Notes on regular, exact and additive categories

Marino Gran Universit´ecatholique de Louvain

Summer School on Theory and Algebraic Topology Ecole Polytechnique F´ed´eralede Lausanne 11-13 September 2014 2 Chapter 1

Regular and exact categories

The notions of and of exact categories (in the sense of Barr [1]) play an important role in modern categorical algebra. They are useful to axiomatise some exactness properties the categories of sets, Grp of groups, Ab of abelian groups, R-Mod of modules on a commutative R, Rng of rings, all have in common. Some other categories, as for instance the category Top of topological spaces, do not share these same properties. In order to understand the notions of regular and exact categories it is useful to first examine various types of : this will be the subject of the first section.

1.1 Epimorphisms

Let f: A → B be an arrow in a category C. One says that f is an if, for any pair of parallel arrows u, v: B → C such that u ◦ f = v ◦ f, one has u = v. In the category Set the epimorphisms are precisely the surjective maps. In the category Ab of abelian groups the epimorphisms are the surjective homomor- phisms. This is also the case for in the category Grp (the proof is less simple than in Ab, see Exercise 1.1.8). In the category Top of topological spaces, the epimorphisms are the surjective continuous functions. There are some algebraic categories where epimorphisms fail to be surjective: for instance, the canonical inclusion Z → Q is an epimorphism in the category Rng of rings, and it is not surjective.

The notion of is defined dually: such is an f: A → B with the property that u = v whenever f ◦u = f ◦v for some parallel arrows u, v: C → A. It is easy to check that the in Set are the injective maps, in Grp and in Ab the injective , in Top the continuous injective maps.

3 4 CHAPTER 1. REGULAR AND EXACT CATEGORIES

Any is both a monomorphism and an epimorphism. The converse does not hold, in general, as the canonical inclusion Z → Q in Rng clearly shows. A category C where the equivalence “mono + epi = iso” holds is usually called a balanced category. For instance, Set, Grp and Ab are balanced, Rng and Top are not balanced. 1.1.1. Definition. f: A → B is a strong epimorphism if, given any com- mutative square f A / B

g h   C m / D where m: C → D is a mono, then there exists a unique arrow t: B → C such that m ◦ t = h and t ◦ f = g.

1.1.2. Lemma. If C has binary products, every strong epimorphism is an epimorphism. Proof. Let f: A → B be a strong epi, u, v: B → C be such that u ◦ f = v ◦ f. Consider the diagonal (1C , 1C ) = ∆: C → C × C, and the commutative square

f A / B

u◦f=v◦f (u,v)   C / C × C. ∆

Since ∆ is a monomorphism, there exists a unique t: B → C such that ∆ ◦ t = (u, v) and t ◦ f = u ◦ f = v ◦ f. It follows that

u = p1 ◦ (u, v) = p1 ◦ ∆ ◦ t = t = p2 ◦ ∆ ◦ t = p2 ◦ (u, v) = v.

 1.1.3. Lemma. An arrow f: A → B is an isomorphism if and only if f: A → B is a monomorphism and a strong epimorphism.

Proof. If f is both a strong epimorphism and a monomorphism, one considers the commutative square f A / B

1A 1B   A / B. f 1.1. EPIMORPHISMS 5

The existence of the t: B → A such that f ◦ t = 1B and t ◦ f = 1A shows that f is an isomorphism. Conversely, assume that f is an iso, and consider a commutative square

f A / B

g h   C m / D. where m is a monomorphism. The arrow t = g ◦ f −1: B → C is the desired factorisation.

1.1.4. Exercise. Show the following properties:

1. if f: X → Y and g: Y → Z are two strong epimorphisms, then g◦f: X → Z is a strong epi.

2. If f: X → Y and g: Y → Z are such that g ◦ f: X → Z is a strong epimorphism, then g: Y → Z is a strong epimorphism.

3. If f = i ◦ g is a strong epimorphism with i a monomorphism, then i is an isomorphism.

1.1.5. Definition. One says that f: A → B is a regular epimorphism if f: A → B is the coequaliser of two arrows. In other words, f is a regular epimorphism if one can find u, v: C → A such that f = Coeg(u, v).

1.1.6. Definition. A split epimorphism is an arrow f: A → B such that there is an i: B → A with f ◦ i = 1B. Observe that in the category Set any epimorphism is split (= axiom of choice). This is not the case in the categories Grp and Ab, for instance. We are now going to prove the following chain of implications:

1.1.7. Proposition. Let C be a category with binary products. One then has the implications split epi ⇒ regular epi ⇒ strong epi ⇒ epi

Proof. If f: A → B is split by an arrow i: B → A, then f = Coeg(1A, i ◦ f). Indeed,

f ◦ (i ◦ f) = f = f ◦ 1A, and, furthermore, if g: A → X is such that g ◦ (i ◦ f) = g ◦ 1A, then φ = g ◦ i is the only arrow with the property that φ ◦ f = g. 6 CHAPTER 1. REGULAR AND EXACT CATEGORIES

Assume that f: A → B is a regular epimorphism. It is then the coequaliser of two arrows u: T → A and v: T → A: consider then the

f A / B

g h   C m / D with m a mono. The equalities

m ◦ g ◦ u = h ◦ f ◦ u = h ◦ f ◦ v = m ◦ g ◦ v imply that g ◦ u = g ◦ v. The universal property of the coequaliser f implies that there is a unique t: B → C such that t ◦ f = g. This arrow t is also such that m ◦ t = h, so that f is a strong epimorphism.

The fact that any strong epimorphism is an epimorphism has been shown in Lemma 1.1.2. 1.1.8. Exercise. The aim of this exercise (from [7]) is to prove that the epimorphisms in Grp are precisely the surjective homomorphisms. Given an epimorphism f: G → H of groups, denote its f(G) by M. Assume that M 6= H. We distinguish between the following two cases:

(a) M is a normal of H;

(b) M is not normal in H.

In case (a), the proof is the same as for abelian groups: show that the quotient H/M must be trivial. In case (b) we can assume the index of M in H to be strictly greater than 2: otherwise, the subgroup would be normal. We consider the S(H) of of the set H, and choose three distinct cosets M, Mu and Mv, with u and v elements of H. Define a σ ∈ S(H) as follows: for z = xu ∈ Mu, we put σ(z) = σ(xu) = xv (for all x ∈ M); for z = xv ∈ Mv, we put σ(z) = σ(xv) = xu (for all x ∈ M), and we let σ act as the identity on all other elements of H (i.e. σ(z) = z if z∈ / Mu ∪ Mv). Let ψ: H → S(H) be the which associates with an element h of H the permutation ψh, defined by ψh(x) = hx, ∀x ∈ H. Let ψ: H → S(H) be the homomorphism which associates, with an element h of H, the permutation −1 ψh = σ ◦ ψh ◦ σ. Show that ψf = ψf, while ψ 6= ψ, a contradiction. 1.2. REGULAR CATEGORIES 7

1.2 Regular categories

Let us begin by considering some examples of quotients in the categories of sets, of groups and of topological spaces. Let f: A → B be a map in Set, and

Eq(f) = {(x, y) ∈ A × A | f(x) = f(y)} its pair, i.e. the on A that identifies the elements of A having the same image by f. This equivalence relation can be obtained by building the pullback of f along f:

p2 Eq(f) / A

p1 f   A / B f

In general, one calls the relation

p1 Eq(f) // A p2 the kernel pair of f.

1.2.1. Exercise. Show that any regular epimorphism f in a category with kernel pairs is the coequaliser of its kernel pair (Eq(f), p1, p2). In the category Set one can see that the canonical quotient π: A → A/Eq(f) defined by π(a) = a actually is the coequaliser of p1 and p2. This yields a unique arrow i: A/Eq(f) → B such that i ◦ π = f:

p1 f Eq(f) // A / B p2 ;

π i # A/Eq(f)

The map i is actually defined by i(a) = f(a) for any a ∈ A/Eq(f). Remark that this gives a factorisation i ◦ π of the arrow f, where π is a regular epimorphism and i is a monomorphism in the category Set. The same construction is possible in the category Grp of groups. Indeed, given a homomorphism f: G → G0, one can consider the kernel pair Eq(f) which is again obtained by the pullback above, but this time computed in the category Grp. The equivalence relation Eq(f) is also a group, as a subgroup of the product G × G of the group G with itself. The canonical quotient π: G → G/Eq(f) is a , and this allows one to build the following commutative 8 CHAPTER 1. REGULAR AND EXACT CATEGORIES diagram in Grp:

p1 f 0 Eq(f) / G / G p2 ;

π i # G/Eq(f), where π is a regular epimorphism and i a monomorphism. Let us then consider the category Grp(Top) of topological groups. It is possible to obtain the same kind of factorisation regular epimorphism-monomorphism also in this category, for any continuous homomorphism of topological groups. 0 Given a continuous homomorphism f:(G, ·, τG) → (G , ·, τG0 ) in Grp(Top), the kernel pair (Eq(f), ·, τi) is a topological group for the topology τi induced by the product topology τG×G of the topological group (G×G, ·, τG×G). The quotients in Grp(Top) are actually computed as in Grp, by using the quotient topology. In this way one gets the following commutative diagram

p1 f 0 (Eq(f), ·, τi) // (G, ·, τG) / (G , ·, τG0 ) p2 7

π i ' (G/Eq(f), ·, τq) where π is the canonical quotient, and τq the quotient topology. It turns out that π is the coequaliser of p1 and p2 in Grp(Top), and the induced morphism 0 i:(G/Eq(f), ·, τq) → (G , ·, τG0 ) is a monomorphism (since it is injective).

There are many other categories where this same construction is possible, for instance in the category Rng of rings, Ab of abelian groups, R-Mod of modules on a R. All these are examples of regular categories in the following sense: 1.2.2. Definition. A finitely C is regular if • coequalisers of kernel pairs exist in C ; • regular epimorphisms are pullback stable.

1.2.3. Examples. The categories Set of sets is regular. We have observed that the coequalisers of kernel pairs exist in Set, and it remains to check the pullback stability of regular epimorphisms. Consider a pullback

π2 E ×B A / A

π1 f   E p / B 1.2. REGULAR CATEGORIES 9 in Set where p is a surjective map (=regular epimorphism), and let us show that π2 is also a surjective map. Let a be an element in A; there exists then an e ∈ E such that p(e) = f(a). This shows that there is an (e, a) ∈ E ×B A such that π2(e, a) = a. The same argument still works in the category Grp, by taking into account the fact that regular epimorphisms therein are precisely the surjective homomorphisms, and that pullbacks are computed in Grp as in Set. For essentially the same reason the categories Rng, Ab and R-Mod are also regular categories. The category Grp(Top) of topological groups is regular. The point here is that the canonical quotient π: H → H/Eq(f) of a topological group (H, θ) by the equivalence relation (Eq(f), τi) of an arrow f in the category Grp(Top) is an open surjective homomorphism. Let us check this latter statement: if we write K = ker (π) for the kernel of π, then for any open V ∈ θ let us show that

π−1(π(V )) = VK.

On the one hand if z = vk, where v ∈ V and k ∈ K, one has

π(z) = π(vk) = π(v)π(k) = π(v) ∈ π(V ), so that z ∈ π−1(π(V )). −1 −1 If z ∈ π (π(V )), then π(z) = π(v1), for a v1 ∈ V , so that v1 z ∈ K, and z = v1k, for a k ∈ K. This implies that [ π−1(π(V )) = V k ∈ θ. k∈K

Indeed, the mk: G → G defined by mk(x) = xk for any x ∈ G (with fixed k ∈ K) is a , hence V k = mk(V ) ∈ θ, since V ∈ θ. Conclusion: ∀V ∈ θ, π(V ) is open, and this implies that π is an open map. It follows that in Grp(Top) the regular epimorphisms are the open surjective homomorphisms. To conclude that Grp(Top) is a regular category it suffices to verify that the open surjective homomorphisms are pullback stable. This is not difficult, and we leave the verification to the reader. Counter-example The category Top of topological spaces is not regular. To see that regular epi- are not pullback-stable, consider the following example, given in [2], of the pullback p2 X ×Y Z / Z

p1 g   X / Y f where X = ({a, b, c, d}, τX }, Y = ({x, y, z}, τY ) and Z = ({l, m, n}, τZ ), and the topologies are: τX = {∅, {a, b, c, d}, {a, b}}, τY = {∅, {x, y, z}} and τZ = 10 CHAPTER 1. REGULAR AND EXACT CATEGORIES

{∅, {l, m, n}, {l, m}}. The continuous maps f and g are defined by:

f(a) = x, f(b) = y = f(c), f(d) = z,

g(l) = x, g(m) = z = g(n). Remark that f: X → Y is a quotient map. The pullback

X ×Y Z = ({(a, l), (d, m), (d, n)}, τ} is equipped with the topology

τ = {∅,X ×Y Z, {(a, l)}, {(a, l), (d, m)}}.

−1 Now, the p2 is not a quotient map, since p2 (l) = {(a, l)} is open, while {l} is not.

We are now going to show that any arrow in a regular category has a canonical factorisation as a regular epimorphism followed by a monomorphism, exactly as in the examples of the categories Set, Grp and Grp(Top) recalled here above. For this, the following simple lemma will be needed:

1.2.4. Lemma. For an arrow p: A → B having a kernel pair (Eq(p), p1, p2) the following conditions are equivalent:

1. p is a monomorphism;

2. the projections p1: Eq(p) → A and p2: Eq(p) → A are equal.

The following elementary properties of pullbacks will also be useful: 1.2.5. Lemma. Consider a commutative diagram

k f A / B / C

a (1)b (2) c

   A0 / B0 / C0. k0 f 0

Then:

1. if the squares (1) and (2) are pullbacks, then also the exterior rectangle (1) + (2) is a pullback;

2. if the exterior rectangle (1) + (2) and the square (2) are pullbacks, then the square (1) is a pullback. 1.2. REGULAR CATEGORIES 11

1.2.6. Theorem. Let C be a regular category. Then

1. any arrow in C has a factorisation as a regular epimorphism followed by a monomorphism;

2. this factorisation is unique (up to isomorphism).

Proof.

1. Let f : A → B be an arrow in C. Consider the diagram here below where (Eq(f), f1, f2) is the kernel pair of f, and q is the coequaliser of (f1, f2). There exists then a unique m such that m ◦ q = f.

f1 / q Eq(f) . / A / I f2 m f  B

We are now going to show that m is a monomorphism: for this, it suffices to show that the projections p1: Eq(m) → I and p2: Eq(m) → I of the kernel pair of m are equal (Lemma 1.2.4). Consider the diagram

b π2 Eq(f) / Eq(m) ×I A / A

a π1 q

   A ×I Eq(m) / Eq(m) / I φ2 p2

φ1 p1 m

   A q / I m / B

where all the squares are pullbacks. By Lemma 1.2.5 we know that the whole square is a pullback, so that f1 = φ1 ◦ a and f2 = π2 ◦ b. The arrow φ2 ◦ a = π1 ◦ b is then an epimorphism, as a composite of epimorphisms. The fact that φ1 ◦ a = f1 and π2 ◦ b = f2 implies that

p1 ◦(φ2 ◦a) = q◦φ1 ◦a = q◦f1 = q◦f2 = q◦π2 ◦b = p2 ◦π1 ◦b = p2 ◦(φ2 ◦a);

since φ2 ◦ a is an epimorphism, it follows that p1 = p2. 12 CHAPTER 1. REGULAR AND EXACT CATEGORIES

2. Since any regular epi is a strong epi, the uniqueness of the factorisation can be easily checked by using the definition of strong epi.



1.2.7. Proposition. Let C be a regular category, then the following properties are satisfied:

1. regular epis coincide with strong epis;

2. if g ◦ f is a regular epi, then g is a regular epi;

3. if g and f are regular epis, then g ◦ f is a regular epi;

4. f is an isomorphism if and only if f is a mono and a regular epi;

5. if f: X → Y and g: X0 → Y 0 are regular epis, then f ×g: X ×X0 → Y ×Y 0 is also a regular epi.

Proof. For 1., we only need to prove that any strong epimorphism f: A → B is regular. If f = m ◦ q with m a monomorphism and q a regular epimorphism

f A / B

q d

  I m / B there is a unique arrow d: B → I such that d ◦ f = q and m ◦ d = 1B. The arrow d is the inverse of m, and f is then a regular epimorphism. 2., 3. and 4. follow from 1. and from the properties of strong epimorphisms. 5. If f: X → Y and g: X0 → Y 0 are two regular epimorphisms, consider the commutative diagrams

f×1 0 X × X0 X / Y × X0

π1 (a) π1

  X / / Y f 1.2. REGULAR CATEGORIES 13

1Y ×g Y × X0 / Y × Y 0

π2 (b) π2

 0  0 X g / / Y .

These squares are easily seen to be pullbacks, so that the arrows f × 1X0 and 1Y × g are regular epimorphisms (by regularity of C), and then

f × g = (1Y × g) ◦ (f × 1X0 ) is a regular epimorphism by 3.  We are now going to give an equivalent formulation of the notion of regular category:

1.2.8. Theorem. Let C be a finitely complete category. Then C is a regular category if and only if

1. any arrow in C factorises as a regular epi followed by a mono;

2. these factorisations are pullback-stable.

Proof. If C is regular, the validity of 1. and 2. follows from Theorem 1.2.6. For the converse, assume that the conditions 1. and 2. hold in C: then, in particular, regular epis are pullback stable. It remains to show that the co-

f1 / equalisers of kernel pairs exist in C. Let Eq(f) / X be the kernel pair of f2 an arrow f : X → Y of C. Consider m ◦ q the regular epi-mono factorisation f1 / of f. The fact that m is a mono implies that Eq(f) / X is also the kernel f2 pair of the regular epi q. The arrow q is then the coequaliser of its kernel pair

f1 / Eq(f) / X (see Exercise 1.2.1). f2

By using Lemma 1.2.5 one can prove the following

1.2.9. Lemma. Let C be a category with pullbacks. One considers the com- 14 CHAPTER 1. REGULAR AND EXACT CATEGORIES mutative diagram here below where the horizontal rows are kernel pairs:

p1 / f Eq(f) / A / X p2

v (1)u (2) w

 p1   Eq(g) / B Y / g / p2

If (2) is a pullback, then each commutative square in (1) is a pullback. Proof. Consider the commutative diagram

p2 Eq(f) / A f

p1 " v A / X f u

 p2  Eq(g) / B w u p1 g !   B g / Y, where the upper face of the cube and the right-hand face are pullbacks by assumption, so that also the rectangle

v p2 Eq(f) / Eq(g) / B

p1 p1 g    A u / B g / Y is a pullback (the diagram is commutative). Since the right-hand square in this latter diagram is a pullback, Lemma 1.2.5.2 tells us that the left square is a pullback as well. A similar proof shows that also the square

v Eq(f) / Eq(g)

p2 p2   A u / B is a pullback.  1.2. REGULAR CATEGORIES 15

1.2.10. Lemma. Consider a commutative diagram

k l A / B / C

a b c    A0 / B0 / C0 k0 l0 in a regular category C, where the left-hand square and the external rectangle are pullbacks. If the arrow k0 is a regular epimorphism, then the right-hand square is a pullback.

Proof. Consider the commutative diagram

k l A / B / C α β b c 0$ 0$ a A ×C0 C / B ×C0 C / C π2 0 π2

π1 π0 c  z  z 1   A0 / B0 / C0 k0 l0

0 0 0 0 0 where (B ×C0 C, π1, π2) is the pullback of l and c, and (A ×C0 C, π1, π2) is 0 0 the pullback of k and π1. The fact that the external rectangle is a pullback implies that the induced arrow α is an isomorphism. The arrow π2 is a regular 0 epimorphism (because k is one), so that π2 ◦α = β ◦k is a regular epimorphism, and then β is a regular epimorphism (see Proposition 1.2.7). The arrow β is a monomorphism: this follows from the fact that the square

k A / B

α β

0  0  A ×C0 C / B ×C0 C π2 is a pullback, so that both the induced commutative squares

Eq(α) / Eq(β)

p1 p2 p1 p2     A / B k are pullbacks, with the dotted arrow a (regular) epimorphism. This implies that the projections p1: Eq(β) → B and p2: Eq(β) → B are equal, so that β is indeed a monomorphism. 16 CHAPTER 1. REGULAR AND EXACT CATEGORIES

We are now ready to prove the following interesting result, often referred to as the Barr-Kock Theorem (although it was first observed by A. Grothendieck):

1.2.11. Theorem. Let C be a regular category, and

p1 / f Eq(f) / A / / X p2

v (1)u (2) w

 p1   Eq(g) / B Y / g / p2 a commutative diagram with f a regular epimorphism. If either of the left-hand commutative squares are pullbacks, then the right-hand square (2) is a pullback.

Proof. Consider the following commutative diagram

p2 Eq(f) / A f

p1 " v A / X f u

 p2  Eq(g) / B w u p1 g "   B g / Y

The assumptions guarantee that, for instance, the left-hand face and the bottom face of the cube are pullbacks. By commutativity it follows that the rectangle

p2 u Eq(f) / A / B

p1 f g    A / X / Y f w is also a pullback, as well as its left-hand square. Since f is a regular epi, by Lemma 1.2.10 it follows that the right-hand square is a pullback.  1.2. REGULAR CATEGORIES 17

1.2.12 Relations in regular categories 1.2.13. Definition. Let C be a finitely complete category. A relation from X to Y in C is a graph R

d1 d2 ~ XY such that the factorisation (d1, d2): R → X × Y is a monomorphism; equiva- lently, the pair (d1, d2) is jointly monomorphic. One usually identifies two relations R → X × Y and S → X × Y when they de- termine the same of X × Y (= equivalence classes of monomorphisms with codomain X × Y ). If X = Y , one says that R is a relation “on X”. 1. A relation R is reflexive when there is an arrow δ : X → R such that d1 ◦ δ = 1X = d2 ◦ δ.

2. R is symmetric if there is an arrow σ : R → R such that d1 ◦ σ = d2 and d2 ◦ σ = d1. 3. Consider the pullback p2 R ×X R / R

p1 d1   R / X d2

The relation R is transitive if there is an arrow τ : R ×X R → R such that d1 ◦ τ = d1 ◦ p1 et d2 ◦ τ = d2 ◦ p2. A relation R on X is an equivalence relation if R is reflexive, symmetric and transitive. Of course, this abstract notion of equivalence relation gives in particular the usual one in the category C = Set. When C = Grp is the , an equivalence relation R ⊂ X × X in the category Grp is an equivalence relation on the set X which is also a subgroup of the group X × X. In general, one has the following:

p1 / 1.2.14. Lemma. In a category with pullbacks the kernel pair Eq(f) / X p2 of an arrow f : X → Y is an equivalence relation on X in C.

Proof. The arrows p1: Eq(f) → X and p2: Eq(f) → X are jointly monomorphic since they are projections of a pullback. The commutativity of the square

1 X X / X

1X f   X / Y f 18 CHAPTER 1. REGULAR AND EXACT CATEGORIES

and the universal property of the kernel pair (Eq(f), p1, p2) implies that there is a unique δ: X → Eq(f) such that p1 ◦ δ = 1X = p2 ◦ δ X 1X δ

" p2 % Eq(f) / X 1X

p1 f    X / Y, f and Eq(f) is then reflexive. Similarly, the commutativity of the external part of the diagram Eq(f) p1 σ

# p2 Eq(f) %/ X p2

p1 f   X / Y f implies that there is a unique arrow σ: Eq(f) → Eq(f) such that p1 ◦ σ = p2 and p2 ◦ σ = p1, hence Eq(f) is symmetric. For the transitivity of Eq(f) one considers the following commutative diagram

π2 Eq(f) ×X Eq(f) / Eq(f)

p2 τ ' ! π1 Eq(f) / X p2

p1

 p2  Eq(f) / X f

p1 p1 f  "  ' X / Y f where the back face is a pullback. The commutativity of the diagram and the universal property of the kernel pair (Eq(f), p1, p2) shows that there is a unique τ such that p1 ◦ τ = p1 ◦ π1 and p2 ◦ τ = p2 ◦ π2, and this completes the proof.



An important aspect of regular categories is that in these categories one can define a composition of relations, which has some nice properties. 1.2. REGULAR CATEGORIES 19

In the category Set, if R → X × Y is a relation from X to Y , and S → Y × Z from Y to Z, one usually defines the relation S ◦ R → X × Z by setting S ◦ R = {(x, z) ∈ X × Z such that ∃ y ∈ Y with xRySz}.

This construction is actually possible in any regular category C, thanks to the existence of regular images (see Theorem 1.2.6). One first builds the pullback

R ×Y S

π1 π2 R { # S

p1 p2 p1 p2 ~ XYZ# { and one then factorises the arrow (p1 ◦π1, p2 ◦π2): R ×Y S → X ×Z as a regular epimorphism q: R ×Y S → I followed by a monomorphism i: I → X × Z:

q i R ×Y S / I / X × Z

In Set, the set I is given by (x, z) ∈ X ×Z such that there is a (u, y, v) ∈ R×Y S with u = x and v = z: this is precisely S ◦ R! This composition is actually associative: 1.2.15. Theorem. Let C be a regular category. If R → A × B, S → B × C and T → C × D are relations in C, one has the equality T ◦ (S ◦ R) = (T ◦ S) ◦ R.

Proof. Consider the diagram obtained by building the following pullbacks: X

x1 x2 { # R ×B S S ×C T

p1 p2 q2 q2

R { # S { # T

r1 r2 s1 s2 t1 t2  ABCD.# { # { The proof consists in showing that the relations T ◦ (S ◦ R) and (T ◦ S) ◦ R are both given by the regular image i: I → A × D in the factorisation

(r1◦p1◦x1,t2◦q2◦x2) X / A × D 7

q i & I 20 CHAPTER 1. REGULAR AND EXACT CATEGORIES as a regular epimorphism folllowed by a monomorphism of the arrow

(r1 ◦ p1 ◦ x1, t2 ◦ q2 ◦ x2). We leave it to the reader the verification of this fact, which uses the pullback stability of regular epimorphisms in a crucial way.  This result allows one to define a new category starting from any regular cat- egory C, the category Rel(C) of relations de C. The objects are the same as the ones in C, an arrow from X to Y is simply a relation from X to Y , and composition is the relational one defined above. For any relation R from X to Y the discrete relation 1X ∆X : X // X 1X is such that R◦∆X = R, and for any relation S from Z to X one has ∆X ◦S = S. It follows that the arrow ∆X in Rel(C) is the identity on the object X. There is actually a faithful Γ: C → Rel(C), associating the graph Γ(f) with any arrow f: X → Y of C, where Γ(f): X → Y in Rel(C) is the graph of f, seen as a relation: X

1X f ~ XY We are not going to develop the theory of relations in a regular category in these notes. The book [6] will be a useful reference for the interested reader. 1.2.16. Exercise. Show that any reflexive relation R in the category Grp of groups is an equivalence relations. Is this result true in the categories Set, or Ab? A finitely complete category C is called a Mal’tsev category if any internal reflexive relation in C is an equivalence relation [4]. As explained in this latter paper, for a regular category being a Mal’tsev category is equivalent to the following property: given any equivalence relations R and S on the same object X, then R ◦S = S ◦R (2-permutability of the composition of equivalence relations).

1.3 Exact categories 1.3.1. Definition. A category C is exact if: 1. C is regular;

2. any equivalence relation (R, d1, d2) in C is effective: this means that (R, d1, d2) is the kernel pair of an arrow q in C.

1.3.2. Examples.The category Set is exact: each equivalence relation R on a set A is the kernel pair of the canonical quotient πR: A → A/R. 1.3. EXACT CATEGORIES 21

Also Grp is exact. Indeed, an equivalence relation R (in the categorical sense) on a group G in Grp is an equivalence relation in Set which is compatible with the group structure: one always has that 1R1, the fact that xRy implies that x−1Ry−1 and, moreover, if xRy and uRv, then necessarily xuRyv. It follows that the canonical quotient π: G → G/R is a group homomorphism, and Eq(π) = R. One can check that, similarly, the categories Ab, R-Mod and Rng are all exact categories (any algebraic variety in the sense of is an exact category, see [2], for instance). One can also show that the category HComp of compact Hausdorff spaces is an exact category. The verification of the following simple fact is left to the reader:

1.3.3. Proposition. Let C be an exact category, and d1, d2: R // X an equivalence relation on X in C. Then the coequaliser of (d1, d2) exists, and d1, d2: R // X is the kernel pair of its coequaliser. Counter-examples • Top is not exact ( it is not even regular!). The effective equivalence rela- tions R ⊂ X × X are the equivalence relations on X having the topology induced by the product space X × X. • Grp(Top) is a regular category which is not exact.

• Let Abt.f. be the category of torsion-free abelian groups. This is the full of Ab whose objects are the abelian groups satisfying the ∗ implication nx = 0 ⇒ x = 0, for any n ∈ N . Limits in Abt.f. are computed as in the category Ab of abelian groups. On the other hand, any homomorphism f: A → B in Abt.f. has a factorisation m ◦ q: A → I → B in Ab, where q is a surjective homomorphism and m a monomorphism. Since B is torsion-free, I is torsion-free as well. The morphism q: A → I is the coequaliser of the projections π1: Eq(f) → A and π2: Eq(f) → A in Abt.f., and it is then a regular epimorphism in Abt.f.. Consequently, the category Abt.f. is regular. However, Abt.f. is not exact. To see this, let us consider the relation R2 on Z defined by mR2n ⇔ ∃k ∈ Z such that m − n = 2k.

This is an equivalence relation in Abt.f. which is not effective. It is easy to see that the coequaliser of the projections π1: R2 → Z and π2: R2 → Z is zero. This implies that R2 can not be effective, since if it were so, it would be Z × Z. 22 CHAPTER 1. REGULAR AND EXACT CATEGORIES Chapter 2

Additive categories

In the category Ab of abelian groups it is possible to define an “” on the set of parallel arrows from one object to another. Indeed, if f: A → B and g: A → B are two arrows in Ab, one defines

(f + g)(x) = f(x) + g(x) ∀x ∈ A, and this function f +g: A → B is again an arrow in Ab. This operation provides Ab(A, B) with an structure, whose neutral element is the 0: A → 0 → B (this latter being the unique morphism that factors through the trivial group 0). Observe that in the category Grp it is not possible to “add” two parallel morphisms, simply because the function f + g is not a group homomorphism, in general. The example of the category Ab of abelian groups (or any category R-Mod of R-modules) will be the guiding one in this chapter, where the notion of is briefly introduced.

2.1 Preadditive categories 2.1.1. Definition. A category C is preadditive if:

1. C(X,Y ) is an abelian group ∀X,Y of C ; 2. ∀X,Y,Z in C the composition mappings ◦ : C(X,Y )×C(Y,Z) → C(X,Z) are group homomorphisms in each variable : for all f, f 0 ∈ C(X,Y ), g, g0 ∈ C(Y,Z) one has g ◦ (f + f 0) = g ◦ f + g ◦ f 0, (g + g0) ◦ f = g ◦ f + g0 ◦ f.

The notion of is self-: C is preadditive if and only if Cop is preadditive.

23 24 CHAPTER 2. ADDITIVE CATEGORIES

2.1.2. Examples.

• Ab is preadditive;

• R-Mod, the category of (left) modules on a commutative unitary ring R is preadditive;

• Abs.t. the category of torsion-free abelian groups is preadditive;

• the category Ab(Top) of topological abelian groups is preadditive;

• any unitary ring R can be seen as a preadditive category C having a unique object, written ∗. The elements r of R are the arrows r: ∗ → ∗, so that C(∗, ∗) is a an abelian group (R, +, 0). On the other hand the composite r ◦ s: ∗ → ∗ of two arrows is the product r · s of r and s as elements of the ring R.

The categories Set, Grp and Top are not preadditive. The proof of the following result is left to the reader:

2.1.3. Lemma. Let C be a preadditive category, and X an object in C. The following conditions are equivalent:

1. X is initial;

2. X is terminal;

3. X is a .

When this is the case, the morphism f: A → B which factors through the zero object X = 0 is the neutral element for the group structure of C(A, B). A category C having a zero object is called a pointed category. For instance, the categories Grp, Ab, R-Mod, Mon (of ), Grp(Top), Ab(Top) and Set∗ (of pointed sets) are pointed, whereas the categories Set and Top are not pointed. The category CRng of commutative unitary rings is not pointed.

If X and Y are two objects in a pointed category, one writes 0X,Y , or simply 0, for the unique arrow X → 0 → Y , called the zero morphism. 2.1.4. Definition. A diagram

p1 p2 X o Z / Y / o i1 i2 is a of X and Y if p1 ◦i1 = 1X , p2 ◦i2 = 1Y and i1 ◦p1 +i2 ◦p2 = 1Z .

2.1.5. Remark. The notion of biproduct is self-dual. 2.1. PREADDITIVE CATEGORIES 25

2.1.6. Example. In the category Ab, one has the biproduct

p1 p2 A o A L B / B / o i1 i2 L where A B = {(a, b), | a ∈ A, b ∈ B}, p1(a, b) = a, p2(a, b) = b, i1(a) = (a, 0), i2(b) = (0A, b). It is clear that pk, ik (for k = 1, 2) are group homomorphisms, and p1 ◦ i1 = 1A, p2 ◦ i2 = 1B, i1 ◦ p1 + i2 ◦ p2 = 1A L B.

2.1.7. Theorem. Let C be a preadditive category. The following conditions are equivalent:

1. the product of X and Y exists in C;

2. the biproduct of X and Y exists in C. When this is the case, given a biproduct

p1 p2 X o Z / Y / o i1 i2

i i the diagram X 1 / ZYo 2 is a .

p1 p2 Proof. 1. ⇒ 2. Let XXo × Y / Y be the product of X and Y . Con- sider the diagram

X

1X 0X,Y i1

p1  p2 XXwo × Y '/ Y g O 7

i2 0Y,X 1Y Y

The factorisations i1 and i2 are such that p1 ◦ i1 = 1X , p2 ◦ i1 = 0X,Y and p2 ◦ i2 = 1Y , p1 ◦ i2 = 0Y,X . One has

p1 ◦ (i1 ◦ p1 + i2 ◦ p2) = p1 ◦ i1 ◦ p1 + p1 ◦ i2 ◦ p2 = p1 + 0 = p1 ◦ 1X×Y and

p2 ◦ (i1 ◦ p1 + i2 ◦ p2) = p2 ◦ i1 ◦ p1 + p2 ◦ i2 ◦ p2 = 0 + p2 = p2 ◦ 1X×Y .

Since (p1, p2) is jointly monomorphic, it follows that i1 ◦ p1 + i2 ◦ p2 = 1X×Y , hence X and Y have a biproduct. 26 CHAPTER 2. ADDITIVE CATEGORIES

Conversely, assume that Z is the biproduct of X and Y . First note that p2 ◦i1 = 0: indeed, p2 ◦ i1 = p2 ◦ (i1 ◦ p1 + i2 ◦ p2) ◦ i1 = p2 ◦ i1 ◦ p1 ◦ i1 + p2 ◦ i2 ◦ p2 ◦ i1 = p2 ◦ i1 + p2 ◦ i1, thus p2 ◦ i1 = 0X,Y . One similarly shows that p1 ◦ i2 = 0Y,X . Let us then prove that Z is the product of X and Y . Let U be an object of C, and let α : U −→ X, β : U −→ Y be any two morphisms of C as depicted in the following diagram:

p1 p2 X o Z / Y f / o 8 i1 O i2

α β U

Let λ = i1 ◦ α + i2 ◦ β : U −→ Z. One has:

p2 ◦ λ = p2 ◦ (i1 ◦ α + i2 ◦ β) = p2 ◦ i1 ◦ α + p2 ◦ i2 ◦ β = p2 ◦ i1 ◦ α + β so that p2 ◦ λ = 0 + β = β.

The proof of the equality p1 ◦λ = α is similar. Let us now show that λ is unique. 0 0 0 If λ : U → Z is such that p1 ◦ λ = α and p2 ◦ λ = β, then

λ = i1 ◦ α + i2 ◦ β 0 0 = i1 ◦ (p1 ◦ λ ) + i2 ◦ (p2 ◦ λ ) 0 = (i1 ◦ p1 + i2 ◦ p2) ◦ λ = λ0. Given a biproduct p1 p2 X o Z / Y, / o i1 i2

i i let us check that the diagram X 1 / ZYo 2 is a coproduct. Consider the diagram

X u i1

& ( 8 Z /6 U

i 2 v Y and let us show that there is a unique θ : Z → U such that θ ◦ i1 = u and θ ◦ i2 = v. One sets θ = u ◦ p1 + v ◦ p2, and then observes that

θ ◦ i1 = (u ◦ p1 + v ◦ p2) ◦ i1 = u ◦ p1 ◦ i1 + v ◦ p2 ◦ i1 = u + 0 = u. 2.2. ADDITIVE CATEGORIES 27

One proceeds similarly to check that θ ◦ i2 = v. The uniqueness of the factori- sation is easy to check, and we leave the verification to the reader.

2.2 Additive categories 2.2.1. Definition. A category C is additive if 1. C is preadditive; 2. C is pointed; 3. C has for any pair of objects X and Y of C. Notation : From now on we shall denote by X L Y the (object part of the) biproduct of X and Y .

2.2.2. Examples. The categories Ab, Abs.t., Ab(Top) and R-Mod are additive. We now recall the notions of kernel and of an arrow. 2.2.3. Definition. Let C be a finitely complete pointed category. If f: A → B is an arrow in C, the kernel of f, denoted by ker (f), is the of f and 0A,B : ker (f) f K[f] A B. / // 0A,B Equivalently, one can define the kernel of f as the arrow ker (f): K[f] → A in the pullback ker (f) K[f] / A

f   0 / B

2.2.4. Exercise. Check that the notion of kernel just defined extends the usual one in the categories Grp, Ab and Grp(Top). The dual notion is the following: 2.2.5. Definition. Let C be a finitely cocomplete pointed category. If f: A → B, one defines the cokernel of f, denoted by coker(f), as the coequaliser of f and 0A,B: f coker(f) A B Cok(f). // / 0A,B Equivalently, one can define the cokernel of f as the arrow coker(f): B → Cok(f) in the pushout f A / B

coker(f)   0 / Cok(f). 28 CHAPTER 2. ADDITIVE CATEGORIES

2.2.6. Definition. A normal epimorphism is an arrow in a pointed category C which is the cokernel of an arrow in C. For instance, in the category Grp any epimorphism is normal: if f: A → B is a surjective homomorphism, then f is the cokernel of the canonical inclusion K[f] → A. In general, one has the implication normal epimorphism ⇒ regular epimorphism. The notion of normal monomorphism is defined dually. Normal monomor- phisms in the category Grp correspond to normal , so that monomor- phisms are not normal in Grp, in general. On the contrary monomorphisms are always normal in the categories Ab and R-Mod, of course. 2.2.7. Proposition. Let C be an additive category. Consider the diagram

f k K[f] / / X / Y o s where f is a split epimorphism, and k its kernel. Then X =∼ K[f] L Y .

Proof. By assumption there is the biproduct

p1 p2 K[f] o K[f] L Y / Y / o i1 i2 L L and, by Theorem 2.1.7, (K[f] Y, p1, p2) is a product and (K[f] Y, i1, i2) a coproduct.  k   k  Let : K[f] L Y → X be the unique morphism such that ◦ i = k s s 1  k   k  and ◦ i = s. We would like to show that is an isomorphism. s 2 s On the one hand one has the equalities

f ◦ (1X − s ◦ f) = f − f ◦ s ◦ f = f − f = 0 and, since k = ker (f), there is then a factorisation g : X → K[f] such that k ◦ g = 1X − s ◦ f. Let us then write (g, f): X → K[f] L Y for the unique morphism such that p1 ◦ (g, f) = g and p2 ◦ (g, f) = f. To show that g ◦ k = 1K[f] we observe that

(k ◦ g) ◦ k = (1X − s ◦ f) ◦ k = k − s ◦ f ◦ k = k − 0 = k and that k is a monomorphism, hence g ◦ k = 1K[f]. We then observe that  k  p ◦ (g, f) ◦ ◦ i = g ◦ k = 1 = p ◦ i 1 s 1 K[f] 1 1 2.2. ADDITIVE CATEGORIES 29 and  k  p ◦ (g, f) ◦ ◦ i = f ◦ k = 0 = p ◦ i 2 s 1 K[f],Y 2 1 and the fact that (p1, p2) is jointly monomorphic yields  k  (g, f) ◦ ◦ i = i (∗) s 1 1

Let us prove that g ◦ s = 0:

(k ◦ g) ◦ s = (1X − s ◦ f) ◦ s = s − s ◦ f ◦ s = s − s = 0 and from the fact that k is a monomorphism one deduces that g ◦ s = 0. It follows that  k  p ◦ (g, f) ◦ ◦ i = g ◦ s = 0 = p ◦ i 1 s 2 1 2 and  k  p ◦ (g, f) ◦ ◦ i = f ◦ s = 1 = p ◦ i , 2 s 2 Y 2 2 therefore  k  (g, f) ◦ ◦ i = i (∗∗) s 2 2 since (p1, p2) is jointly monomorphic. The pair (i1, i2) is jointly epimorphic, and (∗) + (∗∗) give  k  (g, f) ◦ = 1 L . s K[f] Y

Finally,  k   k  ◦ (g, f) = ◦ 1 L ◦ (g, f) s s K[f] Y  k  = ◦ (i ◦ p + i ◦ p ) ◦ (g, f) s 1 1 2 2  k   k  = ◦ i ◦ p ◦ (g, f) + ◦ i ◦ p ◦ (g, f) s 1 1 s 2 2 = k ◦ g + s ◦ f = 1X − s ◦ f + s ◦ f = 1X showing that there is an isomorphism X =∼ K[f] L Y .

The additive structure on a pointed category with biproducts is essentially unique: 2.2.8. Proposition. Let C be a pointed category with biproducts. Then two additive structures on C are isomorphic. 30 CHAPTER 2. ADDITIVE CATEGORIES

Proof. The proof consists in 3 steps. L • Step 1: one shows that p1 − p2 : X X → X is determined by the “-colimit” structure of the biproduct X L X. L Define σX = p1 −p2: X X → X, and let us show that σX is the cokernel L of the diagonal ∆X = (1X , 1X ): X → X X. One has:

σX ◦ ∆X = (p1 − p2) ◦ ∆X = p1 ◦ ∆X − p2 ◦ ∆X = 1X − 1X = 0. L Universal property: let f : X X → Y be such that f ◦∆X = 0. Consider the biproduct p1 p2 X o X L X / X, / o i1 i2

and observe that ∆X can be expressed in terms of i1 and i2:

∆X = 1X L X ◦ ∆X = (i1 ◦ p1 + i2 ◦ p2) ◦ ∆X = i1 ◦ p1 ◦ ∆X + i2 ◦ p2 ◦ ∆X = i1 ◦ 1X + i2 ◦ 1X = i1 + i2.

It follows that f ◦ ∆X = 0 = f ◦ (i1 + i2) = f ◦ i1 + f ◦ i2 so that f ◦ i1 = −f ◦ i2. Define ϕ = f ◦ i1 : X → Y , and one has the equalities

ϕ ◦ σX = f ◦ i1 ◦ (p1 − p2) = f ◦ i1 ◦ p1 − f ◦ i1 ◦ p2 = f ◦ (i1 ◦ p1 + i2 ◦ p2) = f.

The arrow ϕ is the unique arrow from X to Y such that ϕ ◦ σX = f. Indeed, assume that h : X → Y is such that h ◦ σX = f. Then

h = h−0 = h◦1X −h◦0 = h◦p1◦i1−h◦p2◦i1 = h◦(p1−p2)◦i1 = h◦σX ◦i1 = f◦i1,

therefore σX = coker(∆X ).

f • Step 2: let us show that, for any graph Y / X , the difference f-g is g / determined by the “limit-colimit” structure of the biproduct. One has

(p1 − p2) ◦ (f, g) = p1 ◦ (f, g) − p2 ◦ (f, g) = f − g

and since both p1 − p2 and (f, g): Y → X × X are determined by the “limit-colimit” structure of the biproduct, then this is also the case for f-g. • Step 3: one observes that f + g = f − (0 − g), and this completes the proof. 2.2. ADDITIVE CATEGORIES 31

The following notion plays a fundamental role in :

2.2.9. Definition. A category C is abelian if

1. C is pointed; 2. C has binary products and binary ; 3. any arrow in C has a kernel and a cokernel; 4. in C any monomorphism is a kernel and any epimorphism a cokernel.

We conclude with notes stating a well known result due to M. Tierney, clarifying the relationship between exact categories and abelian categories (see the second chapter in [5], or [2], for an elementary introduction to the notion of ): 2.2.10. Theorem. A category is abelian if and only if it is exact and additive. The interested reader will find a detailed proof of this theorem, that uses several useful notions from modern categorical algebra, in the article [3]. 32 CHAPTER 2. ADDITIVE CATEGORIES Bibliography

[1] M. Barr, P. A. Grillet and D. H. van Osdol, Exact categories and categories of sheaves, Springer Lecture Notes in Mathematics 236, 1971.

[2] F. Borceux, Handbook of categorical algebra. 2. Categories and structures. Encycl. of Mathematics and its Applications 51, Cambridge University Press, 1994. [3] D. Bourn and M. Gran, Regular, Protomodular and Abelian Categories, in Categorical Foundations - Special Topics in Order, Topology, Algebra and Theory, Encycl. of Mathematics and its Applications 97, Cambridge University Press, 165-211, 2004. [4] A. Carboni, J. Lambek, and M. C. Pedicchio, Diagram chasing in Mal’cev categories, J. Pure Appl. Alg. 69, 271-284, 1991.

[5] P.J. Freyd, Abelian categories. An introduction the the theory of . Harper’s Series in Modern Mathematics, New York, 1964. [6] P.J. Freyd and A. Scedrov, Categories, Allegories, North Holland Mathem. Library, Vol. 39, 1990. [7] S. Mac Lane, Categories for the working mathematician, Graduate Texts in Mathematics, Second Edition, Springer-Verlag, 1998.

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