Complements and the Krull-Schmidt Theorem in Arbitrary Categories

Bodo Pareigis

Mathematisches Institut der Universität München Germany

and

Helmut Röhrl

9322 La Jolla Farms Rd. La Jolla, CA 92037 USA

Abstract. We study direct product decompositions of objects in a finitely complete and cocomplete category with zero object and certain axioms for a coimage factorization of morphisms. Direct products C = A × B can be characterized by ”inner” properties of C and its subobjects A and B. We also show that the Fitting Lemma and the Krull-Schmidt Theorem hold.

Key words: Complements, internal direct products, summable morphisms and idempotent morphisms, Fitting Lemma, Krull-Schmidt Theorem.

1 Introduction

Let C be a category with finite limits, finite colimits, and a 0-object such that the following axioms are satisfied:

(I) for every morphism f : A ⟶ B, the induced (and uniquely determined) morphism g in the commutative diagram

[figure: A; A ∐ B; j_{A}; B; j_{B}; A; A × B; p_{A}; B; p_{B}; id; id; g; 0; f]

is a difference cokernel;

(II) if, in the commutative diagram

[figure: A; C; f; B; g]

f is a difference cokernel and g has kernel 0 then g is an isomorphism.

Axiom I guarantees that A × B is the ”span” of the subobjects g(A) and g(B). In the category of sets A ∐ B = A ˙∪B is too small for this purpose, but in the presence of algebraic structures very often A ∐ B will become large enough. If f is the zero morphism, then we shall use the notation A ∪ B = A × B to denote that A × B is spanned by A and B.

As we shall see later on axiom I has the consequence that certain morphisms in C can be added. Axiom II means that C is in some sense balanced (cf. a monoand epimorphism is an isomorphism).

Two fundamental examples are Gr the category of groups and Mo the category of monoids. Gr satisfies (I) and (II) as will be seen soon in more general contexts, whereas Mo does not satisfy any of (I) or (II). A counterexample for (I) is A = B = (N_{0}, +) and f = id. It is easy to see that (1, 0) ∉ Im(g), and g is surjective if and only if g is a difference cokernel. To give a counterexample for (II) adjoin an additional element ∞ to (N_{0}, +) such that n+∞ = ∞+n = n+1 and ∞+∞ = 2. Then N_{0} ∪{∞} is a monoid and f : N_{0} ∪{∞} ⟶ N_{0} with f(n) = n and f(∞) = 1 is surjective (a difference cokernel) with kernel 0. But f is not bijective.

To find a larger class of categories C which satisfy the hypotheses, let us consider an (equationally defined) algebraic category C. The final object E is always the set with one element with the unique algebra structure on it. In order to be a 0-object in C it is necessary and sufficient that, for any algebra A in C, there is a unique algebra morphism E ⟶ A; this means that there must be precisely one 0-ary operation in the theory for C (and it must be compatible with all other operations on A in the obvious sense). Let us call the distinguished element 0 for every algebra A. Assume that there is an m-ary operation ω : A × … × A ⟶ A for some m ≥ 2 and i ≤ m, such that for every algebra A

1. ∀a ∈ A : ω(0, …, a, …, 0) = a (a is in the i-th place),

2. ∀a, a′ ∈ A∃a_{1}, …, a_{m} ∈ A : ω(a_{1}, …, a′, …, a_{m}) = a (a′ is in the i-th place).

Then axiom I is satisfied. To prove this let f : A ⟶ B be an algebra morphism and let (a, b) ∈ A×B be given. By definition of g we have gj_{A}(a) = (a, f(a)) ∈ A× B and gj_{B}(b) = (0, b) ∈ A × B. Pick b_{1}, …, b_{m} such that ω(b_{1}, …, f(a), …, b_{m}) = b, then g(ω(j_{B}(b_{1}), …, j_{A}(a), …, j_{B}(b_{m}))) = ω((0, b_{1}), …, (a, f(a)), …, (0, b_{m})) = (ω_{A}(0, …, a, …, 0), ω_{B}(b_{1}, …, f(a), …, b_{m})) = (a, b). Thus g is surjective which in an algebraic category means the same as g is a difference cokernel ([2], 3.4 Cor. 4).

If there is a binary operation ν such that ν(a, b) = 0 if and only if a = b, then C also satisfies (II). To prove this let g : B ⟶ C have kernel zero then g(b) = g(b′) ⇒ 0 = ν(g(b), g(b′)) = g(ν(b, b′)) ⇒ ν(b, b′) = 0 ⇒ b = b′, hence g is injective. However, in an algebraic category a bijective morphism is an isomorphism.

In particular all algebraic categories where the objects have an underlying group structure and no further distinguished elements satisfy our conditions for C, e.g. rings (without unit elements), associative rings, Lie rings etc. The binary operation ν can be chosen to be ν(a, b) = a − b. Similarly the category of loops (”nonassociative” groups) satisfy the conditions for C.

We will show that certain generalizations of the Fitting Lemma and the Krull-Schmidt Theorem hold in our categories. Instead of considering congruence relations as in [3] or a modular lattice of subobjects as in [1] we use specific subobjects together with certain endomorphisms to prove these theorems.

2 Complements and internal direct products

Let C satisfy the axioms discussed in section 1. We are interested in the question, whether inside an object A × B in C there can be other ”subobjects” X of A × B such that X × B = A × B. For this we have to make precise what the equality means and how A and B are subobjects of A × B. A subobject will be used in the sense of [2], that is as a representative of the usual equivalence class of monomorphisms.

In the following definition p_{A} : A × B ⟶ A (resp. p_{B} : A × B ⟶ B) denotes the canonical projection.

Definition 2.1 1. A subobject ι_{X} : X ↪ A × B is called a weak complement of B, if the morphism X ⟶^{ι_{X}} A × B ⟶^{p_{A}} A has kernel 0 ⟶ X.

2. A subobject ι_{X} : X ↪ A × B is called a complement of B, if the morphism X ⟶^{ι_{X}} A × B ⟶^{p_{A}} A is an isomorphism.

3. U is called internal direct product of the subobjects A ↪ U and B ↪ U if

a) A ↪ U and B ↪ U are kernels,

b) the intersection of A and B is A ∩ B = 0,

c) the canonical morphism A ∐ B ⟶ U is a difference cokernel (a fact which we abbreviate by A ∪ B = U).

Let A and B be objects in C. We consider B as a subobject of A × B via the canonical morphism ι̃_{B} : B ⟶ A × B induced by id : B ⟶ B and 0 : B ⟶ A.

Lemma 2.2 B \underrightarrow{_{ι_{B}}} A × B ⟶^{p_{A}} A is a kernel diagram.

Proof: If g : X ⟶ A × B is given with p_{A}g = 0 then p_{A}g = p_{A}ι̃_{B}p_{B}g = 0 and p_{B}g = p_{B}ι̃_{B}p_{B}g implies g = ι̃_{B}p_{B}g, a factorization of g through ι̃_{B}. Since ι̃_{B} is a section this factorization is unique.

Proposition 2.3 ι_{X} : X ↪ A × B is a weak complement of B if and only if X ∩ B = 0.

Proof: Let X be a weak complement of B. In the commutative diagram

[figure: X ∩ B; B; u_{B}; 0; X; A × B; ι_{X}; A,; p_{A}; ν; u_{X}; _{ι_{B}}; 0]

with u_{B} and u_{X} the canonical morphisms, we have p_{A}ι_{X}u_{X} = 0 hence there is a factorization ν of u_{X} through 0 by the property of weak complements. Since u_{X} is a zero-morphism and a monomorphism we get X ∩ B = 0.

Let X ∩ B = 0. In the commutative diagram

[figure: 0; X; B; A × B; _{ι_{B}}; A; p_{A}; Ke(f); g; h; ι_{X}; f]

B \underrightarrow{_{ι_{B}}} A×B ⟶^{p_{A}} A is a kernel diagram by Lemma 2.2. Hence h can be constructed uniquely from g such that ι̃_{B}h = ι_{X}g. Now g can be factored through 0 and thus must be zero. This means that X is a weak complement of B.

Proposition 2.4 Let ι_{X} : X ↪ A×B be a subobject. The following are equivalent:

1. X is a complement of B.

2. There is a unique morphism f : A ⟶ B and an epimorphism g : A ⟶ X such that the diagram

[figure: A; A × B; p_{A}; B; p_{B}; A; id; f; X; g; ι_{X}]

commutes.

3. The induced morphism h in the commutative diagram

[figure: X; X × B; p_{X}; B; p′_{B}; A; A × B; p_{A}; B; p_{B}; A × B; ι_{X}; p_{A}; h; id]

is an isomorphism.

Proof: (1) ⇔ (2): p_{A}ι_{X} is an isomorphism if and only if there is an epimorphism g with (p_{A}ι_{X})g = id. Thus f can be constructed uniquely such that f = p_{B}ι_{X}g.

(1) ⇒ (3): If p_{A}ι_{X} is an isomorphism then, obviously, h is an isomorphism.

(3) ⇒ (1): Let h be an isomorphism. In the commutative diagram

[figure: A; A × B; _{ι_{A}}; X; X × B; p_{X}; A; A × B; p_{A}; p_{X}h^{−1}_{ι_{A}}; A × B; ι_{X}; p_{A}; h^{−1}; h]

we have p_{A}ι̃_{A} = id. Hence (A ⟶ X ⟶ A) = id in the above diagram. Thus it suffices to show that p_{A}ι_{X} is a monomorphism. Suppose that p_{A}ι_{X}f = p_{A}ι_{X}g. Then we have the commutative diagram

[figure: g; f; Y; _{f}; _{g}; 0; X; X × B; p_{X}; B; p′_{B}; A; A × B; p_{A}; B; p_{B}; p_{A}ι_{X}; h; id]

Define f̃ and g̃ by the universal property of the product X ×B. Then p_{A}hf̃ = p_{A}hg̃ and p_{B}hf̃ = 0 = p_{B}hg̃ and hence hf̃ = hg̃ and f̃ = g̃ since h is an isomorphism. So f = g and p_{A}ι_{X} is a monomorphism.

Remark 2.5 1) In some sense 2.4, (3), means that A × B is generated by the subobjects X and B. Observe, however, that there is also a canonical morphism X ∐ B ⟶ A×B which, in general, does not factor through X ×B in the canonical way (e.g. Gr ) nor is it an epimorphism (e.g. commutative monoids).

2) In 2.4, (2), one can consider X as the graph of the morphism f : A ⟶ B. So this part of Proposition 2.4 may be rephrased as:

there is a bijection between the complements of B in A × B and the morphisms f : A ⟶ B.

To show that each f determines a subobject X of A × B, let f̃: A ⟶ A × B be the morphism with p_{A}f̃ = id and p_{B}f̃ = f. Then (A, f̃) is a subobject of A × B, namely the graph of f.

Proposition 2.6 Let ι_{X} : X ⟶ A × B be a complement of B. Then we have X ∩ B = 0 and X ∪ B = A × B.

Proof: Since a complement is a weak complement we get X ∩ B = 0 by Proposition 2.3. In the commutative diagram

[figure: A; A ∐ B; B; X; X ∐ B; j_{X}; B; A; A × B; p_{A}; B; p_{B}; ≅; ≅; ≅; =; =; ι_{X}]

the morphism A ∐ B ⟶ A × B is a difference cokernel by axiom (I) and so is X ∐ B ⟶ A × B. Hence X ∪ B = A × B.

Theorem 2.7 1. A × B is an internal direct product of the subobjects A and B.

2. If A and B are subobjects of U such that U is an internal direct product of A and B, then there is an isomorphism U ≅ A × B such that

[figure: U; A; B; ≅; A × B]

commutes.

Proof: (1) Since p_{A}ι̃_{A} = id is an isomorphism, A is a complement of B in A × B. Thus by Lemma 2.2 and Proposition 2.6 we get that A × B is an internal direct product of A and B.

(2) Given an internal direct product U of A and B. Let U ⟶ X be the cokernel of A ⟶ U. Then A ⟶ U is the kernel of U ⟶ X since it was a kernel. In the commutative diagram

[figure: 0; B; A; U; X; Ke(f); f]

the morphism Ke(f) ⟶ A and Ke(f) ⟶ 0 exist since A ⟶ U is the kernel of U ⟶ X and 0 = A ∩ B is a pull-back. Hence Ke(f) = 0. Now let

[figure: P; U; g; B; X; f]

be a pull-back and consider the commutative diagrams

[figure: (*); P; U; g; B; X; f; B; ι_{B}; id; P; U; g; B; X.; f; A; ι_{A}; 0]

They induce the canonical morphism A ∐ B ⟶ P ⟶ U, which is a difference cokernel. Since Ke(f) = 0 the diagram

[figure: P; U; g; B; X; f; Ke(g); 0 = Ke(f)]

commutes and the canonical morphism Ke(g) ⟶ P is the zero morphism. By axiom (II) g is an isomorphism. Thus we can replace P by U and g by id so that there is a factorization U ⟶ B ⟶^{f} X of U ⟶ X, which is the cokernel of A ⟶ U.

Since (A ⟶ U ⟶ X) = 0 and Ke(f) = 0 we get (A ⟶ U ⟶ B) = 0 and hence a factorization (U ⟶ X ⟶ B) = (U ⟶ B). Since U ⟶ X is the cokernel of A ⟶ U we get (X ⟶ B ⟶^{f} X) = id. Now f has kernel zero and is a difference cokernel of (B ⟶^{f} X ⟶ B, B ⟶^{id} B). Hence by axiom (II) we obtain that f : B ⟶ X is an isomorphism. Therefore we may replace X by B and consider U ⟶ B as cokernel of A ⟶ U. Furthermore we have (B ⟶ U ⟶ B) = id by diagram (∗). Analogously we get a morphism U ⟶ A, which is a cokernel of B ⟶ U, such that (A ⟶ U ⟶ A) = id.

Now we prove U ≅ A×B. Let Y be the kernel of h in the following commutative diagram

[figure: A; Y; B; A; U; B; A; A × B; B; =; =; =; h; =]

To prove the existence of Y ⟶ A observe that A ⟶ U is the kernel of U ⟶ B and that (Y ⟶ U ⟶ B) = (Y ⟶ U ⟶ A × B ⟶ B) = 0. Thus we get a commutative diagram

[figure: 0; A; B; U; Y]

as A ∩ B = 0. This shows that (Y ⟶ U) = 0. Thus h : U ⟶ A × B has kernel 0. On the other hand we have a commutative diagram

[figure: A; U; B; A; A × B; B; A ∐ B; =; h; =]

which implies by axiom (I) that A ∐ B ⟶ U ⟶ A × B is a difference cokernel. By axiom (II) h is an isomorphism and the above diagram proves that the diagram in the theorem commutes.

3 Summable morphisms

In this paragraph we shall introduce an addition of certain morphisms. One of the aims is the proof of a formula id = ι̃_{A}p_{A} + ι̃_{B}p_{B} for A × B. Let C be as in sections 1 and 2.

Definition 3.1 Let σ : A ∐ A ⟶ A × A be the canonical morphism defined by the commutative diagram

[figure: A; A ∐ A; j_{1}; A; j_{2}; A; A × A; p_{1}; A; p_{2}; id; id; σ; 0; 0]

By axiom (I) σ is a difference cokernel. Let f, g ∈ C(A, B). f and g are called summable if the canonical morphism h of

[figure: A; A ∐ A; A; B; f; h; g]

factors (necessarily uniquely) through σ. The factorization morphism A×A ⟶ B will be written as ⟨f, g⟩.

A similar definition can be given for n morphisms. The family (f_{i}|i = 1, …, n) is summable if for all i, j with 1 ≤ i < j ≤ n the morphism induced by the f_{i}, f_{i+1}, …, f_{j} factors through ∐_{k=i}^{j} A ⟶ ∏_{k=i}^{j} A. The factorization is denoted by ⟨f_{i}, f_{i+1}, …, f_{j}⟩ : A × … × A ⟶ B.

If (f_{i}|i = 1, …, n) is summable then the sum f_{1} + … + f_{n} is defined as the

morphism A ⟶^{Δ^{n}} A × … × A ⟶^{⟨f_{1},...,f_{n}⟩} B.

Lemma 3.2 Let f_{i} ∈ C(A, B), i = 1, …, n, h ∈ C(B, C) and k ∈ C(D, A). Let (f_{i}|i = 1, …, n) be summable. Then (f_{i}k|i = 1, …, n) and (hf_{i}|i = 1, …, n) are summable and we have h(f_{1} + … + f_{n}) = hf_{1} + … + hf_{n} and (f_{1} + … + f_{n})k = f_{1}k + … + f_{n}k.

Proof: It is sufficient to prove that factorizations ⟨hf_{1}, …, hf_{n}⟩ resp. ⟨f_{1}k, …, f_{n}k⟩ exist. Observe that the factorization ⟨f_{1}, …, f_{n}⟩ is the only morphism which makes all the diagrams

[figure: A; A^{n}; _{ι_{i}}; f_{i}; B; ⟨f_{1},...,f_{n}⟩]

commute. Thus the diagrams

[figure: A; A^{n}; _{ι_{i}}; A; Δ^{n}; B; f_{i}; f_{1}+...+f_{n}; C; hf_{i}; g; ⟨f_{1},...,f_{n}⟩; h]

commute with ⟨hf_{1}, …, hf_{n}⟩ = h⟨f_{1}, …, f_{n}⟩ and hf_{1} + … + hf_{n} = g = h(f_{1} + … + f_{n}). The second part follows from the commutative diagrams

[figure: D; D^{n}; _{ι_{i}}; D; Δ^{n}; A; A^{n}; _{ι_{i}}; A; Δ^{n}; k; k^{n}; k; f_{i}; C; ⟨f_{1},...,f_{n}⟩; f_{1}+...+f_{n}]

by ⟨f_{1}, …, f_{n}⟩k^{n} = ⟨f_{1}k, …, f_{n}k⟩ and (f_{1} + … + f_{n})k = f_{1}k + … + f_{n}k.

Lemma 3.3 Sums of summable morphisms satisfy the associative law. In particular if (f_{1}, f_{2}, f_{3}) ∈ C(A, B)^{3} is summable then (f_{1} + f_{2}) + f_{3} = f_{1} + f_{2} + f_{3} = f_{1} + (f_{2} + f_{3}).

Proof: We prove only the second statement. The first follows by standard reasoning. Consider the following commutative diagram

[figure: A; A × A; ι_{1}^{2}; A; ι_{2}^{2}; A × A; A × A × A; ι_{1,1}^{3}; A; Δ; Δ; Δ×id; ι_{3}^{3}; ⟨f_{1},f_{2}⟩; f_{1}+f_{2}; f_{3}; B; ⟨f_{1},f_{2},f_{3}⟩]

where the composite vertical morphism denotes (f_{1} +f_{2})+f_{3} (and also f_{1} +f_{2} +f_{3} in case the diagram commutes). The morphisms ι̃^{2} are morphisms into A^{2} and ι̃^{3} into A^{3}. The only commutativity which is not immediately clear is ⟨f_{1}, f_{2}, f_{3}⟩ι̃_{1,1}^{3} = ⟨f_{1}, f_{2}⟩, but we have ⟨f_{1}, f_{2}, f_{3}⟩ι̃_{1,1}^{3}ι̃_{1}^{2} = ⟨f_{1}, f_{2}, f_{3}⟩ι̃_{1}^{3} = f_{1} = ⟨f_{1}, f_{2}⟩ι̃_{1}^{2} and similarly ⟨f_{1}, f_{2}, f_{3}⟩ι̃_{1,1}^{3}ι̃_{2}^{2} = ⟨f_{1}, f_{2}⟩ι̃_{2}^{2}. By the uniqueness of the factorization ⟨f_{1}, f_{2}⟩ we get the required commutativity.

Lemma 3.4 For each morphism f ∈ C(A, B) the morphisms 0 and f are summable and we have 0 + f = f = f + 0.

Proof: It is sufficient to prove 0+id = id for then 0+f = 0f +id f = (0+id)f = id f = f by Lemma 3.2. But the factorization ⟨0, id⟩ is p_{2} : A × A ⟶ A since

[figure: A; A × A; _{ι_{1}}; A; _{ι_{2}}; A; 0; p_{2}; id]

commutes by definition of ι̃_{1} and ι̃_{2}. Hence 0 + id = ⟨0, id⟩Δ = p_{2}Δ = id.

Proposition 3.5 Let U = A_{1} × … × A_{n}. Then id_{U} = ι̃_{1}p_{1} + … + ι̃_{n}p_{n}.

Proof: To begin the proof by induction assume U = A × B. The diagram

[figure: U; U × U; _{ι_{1}}; U; _{ι_{2}}; A; A × B = U; _{ι_{A}}; A; _{ι_{B}}; p_{A}; p_{B}; p_{A}×p_{B}]

commutes, whence p_{A} × p_{B} = ⟨ι̃_{A}p_{A}, ι̃_{B}p_{B}⟩ and id_{U} = (p_{A} × p_{B})Δ = ι̃_{A}p_{A} + ι̃_{B}p_{B}. To indicate the induction step assume V = A × B × C = U × C. Then id_{V} = ι̃_{U}p_{U} + ι̃_{C}p_{C} = ι̃_{U} id_{U} p_{U} + ι̃_{C}p_{C} = ι̃_{U}(ι̃_{A}^{U}p_{A}^{U} + ι̃_{B}^{U}p_{B}^{U})p_{U} + ι̃_{C}p_{C} = ι̃_{U} ι̃_{A}^{U}p_{A}^{U}p_{U} + ι̃_{U} ι̃_{B}^{U}p_{B}^{U}p_{U} + ι̃_{C}p_{C} = ι̃_{A}p_{A} + ι̃_{B}p_{B} + ι̃_{C}p_{C}.

Example 3.6 We want to prove in the case C = Gr , the category of groups, that two morphisms f, g : A ⟶ B are summable if and only if f(x)g(y) = g(y)f(x) for all x, y ∈ A. Given ⟨f, g⟩ we have ⟨f, g⟩(x, y) = ⟨f, g⟩((x, e) · (e, y)) = ⟨f, g⟩(x, e) · ⟨f, g⟩(e, y) = f(x) · g(y). Since (x, e) and (e, y) commute we get f(x)g(y) = g(y)f(x). Conversely it is a well-known exercise that this condition implies that ⟨f, g⟩ is a homomorphism. The sum f + g is then defined by (f + g)(x) = f(x)g(x).

4 Idempotent morphisms and the Fitting Lemma

First we need some facts about coimages. The coimage of f : A ⟶ B is defined as the difference cokernel of the kernel pair of f ([2] p.70, Lemma 4a)). In the canonical factorization A ⟶^{f′} Coim(f) ⟶^{ι} B of f, ι fails to be a monomorphism in general.

Lemma 4.1 Given f : A ⟶ B and g : B ⟶ C, then there is a unique morphism k : Coim(gf) ⟶ Coim(g) such that

[figure: A; B; f; C; g; Coim(gf); Coim(g); k]

commutes. If f is an isomorphism then so is k.

Proof: This is an easy exercise in universal properties of difference cokernels and kernel pairs.

One proves just as easily

Lemma 4.2 Given g : A ⟶ B and f : B ⟶ C, there is a unique morphism k : Coim(g) ⟶ Coim(fg) such that

[figure: A; B; g; C; f; Coim(g); Coim(fg); k]

commutes. If f is an isomorphism then so is k.

Definition 4.3 Let f : A ⟶ A be idempotent. We say that f satisfies condition (G_{f}) if the canonical morphism Ke(f) ∐ Coim(f) ⟶ A is a difference cokernel and Coim(f) ⟶ A is a kernel.

In the category of groups Gr the first condition is always satisfied, indeed a ↦ af(a^{−1}) ∗ f(a) ∈ Ke(f) ∐ Coim(f) is a section for the given morphism. The second condition is in Gr equivalent to f being normal (see 4.7). In Mo, the category of monoids, let M = (Z/3Z)^{×} be the multiplicative monoid of Z/3Z and f : M ⟶ M be given by f(\overline{0}) = \overline{0}, f(\overline{1}) = f(\overline{2}) = \overline{1}. Then f is idempotent and Ke(f) ∐ Coim(f) ⟶ M is surjective. But Coim(f) ⟶ M fails to be a kernel. To see this observe that Coim(f) = \{\overline{0}, \overline{1}\} and the kokernel of Coim(f) ⟶ M is M ⟶ \{\overline{1}\}. The kernel of this morphism is id : M ⟶ M, but not Coim(f) ⟶ M.

Now let C be again as in section 1.

Lemma 4.4 Let f : A ⟶ A be idempotent with (G_{f}). Then A = Ke(f) × Coim(f).

Proof: Ke(f) ⟶ A and Coim(f) ⟶ A are kernels. Furthermore (G_{f}) implies A = Ke(f) ∪ Coim(f). Next we show Ke(f) ∩ Coim(f) = 0. In the diagram

[figure: A; Coim(f); f′; A; ι; ι; A; f]

we have ιf′ = f = f^{2} = fιf′ hence ι = fι, since f′ is an epimorphism. In the commutative diagram

[figure: Ke(f) ∩ Coim(f); Ke(f); Coim(f); A; ι; A; f; j; 0]

we have ιj = fιj = 0 hence Ke(f) ∩ Coim(f) = 0. Hence Theorem 2.7 finishes the proof.

Let f : A ⟶ A be an endomorphism. Then f has a factorization A ⟶^{f′} Coim(f) ⟶^{ι} A. By Lemma 4.1 there is also a canonical morphism ι_{n} : Coim(f^{n}) ⟶ Coim(f^{n−1}) and by Lemma 4.2 there is a canonical morphism f′_{n} : Coim(f^{n−1}) ⟶ Coim(f^{n}).

Definition 4.5 An endomorphism f : A ⟶ A is called bounded, if the families (ι_{n} : Coim(f^{n}) ⟶ Coim(f^{n−1})) and (f′_{n} : Coim(f^{n−1}) ⟶ Coim(f^{n})) become stationary (i.e. there is n_{0} such that for all n ≥ n_{0} both ι_{n} and f′_{n} are isomorphisms), and if for each n there is r ≥ n such that Ke(f^{r}) ∐ Coim(f^{r}) ⟶ A is a difference cokernel and Coim(f^{r}) ⟶ A is a kernel.

Proposition 4.6 (Fitting Lemma) Let f : A ⟶ A be bounded. Then for every n_{0} ∈ N there is an n ≥ n_{0} such that

A = Ke(f^{n}) × Coim(f^{n}).

Proof: The chains (ι_{n}) and (f′_{n}) become stationary for all n ≥ n_{0}. In particular we have an inverse (f′_{n}ι_{n})^{−n} of (f′_{n}ι_{n})^{n}. The morphism

φ : A ⟶ Coim(f^{n}) ⟶^{(f′_{n}ι_{n})^{−n}} Coim(f^{n}) ⟶ A

is idempotent which follows from the commutative diagram

[figure: A; A; f^{n}; Coim(f^{n}); Coim(f^{n}); (f′_{n}ι_{n})^{−n}; A; A; f^{n}; A; f^{n}; Coim(f^{n}); Coim(f^{n}); (f′_{n}ι_{n})^{−n}; A; A; f^{n}; id; id]

in which φ^{2} : A ⟶ A is easily identified. The diagonal Coim(f^{n}) ⟶ Coim(f^{n}) is the identity and gives a commutative lower triangle since A ⟶ Coim(f^{n}) is an epimorphism. Thus A ⟶ Coim(f^{n}) ⟶ Coim(f^{n}) ⟶^{id} Coim(f^{n}) ⟶ A in the diagram is φ.

The only remaining problem is the commutativity of the rectangles. It follows from the diagram

[figure: A; A; f^{n−1}; Coim(f^{n}); Coim(f^{n}); A; A; f^{n−1}; A; f; Coim(f^{n−1}); f′_{n}; Coim(f^{n}); ι_{n}; A; f]

where the left part is an (n − 1)-fold repetition of the right part and the right part commutes by Lemma 4.1 and Lemma 4.2 with g = f^{n−1}.

For suitably large n we have also Coim(f^{n}) ⟶ A a kernel. Define φ′ := (A ⟶

Coim(f^{n}) ⟶^{(f′_{n}ι_{n})^{−n}} Coim(f^{n})). Since (f′_{n}ι_{n})^{−n} is an isomorphism, we have that φ′ : A ⟶ Coim(f^{n}) is a coimage of f^{n} as well as of φ = (A ⟶^{φ′} Coim(f^{n}) ⟶ A). Thus Coim(φ) ⟶ A is a kernel and Ke(φ) = Ke(A ⟶ Coim(φ)) = Ke(A ⟶ Coim(f^{n})) = Ke(f^{n}). Since f is bounded, φ satisfies (G_{φ}) and Lemma 4.4 holds for φ. Translated back into terms of f gives the required result.

Example 4.7 Consider the category of groups Gr . A morphism f : G ⟶ G is called normal if f(aba^{−1}) = af(b)a^{−1} for all a, b ∈ G. If f is normal and G has a.c.c. and d.c.c. then for all n we have f^{n}(aba^{−1}) = af^{n}(b)a^{−1} and thus Coim(f^{n}) = Im(f^{n}) normal in G. The chains of subgroups Ke(f^{n}) ⊆ Ke(f^{n+1}) and Im(f^{n}) ⊇ Im(f^{n+1}) become stationary. Finally a ↦ a(f′_{n}ι_{n})^{−1}f^{n}(a^{−1}) ∗ (f′_{n}ι_{n})^{−1}f^{n}(a) is a section for the canonical map Ke(f^{n}) ∐ Im(f^{n}) ⟶ G. Thus f is bounded and the Fitting Lemma holds.

5 The Krull-Schmidt-Theorem

Definition 5.1 Let A ≠ 0 be in C. We call A indecomposable if A = X ×Y implies X = 0 or Y = 0.

Lemma 5.2 Let A be indecomposable and f : A ⟶ A a bounded endomorphism. Then f is either nilpotent or an automorphism.

Proof: By the Fitting Lemma there is an n ∈ N such that A = Ke(f^{n}) × Coim(f^{n}). If Coim(f^{n}) = 0 then f^{n} = 0 and f is nilpotent. If Ke(f^{n}) = 0 then Coim(f^{n}) ⟶ A must be the identity. Hence f^{n} : A ⟶ A has kernel zero and is a difference cokernel. By axiom (II) for C we get that f^{n} and also f are automorphisms.

Definition 5.3 Given A = B × C. We define the subset X ⊆ End(B) of Aproductive endomorphisms as follows:

1. If A = B′ × C′ then (B ⟶ A ⟶ B′ ⟶ A ⟶ B) ∈ X.

2. If f, g ∈ X, then fg ∈ X.

3. If f ∈ X ∩ Aut(B), then f^{−1} ∈ X.

4. If f, g ∈ X are summable then f + g ∈ X.

5. X = the smallest set satisfying (1), …, (4).

Definition 5.4 A is bounded if for all A = B ×C all A-productive endomorphisms f are bounded.

Lemma 5.5 Let B be indecomposable and A = B × C be bounded. Let f, g be A-productive and summable. If f and g are nilpotent, then so is f + g.

Proof: Without loss of generality we assume f ≠ 0 ≠ g. Since f + g is A-productive it is bounded. By Lemma 5.2 f + g is either nilpotent or an automorphism. Assume f + g ∈ Aut(B). Then h = (f + g)^{−1} is A-productive and so are hf and hg. Thus id = h(f + g) = hf + hg with hf and hg A-productive, hence bounded. Since f is nilpotent, there is an n such that f^{n} ≠ 0 and f ·f^{n} = 0. So f^{n} factors through Ke(f) ⟶ B and we get Ke(f) ≠ 0. This implies that Ke(hf) ≠ 0 and thus hf is not an automorphism. By Lemma 5.2 hf is then nilpotent. Replacing f by hf and g by hg we can assume without loss of generality that f and g are summable, A-productive and nilpotent with id = f + g.

Let f^{n} = 0 = g^{n}. We prove by induction on k that f^{t_{1}}gf^{t_{2}}g … gf^{t_{k}} = 0 for t_{i} ≥ 0 and ∑_{i=1}^{k} t_{i} = n. For k = 1 this is trivial. Using Lemma 3.2 and Lemma 3.4 the induction step is f^{t_{1}}g … gf^{t_{k+1}} = f^{t_{1}}g … gf^{t_{k}}ff^{t_{k+1}} + f^{t_{1}}g … gf^{t_{k}}gf^{t_{k+1}} = f^{t_{1}}g … gf^{t_{k}}(f + g)f^{t_{k+1}} = f^{t_{1}}g … gf^{t_{k}}f^{t_{k+1}} = 0. With this remark we get (f + g)^{2n} = 0 since in the expansion each summand contains at least n factors f or n factors g, so it is zero. Thus f + g = id cannot hold with B ≠ 0.

Corollary 5.6 Let A = B×C be bounded and B be indecomposable. Let f_{1}, …, f_{n} ∈ End(B) be summable and A-productive. If f_{1} + … + f_{n} = id then one of the f_{i} is an automorphism.

Proof: If all f_{i} are nilpotent then a simple induction proof shows that f_{1} + … + f_{n} is nilpotent. So one of the f_{i} cannot be nilpotent, and hence it must be an automorphism.

Theorem 5.7 (Krull-Schmidt) Let

A = A_{1} × … × A_{m} = B_{1} × … × B_{n}

be two decompositions of A into internal direct products of indecomposable subobjects A_{i} resp. B_{j}. Let A be bounded. Then m = n and A_{i} ≅ B_{i} for all i and a suitable reordering of the B_{i}s.

Proof: We prove the following statement by induction for t ≤ min(m, n). P (t): there is a reordering of B_{1}, …, B_{n} such that A_{i} ≅ B_{i} for i = 1, …, t and

A = A_{1} × … × A_{t} × B_{t+1} × … × B_{n}. P (0) holds by hypothesis. Assume that P (t − 1) holds. Then A = A_{1} × … × A_{t−1} × B_{t} × … × B_{n} with suitable indexing and A_{i} ≅ B_{i} for 1 ≤ i ≤ t − 1. Let p′_{i} and ι′_{i} be the corresponding projections resp. injections. Furthermore we have A = A_{1} × … × A_{m} with the projections p_{i} and the injections ι_{i}. Observe that ι_{i} = ι′_{i} for i = 1, …, t − 1, and that we use the same subobjects A_{1}, …, A_{t−1}, but the projections may be different. By Proposition 3.5 we have id = ι′_{1}p′_{1} +…+ι′_{n}p′_{n}. Since p_{t}ι′_{i} = p_{t}ι_{i} = 0 for 1 ≤ i ≤ t − 1 by definition of the injections, we get

id_{A_{t}} = p_{t}ι_{t} = p_{t} id ι_{t} = p_{t}ι′_{1}p′_{1}ι_{t} + … + p_{t}ι′_{n}p′_{n}ι_{t} = p_{t}ι′_{t}p′_{t}ι_{t} + … + p_{t}ι′_{n}p′_{n}ι_{t}.

By the Corollary 5.6 one of the p_{t}ι′_{i}p′_{i}ι_{t} must be an automorphism. After reindexing we may assume that p_{t}ι′_{t}p′_{t}ι_{t} is an automorphism of A_{t}. Since (p_{t}ι′_{t}p′_{t}ι_{t})^{r+1} = p_{t}ι′_{t}(p′_{t}ι_{t}p_{t}ι′_{t})^{r}p′_{t}ι_{t} is also an automorphism, we get (p′_{t}ι_{t}p_{t}ι′_{t})^{r} ≠ 0, hence p′_{t}ι_{t}p_{t}ι′_{t} is also an automorphism due to Lemma 5.2. Thus p′_{t}ι_{t} : A_{t} ⟶ B_{t} is an isomorphism.

It remains to show that A = A_{1} × … × A_{t} × B_{t+1} × … × B_{n}. We have A = B_{t} × (A_{1} × … × A_{t−1} × B_{t+1} × … × B_{n}). Let X = A_{1} × … × A_{t−1} × B_{t+1} × … × B_{n} as subobject of A. Then A_{t} is a complement for X in A = B_{t} × X since p′_{t}ι_{t} : A_{t} ⟶ A ⟶ B_{t} is an isomorphism. By Proposition 2.6 we get A_{t} ∩ X = 0 and A_{t} ∪ X = A. Furthermore A_{t} ⟶ A and X ⟶ A are kernels. Hence A is an internal direct product A = A_{t} × X of A_{t} and X.

References

1. Paul M. Cohn: Universal Algebra. 2nd ed., Reidel, Dordrecht, 1981.

2. Bodo Pareigis: Categories and Functors. Academic Press, 1970.

3. Günther Richter: Krull-Schmidt for arbitrary categories. Contributions to General Algebra, 2 (Klagenfurt, 1982), 319-342, Hölder-Pichler-Tempsky, Vienna, 1983.