How statement and proof provenance work

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.

✓ 31 results · all verified · 12 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 19 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Chain Complexes and Homology

1 · Prerequisites

2 · Summary

Successive differentials squaring to zero are the load-bearing datum of the page. That single equation first turns boundaries into subobjects of cycles, so that homology is a genuine quotient, and then controls how chain maps act on cycles, boundaries, kernels, cokernels, and exactness.

The page keeps the general ambient category abelian and avoids element language except in the module and abelian-group examples on the companion. After the basic homology constructions it builds the category of complexes degreewise, records quasi-isomorphisms and the exact-functor comparison on homology, and ends with the deliberately narrow Euler-Poincare formula for finite complexes of finite-rank free abelian groups.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Chain complex in an abelian category

Definition

Let $A$ be an abelian category (Abelian category). A chain complex in $A$ is a family of objects $(C_{n})_{n \in Z}$ together with morphisms $d_{n} : C_{n} \to C_{n - 1}$ such that $d_{n - 1} \circ d_{n} = 0$ for every $n \in Z$ .

The pair is written $(C_{∙}, d)$ , or just $C_{∙}$ when the differentials are clear. The differential has degree $- 1$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Cochain complex in an abelian category

Definition

Let $A$ be an abelian category. A cochain complex in $A$ is a family of objects $(C^{n})_{n \in Z}$ together with morphisms $d^{n} : C^{n} \to C^{n + 1}$ such that $d^{n + 1} \circ d^{n} = 0$ for every $n \in Z$ .

This page uses the reindexing convention that a cochain complex $C^{∙}$ may also be read as the chain complex with $(C^{♯})_{n} : = C^{- n}$ and $d_{n}^{♯} : = d^{- n}$ , so that lower and upper indices differ only by the sign of the grading.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Bounded, bounded below, and bounded above complexes

Definition

A chain complex $C_{∙}$ is:

bounded below if $C_{n} = 0$ for all $n ≪ 0$ ;
bounded above if $C_{n} = 0$ for all $n ≫ 0$ ;
bounded if $C_{n} = 0$ outside some finite interval of integers.

For a cochain complex $C^{∙}$ the same words mean:

$C^{n} = 0$ for all $n ≪ 0$ ;
$C^{n} = 0$ for all $n ≫ 0$ ;
$C^{n} = 0$ outside a finite interval.

Under the reindexing convention of Cochain complex in an abelian category, bounded below and bounded above exchange roles, while boundedness is unchanged.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Zero complex and stalk complex

Definition

Fix a zero object $0$ of an abelian category (Initial object, terminal object, and zero object).

The zero complex is the chain complex with $C_{n} = 0$ for every $n$ and every differential equal to the zero endomorphism of $0$ .

For an object $A$ and an integer $n$ , the stalk complex $S^{n} (A)$ is the chain complex with $S^{n} (A)_{k} = {\begin{cases} A, & k = n, \\ 0, & k \neq n, \end{cases}$ and every differential zero. It is concentrated in degree $n$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Cycle and boundary subobjects of a complex

Definition

Let $C_{∙}$ be a chain complex in an abelian category and fix $n \in Z$ .

The $n$ th cycle subobject of $C$ is $Z_{n} (C) : = \ker (d_{n}) ↪ C_{n} .$

The $n$ th boundary subobject of $C$ is $B_{n} (C) : = im (d_{n + 1}) ↪ C_{n} .$

Thus both are subobjects of $C_{n}$ , one defined by the kernel of $d_{n}$ and the other by the image of $d_{n + 1}$ .

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The boundary subobject factors through the cycle subobject

Statement

Let $C_{∙}$ be a chain complex in an abelian category. For every $n \in Z$ , the boundary inclusion $B_{n} (C) ↪ C_{n}$ factors uniquely through the cycle inclusion $Z_{n} (C) ↪ C_{n}$ .

Facts & Assumptions

Given: A chain complex $C_{∙}$ and an integer $n$ .

[L1]

In a chain complex, $d_{n} d_{n + 1} = 0$ (Chain complex in an abelian category).

[L2]

A kernel $k : Z_{n} (C) \to C_{n}$ is characterized by $d_{n} k = 0$ and universal factorization among arrows annihilated by $d_{n}$ (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

[L3]

Every morphism admits an epic-monic factorization; in particular $d_{n + 1}$ factors as $C_{n + 1} \overset{e}{\to} B_{n} (C) \overset{i}{\to} C_{n}$ with $e$ epic and $i$ monic (Every morphism factors as an epimorphism followed by a monomorphism, uniquely up to unique isomorphism).

Proof

technique · direct

1.1

By [L3], write $d_{n + 1} = i e$ with $i : B_{n} (C) \to C_{n}$ monic and $e$ epic. Then $d_{n} i e = d_{n} d_{n + 1} = 0$ by [L1], so epicity of $e$ gives $d_{n} i = 0$ .

L1L3givenalgebra

2.1

Since $d_{n} i = 0$ , the kernel property in [L2] gives a unique morphism $β_{n} : B_{n} (C) \to Z_{n} (C)$ with $k β_{n} = i$ , where $k : Z_{n} (C) \to C_{n}$ is the cycle inclusion. That is exactly the required factorization.

L2step 1.1constructdischarge-construct∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Homology object of a chain complex

Definition

Let $C_{∙}$ be a chain complex and let $β_{n} : B_{n} (C) \to Z_{n} (C)$ be the canonical morphism supplied by The boundary subobject factors through the cycle subobject.

The $n$ th homology object of $C$ is the cokernel $H_{n} (C) : = coker (β_{n}) .$

Equivalently, one writes $H_{n} (C) = Z_{n} (C) / B_{n} (C),$ with the understanding that this quotient notation abbreviates the cokernel of the canonical monomorphism from boundaries to cycles.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Cohomology object of a cochain complex

Definition

Let $C^{∙}$ be a cochain complex. Its $n$ th cocycle subobject is $Z^{n} (C) : = \ker (d^{n}) ↪ C^{n},$ and its $n$ th coboundary subobject is $B^{n} (C) : = im (d^{n - 1}) ↪ C^{n} .$

Reading $C^{∙}$ as a chain complex by the reindexing convention of Cochain complex in an abelian category, the morphism $d^{n} d^{n - 1} = 0$ is exactly the chain-complex relation in degree $- n$ . Hence The boundary subobject factors through the cycle subobject supplies a canonical monomorphism $B^{n} (C) \to Z^{n} (C)$ .

The $n$ th cohomology object is its cokernel: $H^{n} (C) : = coker (B^{n} (C) \to Z^{n} (C)) .$

Equivalently, $H^{n} (C) = Z^{n} (C) / B^{n} (C)$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Exactness of a complex at a degree and acyclic complexes

Definition

Let $C_{∙}$ be a chain complex and let $β_{n} : B_{n} (C) \to Z_{n} (C)$ be the canonical map from boundaries to cycles.

The complex is exact at degree $n$ when $β_{n}$ is an isomorphism. Equivalently, the pair $C_{n + 1} \overset{d_{n + 1}}{\to} C_{n} \overset{d_{n}}{\to} C_{n - 1}$ is exact at $C_{n}$ in the sense of Exactness at a node, so that $im (d_{n + 1})$ and $\ker (d_{n})$ represent the same subobject of $C_{n}$ .

The complex is acyclic if it is exact at every degree.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

A complex is exact at n exactly when its nth homology is zero

Statement

Let $C_{∙}$ be a chain complex in an abelian category and let $n \in Z$ . Then $C$ is exact at degree $n$ if and only if $H_{n} (C)$ is a zero object.

Facts & Assumptions

Given: A chain complex $C_{∙}$ and an integer $n$ .

[L1]

Exactness at degree $n$ means that the canonical map $β_{n} : B_{n} (C) \to Z_{n} (C)$ is an isomorphism (Exactness of a complex at a degree and acyclic complexes).

[L2]

The homology object $H_{n} (C)$ is the cokernel of $β_{n}$ (Homology object of a chain complex).

[L3]

In an abelian category, a morphism is epic exactly when its cokernel is zero (In an abelian category, monic means zero kernel and epic means zero cokernel).

[L4]

In an abelian category, a morphism that is both monic and epic is an isomorphism (An abelian category is balanced).

Proof

technique · direct

1.1

Assume $C$ is exact at degree $n$ . Then [L1] says $β_{n}$ is an isomorphism, hence in particular epic. By [L2] and [L3], the cokernel of $β_{n}$ , namely $H_{n} (C)$ , is therefore zero.

L1L2L3

2.1

Conversely, assume $H_{n} (C)$ is zero. By [L2], the cokernel of $β_{n}$ is zero, so [L3] makes $β_{n}$ epic. The map $β_{n}$ is monic because it factors the monic image inclusion of $B_{n} (C)$ through the monic cycle inclusion. Hence [L4] makes $β_{n}$ an isomorphism, and [L1] says that $C$ is exact at degree $n$ .

L1L2L3L4algebra∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

An exact sequence is a complex, and its exactness agrees with the earlier notion

Statement

Every exact sequence in an abelian category is a chain complex. At each object, its exactness as a chain complex is exactly the previously defined exactness of the sequence.

Facts & Assumptions

Given: A composable sequence in an abelian category.

[L1]

An exact sequence is exact at each interior object in the sense of Exact sequence and short exact sequence in an abelian category.

[L2]

Exactness at a node is the equality of image and kernel subobjects (Exactness at a node).

[L3]

For composable morphisms $A \overset{f}{\to} B \overset{g}{\to} C$ , the relation $g f = 0$ is equivalent to the factorization of $im (f)$ through $\ker (g)$ (The arrow-theoretic criterion for exactness).

[L4]

Exactness of a chain complex at degree $n$ is the same equality of boundary and cycle subobjects at that degree (Exactness of a complex at a degree and acyclic complexes).

Proof

technique · direct

1.1

If $A \overset{f}{\to} B \overset{g}{\to} C$ is exact at $B$ , then [L1] and [L2] identify $im (f)$ with $\ker (g)$ . By [L3], this implies $g f = 0$ . Applying this at each consecutive pair shows that any exact sequence is a chain complex.

L1L2L3

2.1

At a chosen object of that exact sequence, the chain-complex notion in [L4] again compares $im (f)$ with $\ker (g)$ . That is exactly the criterion in [L2], so no new notion of exactness is introduced.

L2L4∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Chain map

Definition

Let $C_{∙}$ and $D_{∙}$ be chain complexes. A chain map $f : C_{∙} \to D_{∙}$ is a family of morphisms $f_{n} : C_{n} \to D_{n}$ such that for every $n \in Z$ , $d_{n}^{D} \circ f_{n} = f_{n - 1} \circ d_{n}^{C} .$

Equivalently, each square $\begin{array}{ccc} C_{n} & \overset{f_{n}}{\to} & D_{n} \\ d_{n}^{C} ↓ & ↓ d_{n}^{D} \\ C_{n - 1} & \overset{f_{n - 1}}{\to} & D_{n - 1} \end{array}$ commutes.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Cochain map

Definition

Let $C^{∙}$ and $D^{∙}$ be cochain complexes. A cochain map $f : C^{∙} \to D^{∙}$ is a family of morphisms $f^{n} : C^{n} \to D^{n}$ such that $d_{D}^{n} \circ f^{n} = f^{n + 1} \circ d_{C}^{n}$ for every $n \in Z$ .

Thus the upper-index square $\begin{array}{ccc} C^{n} & \overset{f^{n}}{\to} & D^{n} \\ d_{C}^{n} ↓ & ↓ d_{D}^{n} \\ C^{n + 1} & \overset{f^{n + 1}}{\to} & D^{n + 1} \end{array}$ commutes in each degree.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Identities and composites of chain maps are chain maps

Statement

For every chain complex $C_{∙}$ , the identity family $1_{C_{n}}$ is a chain map $1_{C} : C \to C$ . If $f : C \to D$ and $g : D \to E$ are chain maps, then the componentwise composite $(g_{n} f_{n})_{n}$ is a chain map $g \circ f : C \to E$ .

Facts & Assumptions

Given: Chain complexes $C_{∙}, D_{∙}, E_{∙}$ and chain maps $f : C \to D$ , $g : D \to E$ .

[L1]

A chain map is a degree-zero family commuting with the differentials (Chain map).

Proof

technique · direct

1.1

For the identity family, $d_{n}^{C} \circ 1_{C_{n}} = 1_{C_{n - 1}} \circ d_{n}^{C}$ holds trivially in every degree, so [L1] makes $1_{C}$ a chain map.

L1givenalgebra

2.1

Since $f$ and $g$ are chain maps, [L1] gives $d_{n}^{E} g_{n} = g_{n - 1} d_{n}^{D}$ and $d_{n}^{D} f_{n} = f_{n - 1} d_{n}^{C}$ . Therefore $d_{n}^{E} (g_{n} f_{n}) = g_{n - 1} (d_{n}^{D} f_{n}) = g_{n - 1} f_{n - 1} d_{n}^{C},$ so the composite family again satisfies [L1].

L1givenalgebra∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

The category of chain complexes

Definition

The category of chain complexes in an abelian category $A$ is denoted $Ch (A) .$ Its objects are chain complexes in $A$ , and its morphisms are chain maps.

By Identities and composites of chain maps are chain maps, identities and composition are defined degreewise.

The full subcategories on bounded below, bounded above, and bounded complexes are denoted ${Ch}^{+} (A), {Ch}^{-} (A), {Ch}^{b} (A) .$

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

A chain map carries cycles to cycles and boundaries to boundaries

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map. For every $n \in Z$ there are induced morphisms $Z_{n} (f) : Z_{n} (C) \to Z_{n} (D), B_{n} (f) : B_{n} (C) \to B_{n} (D),$ compatible with the canonical inclusions into $C_{n}$ and $D_{n}$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ and an integer $n$ .

[L1]

A chain map satisfies $d_{n}^{D} f_{n} = f_{n - 1} d_{n}^{C}$ (Chain map).

[L2]

Kernels are universal among arrows killed by the given morphism (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

[L3]

Write $d_{n + 1}^{C} = i_{C} e_{C}$ and $d_{n + 1}^{D} = i_{D} e_{D}$ as epic-monic factorizations, where $i_{C}$ and $i_{D}$ are the boundary inclusions (Every morphism factors as an epimorphism followed by a monomorphism, uniquely up to unique isomorphism).

[L4]

In an abelian category, every monomorphism is the kernel of its cokernel (Every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel).

Proof

technique · direct

1.1

Let $k_{C} : Z_{n} (C) \to C_{n}$ and $k_{D} : Z_{n} (D) \to D_{n}$ be the cycle inclusions. Using [L1], $d_{n}^{D} f_{n} k_{C} = f_{n - 1} d_{n}^{C} k_{C} = 0,$ so [L2] yields a unique map $Z_{n} (f) : Z_{n} (C) \to Z_{n} (D)$ with $k_{D} Z_{n} (f) = f_{n} k_{C}$ .

L1L2givenconstruct

2.1

By [L3], $f_{n} i_{C} e_{C} = f_{n} d_{n + 1}^{C} = d_{n + 1}^{D} f_{n + 1} = i_{D} e_{D} f_{n + 1} .$ Let $c_{D}$ be a cokernel of $i_{D}$ . Then $c_{D} f_{n} i_{C} e_{C} = c_{D} i_{D} e_{D} f_{n + 1} = 0.$ Since $e_{C}$ is epic, this implies $c_{D} f_{n} i_{C} = 0$ . By [L4], the monomorphism $i_{D}$ is a kernel of $c_{D}$ , so [L2] gives a unique morphism $B_{n} (f) : B_{n} (C) \to B_{n} (D)$ with $i_{D} B_{n} (f) = f_{n} i_{C} .$ This is the required boundary map.

L1L2L3L4givenalgebraconstructdischarge-construct∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

A chain map induces a well-defined map on homology

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map. For every $n \in Z$ there is a unique morphism $H_{n} (f) : H_{n} (C) \to H_{n} (D)$ such that the quotient maps from cycles to homology commute with $Z_{n} (f)$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ and an integer $n$ .

[L1]

The maps $Z_{n} (f) : Z_{n} (C) \to Z_{n} (D)$ and $B_{n} (f) : B_{n} (C) \to B_{n} (D)$ exist and are compatible with the boundary and cycle inclusions (A chain map carries cycles to cycles and boundaries to boundaries).

[L2]

$H_{n} (C)$ and $H_{n} (D)$ are the cokernels of the canonical maps $B_{n} (C) \to Z_{n} (C)$ and $B_{n} (D) \to Z_{n} (D)$ (Homology object of a chain complex).

[L3]

A cokernel is universal among arrows that kill the map being quotiented (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1

Let $β_{C} : B_{n} (C) \to Z_{n} (C)$ and $β_{D} : B_{n} (D) \to Z_{n} (D)$ be the boundary-to-cycle maps. Compatibility in [L1] means $Z_{n} (f) β_{C} = β_{D} B_{n} (f) .$ Therefore the composite $q_{D} Z_{n} (f)$ kills $β_{C}$ , where $q_{D} : Z_{n} (D) \to H_{n} (D)$ is the homology quotient.

L1L2givenalgebra

2.1

Since $q_{D} Z_{n} (f)$ annihilates $β_{C}$ , the cokernel property [L3] for $q_{C} : Z_{n} (C) \to H_{n} (C)$ gives a unique morphism $H_{n} (f) : H_{n} (C) \to H_{n} (D)$ with $H_{n} (f) q_{C} = q_{D} Z_{n} (f) .$ By [L2], this is exactly the induced map on homology.

L2L3step 1.1constructdischarge-construct∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Homology respects identities and composition

Statement

For every $n \in Z$ :

$H_{n} (1_{C}) = 1_{H_{n} (C)}$ for every chain complex $C$ .
If $f : C \to D$ and $g : D \to E$ are chain maps, then $H_{n} (g \circ f) = H_{n} (g) \circ H_{n} (f) .$

Facts & Assumptions

Given: Chain maps $f : C \to D$ and $g : D \to E$ .

[L1]

Identities and composites of chain maps are chain maps (Identities and composites of chain maps are chain maps).

[L2]

A chain map induces a unique map on homology compatible with the quotient from cycles (A chain map induces a well-defined map on homology).

Proof

technique · direct

1.1

By [L1], $1_{C}$ is a chain map. The identity on $H_{n} (C)$ satisfies the same compatibility with the quotient from cycles as the map $H_{n} (1_{C})$ from [L2], so uniqueness in [L2] gives $H_{n} (1_{C}) = 1_{H_{n} (C)}$ .

L1L2

2.1

Again by [L1], $g \circ f$ is a chain map. Both $H_{n} (g \circ f)$ and $H_{n} (g) \circ H_{n} (f)$ compose with the cycle quotient to the map induced by the composite on cycles, so [L2] forces them to agree.

L1L2algebra∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Homology is an additive functor

Statement

For each $n \in Z$ , homology defines an additive functor $H_{n} : Ch (A) \to A .$

Facts & Assumptions

Given: An abelian category $A$ and an integer $n$ .

[L1]

A chain map induces a map on homology (A chain map induces a well-defined map on homology).

[L2]

Those induced maps respect identities and composition (Homology respects identities and composition).

[L3]

An additive functor is a functor that is additive on each hom-group (Additive functor).

[L4]

An abelian category is additive, so $Ch (A)$ is also additive and sums of chain maps are defined degreewise (Abelian category, The category of complexes in an additive category is additive).

[L5]

Kernels are universal among arrows annihilated by the displayed map (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1

By [L1] and [L2], the assignment $C \mapsto H_{n} (C)$ and $f \mapsto H_{n} (f)$ is already a functor.

L1L2

1.2

Let $f, g : C \to D$ be chain maps. By [L4], their sum $f + g$ is the chain map with components $f_{n} + g_{n}$ . Let $k_{C} : Z_{n} (C) \to C_{n}, k_{D} : Z_{n} (D) \to D_{n}$ be the cycle inclusions and $q_{C} : Z_{n} (C) \to H_{n} (C), q_{D} : Z_{n} (D) \to H_{n} (D)$ the homology quotients. Then $k_{D} (Z_{n} (f) + Z_{n} (g)) = f_{n} k_{C} + g_{n} k_{C} = (f_{n} + g_{n}) k_{C} = k_{D} Z_{n} (f + g) .$ Since both maps on cycles are killed by $d_{n}^{D}$ , the uniqueness in the kernel property [L5] for $k_{D}$ gives $Z_{n} (f + g) = Z_{n} (f) + Z_{n} (g) .$ Therefore $(H_{n} (f) + H_{n} (g)) q_{C} = H_{n} (f) q_{C} + H_{n} (g) q_{C} = q_{D} Z_{n} (f) + q_{D} Z_{n} (g) = q_{D} Z_{n} (f + g) .$ By the uniqueness clause in [L1], this forces $H_{n} (f + g) = H_{n} (f) + H_{n} (g) .$

L1L4L5algebra

2.1

Steps 1.1 and 1.2 are exactly the functoriality and additivity demanded by [L3]. Hence $H_{n}$ is an additive functor.

L3step 1.1step 1.2∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Quasi-isomorphism

Definition

A chain map $f : C_{∙} \to D_{∙}$ is a quasi-isomorphism if for every $n \in Z$ the induced map $H_{n} (f) : H_{n} (C) \to H_{n} (D)$ is an isomorphism.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Isomorphisms of complexes are quasi-isomorphisms

Statement

Every isomorphism in $Ch (A)$ is a quasi-isomorphism.

Facts & Assumptions

Given: An isomorphism of chain complexes $f : C \to D$ with inverse $g : D \to C$ .

[L1]

A quasi-isomorphism is a chain map inducing isomorphisms on all homology objects (Quasi-isomorphism).

[L2]

Homology respects identities and composition (Homology respects identities and composition).

Proof

technique · direct

1.1

Since $g f = 1_{C}$ and $f g = 1_{D}$ , [L2] gives $H_{n} (g) H_{n} (f) = 1_{H_{n} (C)}, H_{n} (f) H_{n} (g) = 1_{H_{n} (D)}$ for every $n$ . Thus each $H_{n} (f)$ is an isomorphism.

L2givenalgebra

2.1

By [L1], that means $f$ is a quasi-isomorphism.

L1step 1.1∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

A chain map is a quasi-isomorphism exactly when its cochain reindexing is

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map, and read both complexes as cochain complexes by the reindexing convention $(C^{♯})^{- n} = C_{n}$ , $(D^{♯})^{- n} = D_{n}$ . Then $f$ is a quasi-isomorphism if and only if the induced cochain map $f^{♯} : C^{♯} \to D^{♯}$ is a quasi-isomorphism in cohomology.

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ .

[L1]

A cochain complex may be read as a reindexed chain complex (Cochain complex in an abelian category).

[L2]

Cohomology is the cokernel of the coboundary subobject inside the cocycle subobject (Cohomology object of a cochain complex).

[L3]

A quasi-isomorphism is a chain map inducing isomorphisms on every homology object (Quasi-isomorphism).

Proof

technique · direct

1.1

Under the reindexing convention [L1], the cocycles and coboundaries of $C^{♯}$ in degree $- n$ are exactly $Z_{n} (C)$ and $B_{n} (C)$ . Therefore [L2] identifies $H^{- n} (C^{♯}) ≅ H_{n} (C), H^{- n} (D^{♯}) ≅ H_{n} (D) .$

L1L2given

2.1

The map induced by $f^{♯}$ on $H^{- n}$ is therefore the same morphism as $H_{n} (f)$ under the identifications of step 1.1. Hence $H_{n} (f)$ is an isomorphism for every $n$ if and only if $H^{- n} (f^{♯})$ is an isomorphism for every $n$ . By [L3], this is exactly the claimed equivalence.

L3step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-30Open item page →

Subcomplex

Definition

Let $C_{∙}$ be a chain complex. A subcomplex of $C$ is a family of subobjects $S_{n} ↪ C_{n}$ such that for every $n$ the differential $d_{n}^{C} : C_{n} \to C_{n - 1}$ restricts to a morphism $d_{n}^{S} : S_{n} \to S_{n - 1}$ making $\begin{array}{ccc} S_{n} & ↪ & C_{n} \\ d_{n}^{S} ↓ & ↓ d_{n}^{C} \\ S_{n - 1} & ↪ & C_{n - 1} \end{array}$ commute.

Because the structure maps of a subobject are monic, each restricted differential is unique when it exists.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The differential descends to a quotient complex

Statement

Let $S_{∙} ↪ C_{∙}$ be a subcomplex in an abelian category. For every $n$ the differential $d_{n}^{C}$ induces a unique morphism ${\overset{ˉ}{d}}_{n} : C_{n} / S_{n} \to C_{n - 1} / S_{n - 1},$ and these induced morphisms satisfy ${\overset{ˉ}{d}}_{n - 1} {\overset{ˉ}{d}}_{n} = 0$ .

Facts & Assumptions

Given: A subcomplex $S_{∙} ↪ C_{∙}$ with quotient maps $q_{n} : C_{n} \to C_{n} / S_{n}$ .

[L1]

In a subcomplex, the differential restricts to $S_{n} \to S_{n - 1}$ (Subcomplex).

[L2]

A cokernel is universal among arrows vanishing on the given subobject (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

[L3]

In a chain complex, consecutive differentials compose to zero (Chain complex in an abelian category).

Proof

technique · constructive

1.1

By [L1], the composite $q_{n - 1} d_{n}^{C}$ kills $S_{n}$ , because $d_{n}^{C} (S_{n})$ lands in $S_{n - 1}$ and $q_{n - 1}$ kills $S_{n - 1}$ . Hence [L2] gives a unique map ${\overset{ˉ}{d}}_{n} : C_{n} / S_{n} \to C_{n - 1} / S_{n - 1}$ with ${\overset{ˉ}{d}}_{n} q_{n} = q_{n - 1} d_{n}^{C} .$

L1L2givenconstruct

2.1

Composing the identity from step 1.1 twice and using [L3], ${\overset{ˉ}{d}}_{n - 1} {\overset{ˉ}{d}}_{n} q_{n} = q_{n - 2} d_{n - 1}^{C} d_{n}^{C} = 0.$ Since $q_{n}$ is epic as a cokernel map, ${\overset{ˉ}{d}}_{n - 1} {\overset{ˉ}{d}}_{n} = 0$ .

L2L3step 1.1algebradischarge-construct∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Quotient complex

Definition

Let $S_{∙} ↪ C_{∙}$ be a subcomplex. The quotient complex $C / S$ is the chain complex whose degree- $n$ term is the quotient object $C_{n} / S_{n}$ and whose differential is the descended morphism ${\overset{ˉ}{d}}_{n} : C_{n} / S_{n} \to C_{n - 1} / S_{n - 1}$ constructed in The differential descends to a quotient complex.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Short exact sequence of complexes

Definition

A short exact sequence of complexes is a sequence of chain maps $0 \to A_{∙} \to B_{∙} \to C_{∙} \to 0$ that is exact in each degree as a sequence in the ambient abelian category.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The kernel of a chain map is computed degreewise

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map. The kernels $\ker (f_{n}) ↪ C_{n}$ assemble into a chain complex, and this complex is a kernel of $f$ in $Ch (A)$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ .

[L1]

A chain map satisfies $d_{n}^{D} f_{n} = f_{n - 1} d_{n}^{C}$ (Chain map).

[L2]

Kernels are universal among arrows annihilated by the displayed map (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · constructive

1.1

Let $k_{n} : K_{n} \to C_{n}$ be a kernel of $f_{n}$ . By [L1], $f_{n - 1} d_{n}^{C} k_{n} = d_{n}^{D} f_{n} k_{n} = 0,$ so [L2] gives a unique differential $d_{n}^{K} : K_{n} \to K_{n - 1}$ with $k_{n - 1} d_{n}^{K} = d_{n}^{C} k_{n} .$

L1L2givenconstruct

2.1

The family $K_{∙}$ is a chain complex because $k_{n - 2} d_{n - 1}^{K} d_{n}^{K} = d_{n - 1}^{C} d_{n}^{C} k_{n} = 0,$ and $k_{n - 2}$ is monic. The map $k : K \to C$ is a chain map by construction, and its componentwise kernel universal properties from [L2] together give the kernel universal property in $Ch (A)$ .

L2step 1.1algebradischarge-construct∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The cokernel of a chain map is computed degreewise

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map. The cokernels $D_{n} ↠ coker (f_{n})$ assemble into a chain complex, and this complex is a cokernel of $f$ in $Ch (A)$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ .

[L1]

A chain map satisfies $d_{n}^{D} f_{n} = f_{n - 1} d_{n}^{C}$ (Chain map).

[L2]

Cokernels are universal among arrows that kill the displayed map (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · constructive

1.1

Let $q_{n} : D_{n} \to Q_{n}$ be a cokernel of $f_{n}$ . Since $q_{n - 1} d_{n}^{D} f_{n} = q_{n - 1} f_{n - 1} d_{n}^{C} = 0$ by [L1], [L2] gives a unique differential $d_{n}^{Q} : Q_{n} \to Q_{n - 1}$ with $d_{n}^{Q} q_{n} = q_{n - 1} d_{n}^{D} .$

L1L2givenconstruct

2.1

The family $Q_{∙}$ is a chain complex because $d_{n - 1}^{Q} d_{n}^{Q} q_{n} = q_{n - 2} d_{n - 1}^{D} d_{n}^{D} = 0,$ and $q_{n}$ is epic. The quotient map $q : D \to Q$ is a chain map by construction, and the componentwise cokernel universal properties from [L2] assemble to the cokernel universal property in $Ch (A)$ .

L2step 1.1algebradischarge-construct∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Images and coimages of chain maps are computed degreewise

Statement

For a chain map $f : C_{∙} \to D_{∙}$ , the image complex and coimage complex in $Ch (A)$ are obtained degreewise from the images and coimages of the component maps $f_{n}$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ .

[L1]

In any category with kernels and cokernels, the image is the kernel of the cokernel and the coimage is the cokernel of the kernel (Image and coimage in a category with kernels and cokernels).

[L2]

Kernels of chain maps are computed degreewise (The kernel of a chain map is computed degreewise).

[L3]

Cokernels of chain maps are computed degreewise (The cokernel of a chain map is computed degreewise).

Proof

technique · direct

1.1

By [L2] and [L3], the kernel and cokernel complexes of $f$ have $n$ th terms $\ker (f_{n})$ and $coker (f_{n})$ . Applying [L1] inside $Ch (A)$ therefore shows that the coimage and image complexes have $n$ th terms $coker (\ker f_{n})$ and $\ker (coker f_{n})$ .

L1L2L3

2.1

Those are exactly the ordinary coimage and image objects of $f_{n}$ by [L1]. Hence the complex-level image and coimage are computed degreewise, and the canonical coimage-to-image map is the family of the component canonical maps.

L1step 1.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The category of complexes in an additive category is additive

Statement

Let $Ch (A)$ denote the category whose objects are \mathbb Z-graded objects of $A$ equipped with differentials $d_{n} : C_{n} \to C_{n - 1}$ satisfying $d_{n - 1} d_{n} = 0$ , and whose morphisms are chain maps. If $A$ is an additive category, then $Ch (A)$ is an additive category.

Facts & Assumptions

Given: An additive category $A$ .

[L1]

An additive category has a zero object and finite biproducts (Additive category).

[L2]

A biproduct is a common product-coproduct object (Biproduct).

[L3]

A chain map is a degreewise family commuting with the differentials (Chain map).

Proof

technique · direct

1.1

Since $A$ is additive, [L1] includes the preadditive structure that supplies zero morphisms, so the displayed chain-complex condition is meaningful in $A$ . The zero object of $A$ from [L1] therefore gives the zero complex, which is a zero object of $Ch (A)$ because every chain map to or from it is forced degreewise.

L1L3

1.2

For chain complexes $C$ and $D$ , let $(C \oplus D)_{n} : = C_{n} \oplus D_{n}$ using the biproducts from [L1]. Define the differential by $d_{n}^{C \oplus D} : = d_{n}^{C} \oplus d_{n}^{D}$ . Then $(d_{n - 1}^{C} \oplus d_{n - 1}^{D}) (d_{n}^{C} \oplus d_{n}^{D}) = 0$ , so this is a chain complex, and the degreewise injections and projections are chain maps. By [L2], they make $C \oplus D$ a biproduct in $Ch (A)$ .

L1L2L3algebra

2.1

Addition of chain maps is defined degreewise on the additive hom-groups of $A$ , and the chain-map equation is preserved because composition is bilinear. Together with steps 1.1 and 1.2, [L1] shows that $Ch (A)$ is additive.

L1L3step 1.1step 1.2algebra∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The category of complexes in an abelian category is abelian

Statement

If $A$ is an abelian category, then $Ch (A)$ is an abelian category.

Facts & Assumptions

Given: An abelian category $A$ .

[L1]

An abelian category is an additive category in which every morphism has a kernel and a cokernel, and every coimage-to-image comparison is an isomorphism (Abelian category).

[L2]

$Ch (A)$ is additive (The category of complexes in an additive category is additive).

[L3]

Kernels and cokernels of chain maps are computed degreewise (The kernel of a chain map is computed degreewise, The cokernel of a chain map is computed degreewise).

[L4]

Images, coimages, and the coimage-to-image comparison of a chain map are computed degreewise (Images and coimages of chain maps are computed degreewise).

Proof

technique · direct

1.1

By [L2] and [L3], $Ch (A)$ is additive and every chain map has a kernel and a cokernel.

L2L3

1.2

Let $f : C \to D$ be a chain map. By [L4], the coimage-to-image comparison of $f$ is the family of the coimage-to-image comparisons of the component maps $f_{n}$ . Each of those is an isomorphism by [L1], so the family is an isomorphism of complexes. Therefore the defining clauses of [L1] hold in $Ch (A)$ .

L1L4algebra

2.1

Steps 1.1 and 1.2 prove that $Ch (A)$ is abelian.

L1step 1.1step 1.2∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

A sequence of chain maps is exact exactly when it is exact degreewise

Statement

Let $A_{∙} \overset{u}{\to} B_{∙} \overset{v}{\to} C_{∙}$ be composable chain maps in an abelian category. The sequence is exact at $B_{∙}$ in $Ch (A)$ if and only if for every $n \in Z$ the sequence $A_{n} \overset{u_{n}}{\to} B_{n} \overset{v_{n}}{\to} C_{n}$ is exact at $B_{n}$ in $A$ .

Facts & Assumptions

Given: Composable chain maps $A_{∙} \overset{u}{\to} B_{∙} \overset{v}{\to} C_{∙}$ .

[L1]

$Ch (A)$ is abelian (The category of complexes in an abelian category is abelian).

[L2]

Exactness at a node means equality of the image and kernel subobjects (Exactness at a node).

[L3]

Images and kernels of chain maps are computed degreewise (Images and coimages of chain maps are computed degreewise).

Proof

technique · direct

1.1

If the sequence is exact at $B_{∙}$ in $Ch (A)$ , then [L2] says $im (u) = \ker (v)$ as subobjects of $B_{∙}$ . By [L3], their $n$ th components are $im (u_{n})$ and $\ker (v_{n})$ , so [L2] gives exactness of $A_{n} \to B_{n} \to C_{n}$ in every degree.

L1L2L3

2.1

Conversely, if every degreewise sequence is exact, then [L2] gives $im (u_{n}) = \ker (v_{n})$ inside $B_{n}$ for every $n$ . By [L3], that is exactly the degreewise comparison between the image and kernel complexes, so [L2] makes the sequence exact at $B_{∙}$ in $Ch (A)$ .

L2L3∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

A subcomplex is the kernel of its quotient map

Statement

If $S_{∙} ↪ C_{∙}$ is a subcomplex, then the quotient map $C_{∙} \to C_{∙} / S_{∙}$ has kernel $S_{∙}$ . Conversely, every monomorphism of complexes identifies its source, up to unique isomorphism, with a subcomplex of its target.

Facts & Assumptions

Given: A subcomplex $S_{∙} ↪ C_{∙}$ and a monomorphism $m : A_{∙} \to C_{∙}$ .

[L1]

The quotient complex $C / S$ is defined degreewise from the quotients $C_{n} / S_{n}$ (Quotient complex).

[L2]

Kernels of chain maps are computed degreewise (The kernel of a chain map is computed degreewise).

[L3]

In $Ch (A)$ , monomorphisms are morphisms in an abelian category (The category of complexes in an abelian category is abelian).

[L4]

In an abelian category, a morphism is monic exactly when its kernel is zero (In an abelian category, monic means zero kernel and epic means zero cokernel).

Proof

technique · direct

1.1

Let $q : C \to C / S$ be the quotient map. By [L1], each component $q_{n} : C_{n} \to C_{n} / S_{n}$ has kernel represented by $S_{n} ↪ C_{n}$ . Then [L2] says the kernel complex of $q$ is exactly $S_{∙}$ .

L1L2

2.1

Now let $m : A \to C$ be monic. Its kernel in $Ch (A)$ is therefore the zero complex. By [L2], that kernel is assembled degreewise from the kernels of the component maps $m_{n}$ , so each $\ker (m_{n})$ is zero. Then [L4] makes every $m_{n}$ monic in $A$ . Since $m$ is a chain map, the commuting squares $m_{n - 1} d_{n}^{A} = d_{n}^{C} m_{n}$ show that the family of monomorphisms $m_{n} : A_{n} ↪ C_{n}$ is stable under the differentials. Hence these degreewise monomorphisms exhibit $A_{∙}$ as a subcomplex of $C_{∙}$ , unique up to the usual unique isomorphism of represented subobjects.

L2L3L4givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The first isomorphism theorem for complexes

Statement

For every chain map $f : C_{∙} \to D_{∙}$ in an abelian category, there is a canonical isomorphism of complexes $C_{∙} / \ker (f) ≅ im (f) .$

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ .

[L1]

In an abelian category, $C_{n} / \ker (f_{n}) ≅ im (f_{n})$ canonically for every component map $f_{n}$ (First isomorphism theorem in an abelian category).

[L2]

The image and coimage of a chain map are computed degreewise (Images and coimages of chain maps are computed degreewise).

[L3]

The kernel of $f$ is the degreewise kernel complex (The kernel of a chain map is computed degreewise).

Proof

technique · direct

1.1

By [L3], the quotient complex $C / \ker (f)$ has $n$ th term $C_{n} / \ker (f_{n})$ . By [L1], each of these terms is canonically isomorphic to $im (f_{n})$ .

L1L3

2.1

By [L2], the complex $im (f)$ has $n$ th term $im (f_{n})$ , and the component isomorphisms from step 1.1 are the coimage-to-image comparisons of the chain map $f$ . Hence they commute with the differentials and assemble to the claimed isomorphism of complexes.

L1L2step 1.1∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Finite biproducts of complexes are computed degreewise

Statement

Let $A$ be an abelian category. Finite biproducts in $Ch (A)$ are formed degreewise. In particular, $H_{n} (C_{∙} \oplus D_{∙}) ≅ H_{n} (C_{∙}) \oplus H_{n} (D_{∙})$ for every $n$ .

Facts & Assumptions

Given: An abelian category $A$ and chain complexes $C_{∙}$ and $D_{∙}$ in $A$ .

[L1]

$Ch (A)$ is additive, so it has finite biproducts (The category of complexes in an additive category is additive).

[L2]

Homology is an additive functor (Homology is an additive functor).

Proof

technique · direct

1.1

The proof of [L1] constructs the biproduct $C_{∙} \oplus D_{∙}$ degreewise, with term $C_{n} \oplus D_{n}$ and differential $d_{n}^{C} \oplus d_{n}^{D}$ . Thus finite biproducts of complexes are formed degreewise.

2.1

Since $H_{n}$ is additive by [L2], it preserves finite biproducts. Applying it to the biproduct from step 1.1 gives $H_{n} (C_{∙} \oplus D_{∙}) ≅ H_{n} (C_{∙}) \oplus H_{n} (D_{∙}) .$

L2step 1.1∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Products and coproducts of complexes are degreewise when they exist and preserve differentials

Statement

Let $(C_{∙}^{(i)})_{i \in I}$ be a family of chain complexes in an abelian category where the termwise products or coproducts exist.

If every $\prod_{i} C_{n}^{(i)}$ exists, then these products form a chain complex with differential characterized by the component differentials, and it is the product of the family in $Ch (A)$ .
If every $∐_{i} C_{n}^{(i)}$ exists, then these coproducts form a chain complex with differential characterized by the component differentials, and it is the coproduct of the family in $Ch (A)$ .

Facts & Assumptions

Given: A family of chain complexes $(C_{∙}^{(i)})_{i \in I}$ in an abelian category.

[L1]

A chain complex is a graded family with $d_{n - 1} d_{n} = 0$ (Chain complex in an abelian category).

[L2]

Products and coproducts are determined by their universal properties (Products and coproducts as limits and colimits of discrete diagrams, including their existence-and-uniqueness equations).

Proof

technique · constructive

1.1

Suppose the products exist. For each degree $n$ , let $P_{n} : = \prod_{i} C_{n}^{(i)}$ . By [L2] there is a unique map $d_{n}^{P} : P_{n} \to P_{n - 1}$ such that for every $i$ , $π_{i}^{n - 1} d_{n}^{P} = d_{n}^{(i)} π_{i}^{n},$ where $π_{i}^{n} : P_{n} \to C_{n}^{(i)}$ is the $i$ th projection. Then $π_{i}^{n - 2} d_{n - 1}^{P} d_{n}^{P} = d_{n - 1}^{(i)} d_{n}^{(i)} π_{i}^{n} = 0$ by [L1], so $d_{n - 1}^{P} d_{n}^{P} = 0$ because all projections of that composite are zero. Thus $P_{∙}$ is a chain complex and has the required product universal property degreewise.

L1L2givenconstruct

2.1

The coproduct case is dual. If $Q_{n} : = ∐_{i} C_{n}^{(i)}$ exists in each degree, [L2] gives a unique $d_{n}^{Q} : Q_{n} \to Q_{n - 1}$ such that for every $i$ , $d_{n}^{Q} ι_{i}^{n} = ι_{i}^{n - 1} d_{n}^{(i)},$ where $ι_{i}^{n} : C_{n}^{(i)} \to Q_{n}$ is the $i$ th coproduct injection. By [L1], $d_{n - 1}^{Q} d_{n}^{Q} ι_{i}^{n} = d_{n - 1}^{Q} ι_{i}^{n - 1} d_{n}^{(i)} = ι_{i}^{n - 2} d_{n - 1}^{(i)} d_{n}^{(i)} = 0,$ so $d_{n - 1}^{Q} d_{n}^{Q} = 0$ because its composites with every injection vanish. Hence $Q_{∙}$ is the coproduct complex.

L1L2givenconstructdischarge-construct∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

An additive functor applies degreewise to complexes and chain maps

Statement

Let $F : A \to B$ be an additive functor between abelian categories. Applying $F$ degreewise to a chain complex $C_{∙}$ and to a chain map $f : C_{∙} \to D_{∙}$ produces a chain complex $F (C)_{∙}$ and a chain map $F (f) : F (C)_{∙} \to F (D)_{∙}$ .

Facts & Assumptions

Given: An additive functor $F : A \to B$ between abelian categories.

[L1]

Additive functors preserve zero morphisms (An additive functor preserves zero morphisms).

[L2]

A chain complex satisfies $d_{n - 1} d_{n} = 0$ (Chain complex in an abelian category).

[L3]

A chain map commutes with differentials (Chain map).

Proof

technique · direct

1.1

If $C_{∙}$ is a chain complex, then $F (d_{n - 1}^{C}) F (d_{n}^{C}) = F (d_{n - 1}^{C} d_{n}^{C}) = F (0) = 0$ by [L1] and [L2]. Hence the objects $F (C_{n})$ with differentials $F (d_{n}^{C})$ form a chain complex.

L1L2givenalgebra

2.1

If $f : C \to D$ is a chain map, then [L3] gives $d_{n}^{D} f_{n} = f_{n - 1} d_{n}^{C}$ . Applying $F$ yields $F (d_{n}^{D}) F (f_{n}) = F (f_{n - 1}) F (d_{n}^{C}),$ so the family $F (f_{n})$ is a chain map $F (C) \to F (D)$ .

L3step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30‡ 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.

An exact functor commutes with homology

Statement

Let $F : A \to B$ be an exact functor between abelian categories. For every chain complex $C_{∙}$ and every $n \in Z$ , there is a canonical isomorphism $F (H_{n} (C)) \tilde{\to} H_{n} (F (C)) .$ These isomorphisms are natural in chain maps.

Facts & Assumptions

Given: An exact functor $F : A \to B$ , a chain complex $C_{∙}$ , and an integer $n$ .

[L1]

An exact functor preserves kernels and cokernels (An additive functor is exact exactly when it preserves kernels and cokernels).

[L2]

An additive functor applies degreewise to complexes and chain maps (An additive functor applies degreewise to complexes and chain maps).

[L3]

$H_{n} (C)$ is the cokernel of the canonical map $β_{n} : B_{n} (C) \to Z_{n} (C)$ (Homology object of a chain complex).

Proof

technique · direct

1.1

By [L2], $F (C)$ is a chain complex. Because $F$ preserves kernels and cokernels by [L1], it carries $\ker (d_{n}^{C})$ to $\ker (d_{n}^{F (C)})$ and $im (d_{n + 1}^{C})$ to $im (d_{n + 1}^{F (C)})$ . Thus it identifies $F (Z_{n} (C))$ with $Z_{n} (F (C))$ and $F (B_{n} (C))$ with $B_{n} (F (C))$ .

L1L2L3

2.1

Applying $F$ to the cokernel description in [L3] and using [L1], the object $F (H_{n} (C))$ is the cokernel of the image of $β_{n}$ under $F$ . Under the identifications of step 1.1, that cokernel is exactly $H_{n} (F (C))$ . This gives the canonical isomorphism $F (H_{n} (C)) ≅ H_{n} (F (C))$ .

L1L3step 1.1

3.1

Naturality follows because the cycle maps, boundary maps, and quotient maps used in steps 1.1 and 2.1 are all functorial in the chain map input.

L2step 1.1step 2.1∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30‡ rests on unproved material (inherited)Open item page →

An exact functor preserves quasi-isomorphisms and reflects them when it is conservative

Statement

Let $F : A \to B$ be an exact functor between abelian categories. Then $F$ preserves quasi-isomorphisms. If $F$ is also conservative, then it reflects quasi-isomorphisms.

Facts & Assumptions

Given: An exact functor $F : A \to B$ and a chain map $f : C_{∙} \to D_{∙}$ .

[L1]

A quasi-isomorphism is a chain map inducing isomorphisms on all homology objects (Quasi-isomorphism).

[L2]

Exact functors commute with homology (An exact functor commutes with homology).

[L3]

A conservative functor reflects isomorphisms (Conservative functor).

Proof

technique · direct

1.1

If $f$ is a quasi-isomorphism, then [L1] says each $H_{n} (f)$ is an isomorphism. By [L2], the induced map on the homology of $F (f)$ is identified with $F (H_{n} (f))$ , hence is an isomorphism. Therefore $F (f)$ is a quasi-isomorphism by [L1].

L1L2

2.1

Now assume $F$ is conservative and $F (f)$ is a quasi-isomorphism. Then [L1] makes each induced map on $H_{n} (F (f))$ an isomorphism. By [L2], those are the morphisms $F (H_{n} (f))$ . Since $F$ reflects isomorphisms by [L3], each $H_{n} (f)$ is an isomorphism, so $f$ is a quasi-isomorphism by [L1].

L1L2L3∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-30Open item page →

Euler characteristic of a finite complex of finite-rank free abelian groups

Definition

Let $C_{∙}$ be a bounded chain complex of abelian groups such that every $C_{n}$ is free of finite rank. Its Euler characteristic is $χ (C) : = \sum_{n \in Z} (- 1)^{n} rank (C_{n}) .$

Boundedness makes the sum finite, and Invariant basis number and the rank of a free module supplies the meaning of $rank$ for each term.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30‡ rests on unproved material (inherited)Open item page →

Euler-Poincare formula for finite free complexes

Statement

Let $C_{∙}$ be a bounded chain complex of finite-rank free abelian groups. Assume that every homology group $H_{n} (C)$ is free of finite rank. Then $χ (C) = \sum_{n \in Z} (- 1)^{n} rank (H_{n} (C)) .$

Facts & Assumptions

Given: A bounded chain complex $C_{∙}$ of finite-rank free abelian groups whose homology groups are free of finite rank.

[L1]

The Euler characteristic is the finite alternating sum $\sum_{n} (- 1)^{n} rank (C_{n})$ (Euler characteristic of a finite complex of finite-rank free abelian groups).

[L2]

Homology is the quotient $Z_{n} (C) / B_{n} (C)$ (Homology object of a chain complex).

[L3]

Abelian groups and $Z$ -modules are the same objects and morphisms (Abelian groups and $Z$ -modules have the same objects and morphisms).

[L4]

The ring $Z$ is a principal ideal domain: it is a commutative ring (The integers form a commutative ring), it has no zero divisors (The integers have no zero divisors; multiplicative cancellation), and every ideal is generated by one integer because every subgroup of $(Z, +)$ is cyclic (Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$ , Principal ideal domain).

[L5]

A submodule of a finite-rank free module over a PID is again finite free (A submodule of a free module of finite rank over a PID is free of no larger rank).

[L6]

A finite free module over a commutative ring is projective (Under the stated choice boundary, free modules are projective and hence flat).

[L7]

A short exact sequence ending in a projective module splits (Equivalent characterizations of projective modules).

[L8]

A free module of rank $r$ has a basis of $r$ elements (The free module on a set and its standard basis, Invariant basis number and the rank of a free module).

Proof

technique · direct

1.1

By [L3] and [L4], each $C_{n}$ is a finite-rank free $Z$ -module over a PID. Therefore [L5] makes every cycle group $Z_{n} (C) \leq C_{n}$ and every boundary group $B_{n} (C) \leq C_{n}$ finite free. Since the homology groups are finite free by the hypothesis, all ranks appearing below are defined.

L3L4L5given

2.1

By [L2], the short exact sequence $0 \to Z_{n} (C) \to C_{n} \to B_{n - 1} (C) \to 0$ has projective quotient $B_{n - 1} (C)$ by steps 1.1 and [L6], so [L7] splits it. Likewise $0 \to B_{n} (C) \to Z_{n} (C) \to H_{n} (C) \to 0$ splits because $H_{n} (C)$ is finite free by hypothesis and hence projective by [L6]. Using [L8], split exactness gives $rank (C_{n}) = rank (Z_{n} (C)) + rank (B_{n - 1} (C))$ and $rank (Z_{n} (C)) = rank (B_{n} (C)) + rank (H_{n} (C)) .$

L2L6L7L8step 1.1algebra

3.1

Substitute the second identity of step 2.1 into the first: $rank (C_{n}) = rank (B_{n} (C)) + rank (H_{n} (C)) + rank (B_{n - 1} (C)) .$ Multiply by $(- 1)^{n}$ and sum over all $n$ . The boundedness in [L1] makes this a finite sum, and the two boundary sums cancel after the index shift $m = n - 1$ : $\sum_{n} (- 1)^{n} rank (B_{n - 1} (C)) = - \sum_{n} (- 1)^{n} rank (B_{n} (C)) .$ Hence only the homology contribution remains.

L1step 2.1algebra

4.1

The resulting identity is $\sum_{n} (- 1)^{n} rank (C_{n}) = \sum_{n} (- 1)^{n} rank (H_{n} (C)),$ which is exactly the formula in the statement by [L1].

L1step 3.1∎

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30‡ rests on unproved material (inherited)Open item page →

FALSE: any sequence of morphisms is a chain complex

Statement

Any sequence of morphisms in an abelian category is a chain complex.

Facts & Assumptions

Given: The sequence $Z \overset{1_{Z}}{\to} Z \overset{1_{Z}}{\to} Z$ in $A b$ .

[L1]

$A b$ is an abelian category (Abelian groups form an abelian category).

[L2]

A chain complex requires consecutive composites to be zero (Chain complex in an abelian category).

Refutation

technique · direct

1.1

By [L1], the displayed sequence is a sequence of morphisms in an abelian category.

L1given

2.1

Its consecutive composite is $1_{Z} \circ 1_{Z} = 1_{Z}$ , which is not the zero endomorphism of $Z$ . Therefore [L2] shows that this sequence is not a chain complex.

L2step 1.1algebra∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

FALSE: the boundaries of a complex are a quotient of its cycles

Statement

For every chain complex $C_{∙}$ and degree $n$ , the boundary object $B_{n} (C)$ is a quotient of the cycle object $Z_{n} (C)$ .

Facts & Assumptions

Given: In the category $A b$ , the two-term complex $0 \to Z \overset{ι}{\to} Q \to 0,$ with $Z$ in degree $1$ , $Q$ in degree $0$ , and $ι$ the inclusion.

[L1]

Boundaries are the images of the incoming differentials, and cycles are the kernels of the outgoing differentials (Cycle and boundary subobjects of a complex).

Refutation

technique · direct

1.1

In degree $0$ , the outgoing differential is $0 : Q \to 0$ , so [L1] gives $Z_{0} (C) = \ker (0) = Q .$ The incoming differential is $ι : Z \to Q$ , so $B_{0} (C) = im (ι) = Z .$

L1given

1.2

Let $ϕ : Q \to Z$ be any group homomorphism. Then for every positive integer $n$ , $ϕ (1) = ϕ (n \cdot \frac{1}{n}) = n ϕ (\frac{1}{n}),$ so $ϕ (1)$ is divisible by every positive integer. The only integer with that property is $0$ , hence $ϕ (1) = 0$ . Therefore $ϕ (\frac{m}{n}) = m ϕ (\frac{1}{n}) = 0$ for every rational number $\frac{m}{n}$ , so $ϕ = 0$ . Thus no epimorphism $Q \to Z$ exists.

algebra

2.1

Step 1.1 shows that in this complex $B_{0} (C) = Z$ and $Z_{0} (C) = Q$ , while step 1.2 shows that $Z$ is not a quotient of $Q$ . Therefore the statement is false.

step 1.1step 1.2∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30‡ rests on unproved material (inherited)Open item page →

FALSE: a chain map is determined by its maps on homology

Statement

If two chain maps induce the same morphism on every homology object, then the two chain maps are equal.

Facts & Assumptions

Given: The two-term complex $0 \to Z \overset{1_{Z}}{\to} Z \to 0$ and the endomorphisms $f = 1_{C}$ and $g = 0_{C}$ of this complex.

[L1]

$A b$ is an abelian category (Abelian groups form an abelian category).

[L2]

Chain maps induce homology maps (A chain map induces a well-defined map on homology).

Refutation

technique · direct

1.1

The complex is acyclic: in degree $1$ the kernel of $1_{Z}$ is $0$ , and in degree $0$ the cokernel of $1_{Z}$ is also $0$ . Hence all its homology objects are zero. The maps $f$ and $g$ are distinct because $f_{1} = 1_{Z}$ while $g_{1} = 0$ .

L1givenalgebra

2.1

By [L2], both $f$ and $g$ induce the zero endomorphism on every homology object, since those homology objects are zero by step 1.1. Therefore equal homology maps do not force equality of chain maps.

L2step 1.1∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30‡ rests on unproved material (inherited)Open item page →

FALSE: every quasi-isomorphism is an isomorphism of complexes

Statement

Every quasi-isomorphism of chain complexes is an isomorphism in $Ch (A)$ .

Facts & Assumptions

Given: The zero map from the acyclic complex $0 \to Z \overset{1_{Z}}{\to} Z \to 0$ to the zero complex.

[L1]

A quasi-isomorphism is a chain map inducing isomorphisms on homology (Quasi-isomorphism).

[L2]

$A b$ is an abelian category (Abelian groups form an abelian category).

Refutation

technique · direct

1.1

As in the standard identity-complex computation, the source complex is acyclic and the target zero complex is also acyclic, so every homology object on both sides is zero. Therefore the zero chain map induces isomorphisms on all homology objects and is a quasi-isomorphism by [L1].

L1L2givenalgebra

2.1

The source complex is nonzero while the target is zero, so no inverse chain map can exist. Thus this quasi-isomorphism is not an isomorphism of complexes.

step 1.1algebra∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30‡ rests on unproved material (inherited)Open item page →

FALSE: an additive functor commutes with homology

Statement

Every additive functor between abelian categories commutes with homology.

Facts & Assumptions

Given: The additive functor $F : A b \to A b$ defined by $F (A) = A / 2 A$ , and the complex $0 \to Z \overset{\times 2}{\to} Z \to 0.$

[L1]

Additive functors apply degreewise to complexes and chain maps (An additive functor applies degreewise to complexes and chain maps).

[L2]

Exactness is the hypothesis that makes a functor commute with homology (An exact functor commutes with homology).

[L3]

Additivity means preservation of sums on hom-groups (Additive functor).

Refutation

technique · direct

1.1

The functor $F$ is additive in the sense of [L3], and by [L1] it sends the displayed complex to $0 \to Z / 2 \overset{0}{\to} Z / 2 \to 0,$ because multiplication by $2$ becomes zero after quotienting by $2 A$ . The original complex has $H_{1} = 0$ and $H_{0} ≅ Z / 2$ , while the new complex has $H_{1} ≅ Z / 2$ .

L1L3givenalgebra

2.1

Therefore $F (H_{1} (C)) = 0 but H_{1} (F (C)) ≅ Z / 2,$ so $F$ does not commute with homology. The contrast with [L2] shows that exactness is genuinely load-bearing.

L2step 1.1∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30‡ rests on unproved material (inherited)Open item page →

FALSE: every infinite coproduct of complexes has homology equal to the coproduct of their homologies

Statement

In every abelian category, whenever an infinite coproduct of chain complexes exists, its homology is the coproduct of the homologies.

Facts & Assumptions

Given: An abelian category with small coproducts and a family of monomorphisms $u_{i} : A_{i} \to B_{i}$ .

[A1]

There exist abelian categories with small coproducts that do not satisfy AB4.

[L1]

AB4 means that every small coproduct of monomorphisms is again monic (The axioms AB4 and AB4*).

[L2]

Degreewise coproducts of complexes, when they exist, are computed degreewise (Products and coproducts of complexes are degreewise when they exist and preserve differentials).

[L3]

A complex is exact at degree $n$ exactly when its $n$ th homology is zero (A complex is exact at n exactly when its nth homology is zero).

[L4]

In an abelian category, a morphism is monic exactly when its kernel is zero (In an abelian category, monic means zero kernel and epic means zero cokernel).

Refutation

technique · direct

1.1

For each $i$ , form the two-term complex $C_{∙}^{(i)} : 0 \to A_{i} \overset{u_{i}}{\to} B_{i} \to 0,$ with $A_{i}$ in degree $1$ and $B_{i}$ in degree $0$ . Because $u_{i}$ is monic, [L4] gives $\ker (u_{i}) = 0$ , so [L3] implies $H_{1} (C^{(i)}) = 0$ for every $i$ .

L3L4given

2.1

If the statement were true, then by [L2] the coproduct complex $∐_{i} C^{(i)}$ would satisfy $H_{1} ⁣ (∐_{i} C^{(i)}) ≅ ∐_{i} H_{1} (C^{(i)}) = 0.$ Its degree- $1$ differential is exactly $∐_{i} u_{i}$ , so [L3] and [L4] would force $∐_{i} u_{i}$ to be monic. By [L1], that would make every abelian category with small coproducts satisfy AB4.

L1L2L3L4step 1.1

3.1

This contradicts [A1]. Therefore the statement is false.

A1step 2.1∎