In this section we introduce abelian categories and develop some of the most elementary properties about them.
These are the image factorization theorem 2.2.10 and the fact that abelian categories is additive by theorem 2.3.3.
Definition 2.2.1. A categoryAisabelian if the following conditions hold.
AB 1 Ahas the zero object.
AB 2 Every pair of objects ofAhas the product and coproduct.
AB 3 Every morphism ofAhas the kernel and the cokernel.
AB 4 Every monomorphism is the kernel of some morphism ofAand every epimorphism is the cokernel of some morphism ofA.
Remark 2.2.2. From the definition of an abelian category it is clear that the opposite category of an abelian category is an abelian category. This allows us to prove results about abelian categories by duality. More precisely, if we give some proof about an abelian category Aonly using the axioms of the abelian category, then the same proof applies for Aop. Since limit (resp. colimit) in Acorresponds to colimit (resp. limit) in Aop, constructions using limits and colimits have also their dual version inA.
Proposition 2.1.3 shows that the condition AB2 can be replaced by the condition that A has all biproducts.
Proposition 1.2.6 implies that an abelian category has all finite products, and by duality all finite coproducts.
First, let us note that, in an abelian category a morphism which is both a monomorphism and an epimorphism, is an isomorphism. Indeed, let f :X ÑY be a morphism which is a monomorphism and an epimorphism. Then by AB4 f is the kernel of some morphism g :Y ÑZ, sogf “0. Since f is an epimorphism, we getg “0. Now gIdY “0, so there exists a unique morphism h1 :Y ÑX such that f h1 “IdY. By duality there exists a unique morphismh2:Y ÑX such that h2f “IdX. Putting these together yields
h1“ ph2gqh1“h2pgh1q “h2. This shows thatf is an isomorphism.
Example 2.2.3. The categoriesAb, and RModare abelian categories. Moreover, we will see that every abelian category has all finite limits and finite colimits 2.2.5, but the categoryRModhas more. It has all small limits and all small colimits, [Bor94b, Example 1.4.6a].
Let A be an abelian category and consider an object A of A. The monomorphisms in A with codomain the object A are called the subobjects under the following equivalence relation: let f : X Ñ A and g : Y Ñ A be monomorphisms. They are equivalent as subobjects if there exists an isomorphism h:X ÑY such thatf “gh.
Sinceg is a monomorphism, one sees that the morphismhis unique.
Lemma 2.2.4. An abelian categoryAhas the pullback of two subobjects of any object.
Proof. Leta, b be two monomorphisms with equal codomains, and fix morphismsf and g so that a“kerf and b “ kerg by AB4. Denote by k the morphism of kerpf, gq. Now, f k “ p1pf, gqk “ 0, so there exists a unique morphisma1 such thatk“aa1. Similarly, fromgk“p2pf, gqk“0, we get a unique morphismb1 withk“bb1. We
have the following commutative diagram uniqueness of the product we have thatpf, gqbv“ pf, gqau“0. Hence there exists a unique morphismwsuch that kw“bv“au. Sinceaandbare monomorphisms, by commutativity we have v“b1wandu“a1w. Uniqueness of wfollows from commutativity and the fact thatkis a monomorphism.
The following lemma shows that all constructions like equalizers, coequalizers, pullbacks and pushouts con-structed by finite limits and finite colimits exist in abelian categories.
Lemma 2.2.5. An abelian categoryAis finitely complete and finitely cocomplete.
Proof. By duality for abelian categories, it suffices to show thatAis finitely complete. By proposition 1.2.6Ahas finite products, and by proposition 1.2.7 it remains to show thatAhas equalizers for all pairs of parallel morphisms.
Let f, g:AÑB be morphisms. NowpIdA, fqand pIdA, gqare monomorphisms, so by lemma 2.2.4, they have a pullbackpP, u, vq. From
#u“p1pIdA, fqu“p1pIdA, gqv“v f u“p2pIdA, fqu“p2pIdA, gqv“gv,
one gets that f u “gu. Suppose f x “gx for some morphism x. Then pIdA, fqx“ pIdA, gqx, by the uniqueness property of the product. Hence, by the uniqueness property of the pullback P there exists a unique morphism y such thatuy“x“vy. This shows thatpP, u, vqis the equalizer off andg.
In a general category, a morphism with kernel 0 need not be a monomorphism, but in abelian categories vanishing of the kernel implies that the morphism is a monomorphism.
Lemma 2.2.6. Let A be an abelian category. A morphism f is a monomorphism (resp. an epimorphism) if and only ifkerf “0 (resp. cokerf “0).
Proof. By duality it suffices to prove that f is a monomorphism if and only if kerf “0. Supposef :X ÑY is a monomorphism. Iff g“0, for some morphismg, theng“0. Thus 0 is the kernel off.
Conversely, let kerf “0. Letu, v be morphisms such thatf u“f v. Letqbe the coequalizer ofuandv andm the unique morphism such thatf “mq. Since a coequalizer is an epimorphism,q“cokerwfor some morphismw.
Letkbe the kernel off. Fromf w“mqw“0 we get a unique morphismnsuch thatkn“w. Thusw“0 because kerf “0. The morphismqis an isomorphism because the cokernel of the zero morphism is an isomorphism. Thus qu“qv impliesu“v which shows thatf is a monomorphism.
To prove factorization of morphisms in abelian categories, we introduce strong epimorphisms. We will see that in an abelian category every epimorphism is strong.
Definition 2.2.7(Strong and regular epimorphisms). An epimorphismf :AÑBisstrongif for every commutative
wheref1 is a monomorphism, there exists a unique morphismh:BÑC making the diagram commutative.
An epimorphism is said regular if it is the coequalizer of some pair of morphisms.
Lemma 2.2.8. A regular epimorphism is a strong epimorphism.
Proof. Letf be the coequalizer ofa, b:AÑB such that the following diagram is commutative
A B C we get ga “ gb. Hence there exists a unique morphism h such that f h “ g. Since f is an epimorphism, by commutativity g1“f1h. From the fact f1 is a monomorphism, the morphism his unique. This shows thatf is a strong epimorphism.
Since all epimorphisms in an abelian category are regular by AB4, the above lemma implies that all epimorphisms in an abelian category are strong.
Lemma 2.2.9. Letf “ipfor some strong epimorphismpand some monomorphismi. This factorization is unique up to unique isomorphism, more precisely, ifip“i1p1 wherep1 is a strong epimorphism andi1 is a monomorphism, then there exists a unique isomorphism hsuch thatp“hp1 andi1“ih.
Proof. Letf “ip“i1p1, withpandp1 strong epimorphisms, andiandi1 monomorphisms. By definition of strong epimorphism there exist unique morphismshandh1such thathi“i1,h1i1“i,hp“p1, andh1p1 “p. By uniqueness hh1“Id andh1h“Id. Thus his an isomorphism.
We are ready to prove the epimorphism monomorphism factorization of morphisms in an abelian category.
Define theimage of a morphism f to be kerpcokerfqand thecoimage of a morphismf to be cokerpkerfq.
Theorem 2.2.10 (Factorization of morphisms). Let A be an abelian category. Every morphism f : X Ñ Y in A can be factorized uniquely, up to unique isomorphism, as f “ me, where e is an epimorphism and m is a monomorphism. In particular, we have f “ip, wherei“kerpcokerfqande“cokerpkerfq.
Moreover, if we have the following commutative diagram
X1 X2 X3
where e1 and e2 are epimorphisms and m1 and m2 are monomorphisms, then there exists a unique morphism h which makes the diagram commutative.
Proof. Factorization: Fix notation by the following diagram
X Y
kerf Coimf Imf cokerf
f
f2 f4
f1
g
f3
(2.2)
Since f f1 “0, there exists a unique morphism φ such that f “φf2. Now f4φf2 “ f4f “ 0, andf2 is an epimorphism, so f4φ“0. Thus there exists a unique morphismg such thatf “φf2“f4gf2.
To show thatgis an isomorphism it suffices to show thatgis an epimorphism and a monomorphism. Letvbe a morphism such thatf3gv“0 and letqbe the cokernel ofv. Hence there exists a unique morphismrsuch that rq “f3g. Nowqf2 is an epimorphism, so by AB4 there exists a morphism usuch thatqf2 “cokeru.
Fromf u“f3gf2u“rqf2u“0 one obtains thatu“f1lfor some unique morphisml. Next,f2u“f2f1l“0, so f2“sqf2 for some unique morphisms andsq “Id by the fact that f2 is an epimorphism. From this it follows that q is a monomorphism, soqv “0 implies v “0. This shows that f3g is a monomorphism. But f3 as a kernel is a monomorphism, sog is a monomorphism. Dually one gets that g is an epimorphism and hence an isomorphism.
Lemma 2.2.9 implies that the factorization is unique up to unique isomorphism.
Morphism between factorizations: Since all epimorphisms in an abelian category are strong, by definition of strong epimorphism there exists a unique morphism hmaking the following diagram commutative.
X1 X2
Y2 Y3 e1
e2f gm1 h m2
The following are useful corollaries of the above result.
Corollary 2.2.11. In an abelian categoryA, every monomorphism (resp. epimorphism) is the kernel of its cokernel (resp. cokernel of its kernel).
Proof. Letf :AÑBbe a monomorphism. By theorem 2.2.10 we havef “kerpcokerpfqqefor some epimorphisme.
Sincef is a monomorphism,eis a monomorphism, and hence an isomorphism. Thusf is the kernel of its cokernel.
By duality one obtains that an epimorphism is the cokernel of its kernel.
Corollary 2.2.12. Let A be an abelian category andf “me a morphism of Awhere mis a monomorphism and eis an epimorphism. Then kerf –kere andcokerf –cokerm.
Proof. By duality it suffices to show that kerf –kere. Letk1: kerf Ñf andk2: kereÑebe the corresponding morphisms. Then f k1 “0 impliesek1 “0 andek2 “0 impliesf k2 “0. From the universal properties if follows that kerf –kere.
The rest of this section is devoted for a quick introduction to exact sequences in abelian categories.
Definition 2.2.13 (Exact sequence). LetAbe an abelian category. A sequence of objects and morphisms of the form
. . . φ Xi´1 Xi Xi`1 . . .
i´2 φi´1 φi φi`1
is anexact sequence if Imφi´1–kerφi, where Imφi´1“kerpcokerφi´1q, for alliPZ.
In case an exact sequence consists of at most 3 consecutive nonzero objects the sequence is called a short exact sequence. In this case we do not write all the 0 objects but only
0 A f B g C 0 .
We have the following useful criterion for exact sequences Proposition 2.2.14. Let Abe an abelian category. Then
(i) 0ÑAÑf B is an exact sequence if and only iff is a monomorphism.
(ii) B Ñf AÑ0 is an exact sequence if and only iff is an epimorphism.
(iii) 0ÑAÑf BÑg C is an exact sequence if and only iff “kerg.
(iv) CÑg B Ñf AÑ0 is an exact sequence if and only iff “cokerg.
Proof. Let us prove (i). The morphism 0ÑAis a monomorphism because 0 is the terminal object inA. Hence by corollary 2.2.11, the morphism 0ÑAis image of itself. Thus by exactness kerf “0. By corollary 2.2.11 it follows thatf is a monomorphism. Conversely, iff is a monomorphism, then by corollary 2.2.11 0ÑAis its kernel. Thus the sequence is exact. By duality (ii) is also proven.
The statement of (iii) is a direct consequence of (i) and corollary 2.2.11. By duality (iv) holds.
The following lemma will be used later in derived functors.
Lemma 2.2.15. Consider the following diagram in an abelian category A.
X Y Z
kerpcokerfq kerpcokergq
f e1
g e2
m1 m2
.
Thenpf, gqis exact if and only if pm1, e2q is exact. If this is true, we have the following short exact sequence 0 kerg m1 Y e2 kerpcokergq 0 .
Proof. Follows from 2.2.14, 2.2.12, and 2.2.6.
Definition 2.2.16(Split short exact sequence). LetAbe an abelian category. We say that a short exact sequence
0 X f Y g Z 0
splits, or that it is a split short exact sequence, if there exist morphisms i : Y Ñ X and j : Z Ñ Y such that if “IdX andgj“IdZ.
Example 2.2.17. Not every short exact sequence splits. Consider the following short exact sequence
0 Z{2Z f Z{4Z g Z{2Z 0
in the categoryAb, wherefp1q “2 andgp1q “1. The only nonzero morphism fromZ{2ZtoZ{4Zsends 1 to 2, so there cannot be a morphismj:Z{2ZÑZ{4Zwithgj“IdZ{2Z.
Definition 2.2.18(Semisimple). An abelian categoryAwhere all the short exact sequences split is called semisim-ple.
Example 2.2.19. The category of finite dimensional vector spaces over a field is semisimple.