutation

technique · direct
1.1

Precomposition sends f:MIq(N) to fa:MIq(N), hence gives Extq(M,N)Extq(M,N); its direction is opposite to a.

givenalgebra
2.1

Postcomposition by the comparison induced by b gives Extq(M,N)Extq(M,N). Thus only the second variable is covariant, while the first is contravariant, disproving covariance in both.

step 1.1algebra
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

FALSE: positive Ext vanishes whenever either variable is injective

Statement

FALSE: positive Ext vanishes whenever either variable is injective

Facts & Assumptions

Given: The category of abelian groups and 0ZQQ/Z0.

Refutation

technique · direct
1.1

The group Q/Z is divisible and therefore injective, but it occurs as the quotient, hence as the first argument of the extension. If the sequence split, a retraction QZ would restrict to the identity on Z.

givenalgebra
2.1

Every homomorphism QZ is zero, so the extension is nonsplit and yields 0Ext1(Q/Z,Z). Injectivity instead forces positive Ext to vanish in the second variable, and projectivity does so in the first.

step 1.1algebra
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

FALSE: double-complex totalisation is unambiguous with infinite diagonals

Statement

FALSE: double-complex totalisation is unambiguous with infinite diagonals

Facts & Assumptions

Given: The zero-differential double complex Kp,q=Z when p+q=0 and Kp,q=0 otherwise, indexed over all p,qZ.

Refutation

technique · direct
1.1

Its degree-zero diagonal has one copy of Z for every pZ. The direct-sum totalisation has Tot0=pZZ, whose elements have finite support.

givenalgebra
2.1

The product totalisation instead has Tot0=pZZ, which contains the all-ones family and is strictly larger. Hence infinite diagonals leave a genuine sum/product choice; first-quadrant finite diagonals are what remove it.

step 1.1algebra
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06 rests on unproved material (inherited)Open item page →
Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

FALSE: balance of Ext requires spectral-sequence pages

Statement

FALSE: balance of Ext requires spectral-sequence pages

Facts & Assumptions

Given: The first-quadrant double complex Kp,q=Hom(Pp,Iq) associated with supplied projective and injective resolutions.

Refutation

technique · direct
1.1

For each p, Hom(Pp,) is exact because Pp is projective, so the augmented q-columns are exact; the finite-diagonal acyclic-assembly lemma gives a quasi-isomorphism from the projective edge complex to TotK.

givenalgebra
2.1

Dually, each Hom(,Iq) is exact, so exact rows give a quasi-isomorphism from the injective edge complex to the same total complex. The resulting cohomology isomorphism balances Ext without constructing any spectral-sequence page.

step 1.1algebra

Sources