An abelian category is an additive category in which every morphism has a compatible kernel, quotient, and image. We formulate this entirely through arrows. The final section introduces the hypotheses that permit reconstruction from a ring.
4.1. Kernels and cokernels
Definition 4.1. Universal zero composites
In an additive category, a kernel of \(f:X\to Y\) is an arrow \(k:K\to X\) with \(fk=0\), such that every \(t:T\to X\) satisfying \(ft=0\) factors uniquely as \(t=ku\). A cokernel is \(q:Y\to Q\) with \(qf=0\), such that every \(t:Y\to T\) satisfying \(tf=0\) factors uniquely as \(t=uq\). They are the equalizer and coequalizer of \(f,0\); their arrows are monic and epic by Proposition 2.7.
We sometimes write \(\ker f\) or \(\operatorname{coker}f\) for the object, and sometimes for the structure arrow, making the domain and codomain explicit when needed.
Alphabeta Math: Kernels and cokernels
Proposition 4.2. Cancellation and zero kernels
If kernels and cokernels exist, \(f\) is monic exactly when its kernel is a zero object, and epic exactly when its cokernel is a zero object. Equalizers of \(f,g\) are kernels of \(f-g\); coequalizers are cokernels of \(f-g\).
Alphabeta Math: Zero-kernel criterion · Zero-cokernel criterion · Equalizers of differences · Coequalizers of differences
Proof.
If \(f\) is monic, \(fk=0=f0\) implies \(k=0\). Since \(k\) is monic, \(k\mathrm{id}_K=k0\) implies \(\mathrm{id}_K=0\), which makes \(K\) a zero object as in Proposition 3.9. Conversely, if the kernel object is zero and \(fa=fb\), then \(f(a-b)=0\). Factoring through the zero kernel gives \(a-b=0\), so \(a=b\). If \(f\) is epic, its cokernel arrow \(q\) is zero because \(qf=0f\), and epicness of \(q\) gives \(\mathrm{id}_Q=0\). Conversely, \(af=bf\) implies \((a-b)f=0\), so \(a-b\) factors through the zero cokernel and vanishes. Finally \(ft=gt\) is equivalent to \((f-g)t=0\), and \(tf=tg\) to \(t(f-g)=0\); their universal properties agree.
□4.2. Image, coimage, and the abelian axiom
Definition 4.3. The canonical comparison
Assume every morphism has a kernel and cokernel. For \(f:X\to Y\), choose
\[k:K\to X=\ker f,\quad q:X\to C=\operatorname{coker}k,\quad c:Y\to Q=\operatorname{coker}f,\quad i:I\to Y=\ker c.\]Call \(C\) the coimage of \(f\) and \(i:I\to Y\) its image. There is a unique arrow \(\bar f:C\to I\) with \(f=i\bar f q\), called the canonical coimage-to-image comparison.
Alphabeta Math: Image and coimage · Canonical comparison
Worked example for Definition 4.3
Counterexample: An additive category can fail the coimage–image condition
Proof.
Since \(fk=0\), the cokernel property of \(q\) gives a unique \(g:C\to Y\) with \(gq=f\). Then \(cgq=cf=0\); \(q\) is epic, so \(cg=0\). The kernel property of \(i\) gives a unique \(\bar f\) with \(i\bar f=g\). If another comparison has the same composite with \(i\) and \(q\), cancellation first by the epic \(q\) and then by the monic \(i\) proves equality.
□Definition 4.4. Abelian categories
An abelian category is an additive category with a kernel and cokernel for every morphism, in which every canonical comparison \(\bar f:\operatorname{coim}f\to\operatorname{im}f\) is an isomorphism. Thus each \(f\) factors as an epimorphism \(q\), an isomorphism \(\bar f\), and a monomorphism \(i\).
Alphabeta Math: Abelian categories
Worked example for Definition 4.4
Counterexample: An additive category can fail the coimage–image condition
Proposition 4.5. Normal morphisms and cancellation in both directions
In an abelian category every monomorphism is a kernel of its cokernel, and every epimorphism is a cokernel of its kernel. Consequently a morphism which is both monic and epic is an isomorphism.
Alphabeta Math: Normal monomorphisms · Balanced categories
Worked example for Proposition 4.5
Counterexample: An additive category can fail the coimage–image condition
Proof.
If \(f\) is monic, its kernel object is zero by Proposition 4.2. The identity of \(X\) is a cokernel of \(0\to X\), since every map from \(X\) factors uniquely through identity. Hence \(q:X\to C\) is an isomorphism by uniqueness. Since \(\bar f\) is an isomorphism, \(f=i(\bar f q)\) is the kernel \(i\) precomposed with an isomorphism. This preserves the kernel universal property: factor through \(i\), then compose with \((\bar f q)^{-1}\).
If \(f\) is epic, its cokernel object is zero. The identity of \(Y\) is a kernel of \(Y\to0\), so \(i:I\to Y\) is an isomorphism. Then \(f=(i\bar f)q\) is the cokernel \(q\) postcomposed with an isomorphism, again with the same universal property. If \(f\) is both monic and epic, both \(q\) and \(i\), as well as \(\bar f\), are isomorphisms, so their composite is an isomorphism.
□Proposition 4.6. A kernel is unchanged by taking its cokernel and kernel
If \(k:K\to X\) is the kernel of some \(f:X\to Y\), then \(k\) is also a kernel of its cokernel. If \(q:Y\to Q\) is a cokernel, then it is also a cokernel of its kernel. These assertions already hold in any additive category with kernels and cokernels.
Alphabeta Math: Kernel and cokernel universal properties · Kernel–cokernel correspondence in abelian categories
Proof.
Write \(c=\operatorname{coker}k\). Since \(fk=0\), factor \(f=hc\). If \(ct=0\), then \(ft=hct=0\), so \(t\) factors uniquely through \(k\). Since \(ck=0\), this is exactly the kernel property for \(c\). For the second assertion, let \(q=\operatorname{coker}f\) and \(k=\ker q\). Because \(qf=0\), factor \(f=ku\). If \(tk=0\), then \(tf=tku=0\), so \(t\) factors uniquely through \(q\). Since \(qk=0\), this is exactly the cokernel property for \(k\).
□Example 4.7. The category of abelian groups
For \(f:A\to B\), its kernel is the subgroup \(\{a:f(a)=0\}\) with inclusion, and its cokernel is \(B/f(A)\) with quotient map. The factorization arguments in Example 2.8 verify both universal properties. Its coimage is \(A/\ker f\), and its image is the subgroup \(f(A)\), because \(\ker(B\to B/f(A))=f(A)\). The canonical map \([a]\mapsto f(a)\) is well-defined, surjective, and injective: \(f(a)=f(a')\) means \(a-a'\in\ker f\). Its inverse is a homomorphism because a bijective group homomorphism has homomorphic inverse, as follows by applying it to a sum and using injectivity. Thus \(\mathrm{Ab}\) is abelian. Chapter 5 gives the same proof for modules over an arbitrary unital ring.
Alphabeta Math: Abelian groups · Explicit kernels and quotients
4.3. Exactness and preservation
Definition 4.8. Exact sequences
A pair \(X\xrightarrow{f}Y\xrightarrow{g}Z\) is exact at \(Y\) when its image arrow \(i:\operatorname{im}f\to Y\) and kernel arrow \(k:\ker g\to Y\) represent the same subobject: there is an isomorphism \(u\) with \(ku=i\). In particular \(gf=0\), since \(f=i\bar f q\). A sequence is exact when this holds at every displayed intermediate object. A short exact sequence
\[0\longrightarrow K\xrightarrow{k}X\xrightarrow{q}Y\longrightarrow0\]is equivalently a kernel-cokernel pair: \(k\) is a kernel of \(q\) and \(q\) a cokernel of \(k\).
Alphabeta Math: Exact sequences · Kernel-cokernel pairs
Proof.
Exactness at \(K\) says its outgoing map has zero kernel, so it is monic. Exactness at \(Y\) says the image of \(q\) is identity, since identity is the kernel of \(Y\to0\); its factorization makes \(q\) epic. For monic \(k\), its image is represented by \(k\) itself, by the proof of Proposition 4.5. Exactness at \(X\) therefore says \(k\) is a kernel of \(q\). Because \(q\) is epic, Proposition 4.5 makes it a cokernel of that kernel. Conversely a kernel-cokernel pair has monic \(k\), epic \(q\), and \(\operatorname{im}k\) represented by \(k=\ker q\), proving all three exactness assertions.
□Proposition 4.9. Presentations are cokernels
A sequence \(A\xrightarrow{f}B\xrightarrow{q}C\to0\) is exact if and only if \(q\) is a cokernel of \(f\).
Alphabeta Math: Exactness criterion · Cokernel universal property
Proof.
If \(q=\operatorname{coker}f\), its kernel is the image of \(f\) by definition, and \(q\) is epic. Conversely exactness makes \(q\) epic and identifies \(\ker q\) with \(i=\operatorname{im}f\). By Proposition 4.5, \(q\) is a cokernel of \(i\). In \(f=i\bar f p\), the composite \(\bar f p\) is epic. Thus \(tf=0\) if and only if \(ti=0\), by cancellation of that epimorphism. The cokernel property for \(i\) is therefore exactly the cokernel property for \(f\).
□Definition 4.10. Right-exact and exact functors
An additive functor between abelian categories is right exact if it preserves cokernels, or equivalently, by Proposition 4.9, sends each exact \(A\to B\to C\to0\) to an exact sequence of the same form. It is left exact if it preserves kernels. It is exact if it preserves both kernels and cokernels. Such a functor preserves every exact pair, since it preserves \(\operatorname{im}f=\ker(\operatorname{coker}f)\) and \(\ker g\).
Alphabeta Math: Left and right exactness · Exact functors · Preservation criterion
Every abelian category has all finite limits and colimits. Finite products and coproducts come from biproducts, and equalizers and coequalizers from differences and Proposition 4.2; apply Theorems 2.9 and 2.10 to finite diagrams. If it also has all set-indexed coproducts, it has every small colimit. We impose this last hypothesis later; it does not follow just from being abelian.
4.4. Projective objects and generators
Definition 4.11. Projective, generator, and small
Let \(\mathcal A\) be locally small and abelian. An object \(P\) is projective if for every epimorphism \(e:X\to Y\) and every \(u:P\to Y\), there is \(v:P\to X\) with \(ev=u\). It is a generator if the functor \(H=\operatorname{Hom}(P,-):\mathcal A\to\mathrm{Ab}\) is faithful: whenever \(a,b:X\to Y\) satisfy \(ah=bh\) for every \(h:P\to X\), then \(a=b\).
Assume \(\mathcal A\) has all set-indexed coproducts. We call \(P\) small if for every family \((X_i)\) the group homomorphism
\[\bigoplus_i\operatorname{Hom}(P,X_i)\longrightarrow\operatorname{Hom}(P,\coprod_iX_i),\qquad (h_i)\longmapsto\sum_i i_i h_i\]is an isomorphism. Its sum is finite because its input has finite support. Smallness thus says precisely that arrows from \(P\) into a coproduct are uniquely finite sums of arrows into its summands.
Alphabeta Math: Projective objects · Generators · Small projective generators
Proposition 4.12. Hom from a projective object is exact
The functor \(H=\operatorname{Hom}(P,-)\) is additive and preserves kernels for every \(P\). If \(P\) is projective, it also preserves cokernels, and sends every short exact sequence to a short exact sequence of abelian groups.
Alphabeta Math: Exactness of Hom
Proof.
The map on morphisms is postcomposition, so bilinearity gives additivity. If \(k:K\to X\) is a kernel of \(f\), then \(H(k)\) is injective by monicity. For every arrow \(h:P\to X\) killed by \(f\), the kernel property gives a unique \(a:P\to K\) with \(ka=h\). Thus \(H(k)\) identifies \(H(K)\) with the kernel subgroup of \(H(f)\).
Suppose now \(P\) is projective and \(q:Y\to Q\) is a cokernel of \(f:X\to Y\). The arrow \(q\) is epic, so projectivity says every \(P\to Q\) lifts through \(q\): \(H(q)\) is surjective. Write \(f=i e\), with \(i=\ker q\) and \(e:X\to\operatorname{im}f\) epic, using the abelian factorization. If \(h:P\to Y\) satisfies \(qh=0\), the kernel property gives \(h=i a\). Projectivity lifts \(a\) to \(b:P\to X\) through \(e\), so \(h=fb\). Conversely \(qfb=0\). Hence the kernel subgroup of \(H(q)\) is the image subgroup of \(H(f)\). A surjective group map with that kernel is a cokernel: a homomorphism vanishing on the subgroup descends uniquely to the quotient, and the induced quotient map to \(H(Q)\) is bijective. This proves cokernel preservation. Kernel preservation and projective lifting likewise give the short exact assertion.
□Proposition 4.13. A generator detects zero objects; a projective generator detects isomorphisms
If \(P\) is a generator and \(H(X)=0\), then \(X\) is a zero object. If \(P\) is also projective and \(H(f)\) is an isomorphism, then \(f\) is an isomorphism.
Alphabeta Math: Isomorphism detection
Proof.
If \(H(X)=0\), every arrow \(P\to X\) is zero. Thus \(\mathrm{id}_X\) and \(0:X\to X\) have equal composites with all such arrows. Faithfulness gives \(\mathrm{id}_X=0\), making \(X\) a zero object. If \(H(f)\) is an isomorphism, preservation of kernels and cokernels by Proposition 4.12 gives \(H(\ker f)=0\) and \(H(\operatorname{coker}f)=0\). The first assertion makes both objects zero, so Proposition 4.2 makes \(f\) monic and epic, and Proposition 4.5 makes it an isomorphism.
□Proposition 4.14. Every object has a presentation by copies of a generator
Suppose \(P\) is a generator and all set-indexed coproducts exist. For each \(X\), put \(I=\operatorname{Hom}(P,X)\) and let \(e_X:P^{(I)}\to X\) have restriction \(h\) on the summand labelled by \(h\in I\), where \(P^{(I)}=\coprod_{I}P\). Then \(e_X\) is epic. If \(k:K\to P^{(I)}\) is its kernel and \(J=\operatorname{Hom}(P,K)\), the sequence
\[P^{(J)}\xrightarrow{k e_K}P^{(I)}\xrightarrow{e_X}X\longrightarrow0\]is exact. Projectivity and smallness are not needed for this assertion.
Alphabeta Math: Generator epimorphisms · Generator presentations
Proof.
If \(a e_X=b e_X\), restriction to the summand labelled \(h\) gives \(ah=bh\) for every \(h:P\to X\). The generator property gives \(a=b\), so \(e_X\) is epic. The same argument makes \(e_K\) epic. Proposition 4.5 says \(e_X\) is a cokernel of \(k\). Since \(e_K\) is epic, \(t k e_K=0\) is equivalent to \(t k=0\), so \(e_X\) is also a cokernel of \(k e_K\). Proposition 4.9 proves exactness. Local smallness makes \(I,J\) sets, so the required coproducts exist.
□Example 4.15. The integer group as a small projective generator
In \(\mathrm{Ab}\), \(\operatorname{Hom}(\mathbb Z,M)\cong M\) by evaluating a homomorphism at \(1\); its inverse sends \(m\) to \(n\mapsto nm\). Thus it is faithful. Epimorphisms in \(\mathrm{Ab}\) are surjective: a cokernel is \(B/f(A)\), so Proposition 4.2 gives exactly that criterion. For a surjection \(A\to B\), a homomorphism \(\mathbb Z\to B\) lifts by choosing a preimage of its value at \(1\). Thus \(\mathbb Z\) is projective. A homomorphism from \(\mathbb Z\) to a direct sum is determined by its value at \(1\), which has finite support, so the comparison in Definition 4.11 is an isomorphism. Thus \(\mathbb Z\) is small. For \(M=\mathbb Z/2\mathbb Z\), the shorter presentation \(\mathbb Z\xrightarrow{\times2}\mathbb Z\to M\to0\) already suffices.
Alphabeta Math: Small projective generators · Abelian groups as integer modules
Chapter 5 supplies the corresponding facts for the regular module over a ring. Chapter 8 turns these presentations into the first equivalence in Theorem 1.1.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Multiplication by two on a finite group
Applies to: Definition 4.1, Proposition 4.2, Definition 4.3, Definition 4.4.
In \(\mathrm{Ab}\) take \(f:\mathbb Z/6\to\mathbb Z/6\), \(f(\bar n)=\overline{2n}\). Its kernel is \(K=\{\bar0,\bar3\}\); any map killed by \(f\) lands uniquely in \(K\). Its image is \(I=\{\bar0,\bar2,\bar4\}\); its cokernel is \((\mathbb Z/6)/I\cong\mathbb Z/2\), since a map killing \(f\) kills precisely the generators of \(I\). The coimage is \((\mathbb Z/6)/K\cong\mathbb Z/3\); the comparison sends \(\bar n+K\) to \(\overline{2n}\), a bijection with inverse \(\overline{2n}\mapsto\bar n+K\). This is the required epi-isomorphism-mono factorization.
Here neither kernel nor cokernel is zero, and \(f\) is neither monic nor epic: it kills \(\bar3\), and the nonzero quotient map kills its image. Kernels of \(f-0\) and cokernels of \(f-0\) are also its equalizer and coequalizer with zero. More generally the same coset formula \(M/\ker f\to\operatorname{im}f\) is bijective for any group homomorphism in \(\mathrm{Ab}\), which, together with finite direct sums, verifies the abelian-category axiom in this category.
Further proof details
The general comparison used in this example has a direct proof. If \(m-m'\in\ker f\), then \(f(m)=f(m')\), so \([m]\mapsto f(m)\) is well defined. It preserves sums (and scalars in the module setting). It is onto \(\operatorname{im}f\) by the definition of image. If \(f(m)=0\), then \(m\in\ker f\), so its kernel is zero and it is injective. Its inverse sends \(f(m)\) to \([m]\); equality of two image representatives makes their difference a kernel element, proving the inverse well defined and additive, and linear when relevant. Kernels are closed under addition and scalars by applying \(f\); quotient operations are well defined because changing representatives changes a sum or scalar multiple by an element of the submodule. These formulas, together with the displayed direct sums, establish the asserted abelian-category comparison rather than merely naming an isomorphism theorem.
Worked example 2. A nonsplit short exact sequence
Applies to: Proposition 4.5, Proposition 4.6, Definition 4.8, Proposition 4.9.
Take \(k:\mathbb Z\to\mathbb Z\), \(k(n)=2n\), and \(q:\mathbb Z\to\mathbb Z/2\). Then \(\ker q=2\mathbb Z=\operatorname{im}k\); \(k\) identifies its domain with that subgroup, and every map killing it descends uniquely modulo two. Thus \(0\to\mathbb Z\xrightarrow{2}\mathbb Z\xrightarrow q\mathbb Z/2\to0\) is short exact and \(q\) is the cokernel of \(k\). Taking the cokernel and then kernel recovers \(k\) up to its displayed identification, and taking the kernel and then cokernel recovers \(q\). There is no section of \(q\), since a homomorphism \(\mathbb Z/2\to\mathbb Z\) must send its generator to an element killed by 2, hence to zero. Exact does not mean split. An abelian-group map both monic and epic is bijective, and its inverse preserves addition by applying the original injective map to the addition identity; this checks the last assertion of Proposition 4.5 as well.
Worked example 3. Right exactness and left exactness differ
Applies to: Definition 4.10.
Applying \(F(M)=M/2M\) to the preceding presentation gives \(\mathbb F_2\xrightarrow0\mathbb F_2\xrightarrow1\mathbb F_2\to0\), still exact. For any presentation \(A\to B\to C\to0\), the quotient after applying \(F\) is \(B/(\operatorname{im}A+2B)\cong C/2C\), so \(F\) is right exact. It is not left exact: the original injection by 2 becomes zero on \(\mathbb F_2\). Conversely \(\operatorname{Hom}(\mathbb Z/2,-)\) preserves kernels by factoring maps through them, but fails to preserve this surjection: \(\operatorname{Hom}(\mathbb Z/2,\mathbb Z)=0\) cannot surject onto \(\operatorname{Hom}(\mathbb Z/2,\mathbb Z/2)\). The identity functor is exact, since its kernel and cokernel maps are unchanged.
Worked example 4. One integer detects everything
Applies to: Definition 4.11, Proposition 4.12, Proposition 4.13.
Let \(P=\mathbb Z\) in \(\mathrm{Ab}\). Evaluation at 1 identifies \(\operatorname{Hom}(P,M)\) with \(M\), with inverse \(m\mapsto(n\mapsto nm)\). Given a surjection \(e:X\to Y\), choose a preimage \(x\) of \(u(1)\); \(n\mapsto nx\) is a lift of \(u\), proving projectivity. If \(a\ne b\), choose \(m\) with \(a(m)\ne b(m)\) and test on \(1\mapsto m\), proving the generator condition. A map into a direct sum is determined by a finitely supported tuple \(h(1)\), proving smallness. Under evaluation, Hom sends the short exact sequence above to that very sequence; the same identification for every sequence proves exactness. If this Hom group is zero, every element is zero. If Hom applied to a map is bijective, the original map is bijective and its additive inverse is an isomorphism. Thus the detection claims hold explicitly.
Worked example 5. The canonical presentation has more generators than necessary
Applies to: Proposition 4.14.
For \(X=\mathbb Z/2\) and \(P=\mathbb Z\), the set \(I=\operatorname{Hom}(P,X)\) consists of two maps, labelled 0 and 1. Hence \(e_X:\mathbb Z^2\to\mathbb Z/2\) is \((a,b)\mapsto\bar b\), a surjection. Its kernel is \(K=\{(a,2c):a,c\in\mathbb Z\}\). Every element of \(K\) labels a map from \(P\), so \(e_K:\mathbb Z^{(K)}\to K\) sends its corresponding basis vector to that element and is onto. The composite into \(\mathbb Z^2\) therefore has image exactly \(K=\ker e_X\). This is the theorem’s canonical exact presentation, even though the shorter presentation by multiplication by 2 also exists.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. An additive category can fail the coimage–image condition
Related statements: Definition 4.3, Definition 4.4, Proposition 4.5.
Claim refuted. Additivity, or cancellation in an additive category, already gives the abelian-category conclusions.
Proof.
Use finite-rank free abelian groups with their homomorphisms. Integer-matrix addition is bilinear under composition, and finite direct sums and the zero group make the category additive. The arrow \(f=2:\mathbb Z\to\mathbb Z\) is monic by injectivity. It is epic within this category: if \(a,b:\mathbb Z\to\mathbb Z^n\) obey \(af=bf\), then \(2(a(1)-b(1))=0\); each integer coordinate is zero, so \(a=b\). But it is not invertible, since an integer inverse would satisfy \(2m=1\).
Its kernel is \(0\to\mathbb Z\). Its categorical cokernel here is \(\mathbb Z\to0\): any map from \(\mathbb Z\) to a free group killed by 2 is zero, and factors uniquely through zero. Hence its coimage and image objects are both \(\mathbb Z\), with the identity quotient and inclusion respectively; the comparison is multiplication by 2, not an isomorphism. In \(\mathrm{Ab}\) the cokernel instead is \(\mathbb Z/2\), an object excluded from this smaller category. This proves why the abelian axiom and the ambient category matter.
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