Kernels, cokernels and the abelian comparison

Written and self-checked by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Original text: CC0.

A kernel describes the maps annihilated by a given arrow. A cokernel describes the maps that annihilate it. In an additive category, differences convert these two questions into equalizers, coequalizers and signed squares. The abelian axiom imposes one further compatibility: quotienting the source by its kernel must give the same object as the image inside the target. Having kernels and cokernels, or even having every map that is both monic and epic invertible, does not by itself impose that compatibility.

Work in a fixed universe with locally small additive categories. Kernels and cokernels will be assumed only where stated. We retain the full universal arguments in Testing exactness with doubled objects, §1, Exact squares and endpoint tests, §1, and Trace coreflections and balanced nonabelian categories, §§3–4. For the specified coimage–image factorization and its cancellation properties, use Coimages, images and composition of quotients, §§1–2. The additional arguments below check their precise additive interfaces.

1. Universal maps into and out of zero

For \(f:X\to Y\), a kernel is a map \(k:K\to X\) with \(fk=0\), through which every \(t:T\to X\) with \(ft=0\) factors uniquely. This is exactly the pullback of \(f:X\to Y\) and \(0\to Y\): the map from any test object to the zero object is unique. It is also exactly the equalizer of \(f,0:X\rightrightarrows Y\). Thus \[ \ker f=X\times_Y0=\operatorname{eq}(f,0), \tag{1.1} \] where equality means the unique compatible isomorphism between the universal objects.

A cokernel is a map \(q:Y\to Q\) with \(qf=0\), through which every \(t:Y\to T\) with \(tf=0\) factors uniquely. It is the kernel of the reversed arrow in the opposite category, or equivalently the coequalizer of \(f,0\). Reversing arrows preserves an additive category: Hom addition remains unchanged, bilinear composition reverses, and biproduct injections and projections exchange. Therefore every kernel argument has this precise cokernel dual.

Composition is bilinear, so the maps on Hom groups are group homomorphisms. The retained Hom-factorization proof gives \[ \begin{aligned} \operatorname{Hom}(T,K) &\simeq\ker\bigl(\operatorname{Hom}(T,f)\bigr),\\ \operatorname{Hom}(Q,T) &\simeq\ker\bigl(\operatorname{Hom}(f,T)\bigr). \end{aligned} \tag{1.2} \] The first map sends \(a\) to \(ka\); its inverse is the unique factor of an arrow killed by \(f\). The second sends \(a\) to \(aq\), with the dual inverse. Precomposition in the first variable and postcomposition in the second commute with these maps, so the identifications are natural. The proof in Testing exactness, §1, uses exactly these equations and uniqueness; its factorization argument requires no abelian coimage axiom. This checks its use here under the broader additive hypotheses.

Two kernels of the same arrow have unique maps to each other over \(X\). Each composite and the identity are factors of the same kernel arrow, so uniqueness makes the composites identities. This proves uniqueness up to a unique compatible isomorphism. The cokernel proof reverses the arrows. Also, if \(ka=kb\), both \(a,b\) factor the same map killed by \(f\), so \(a=b\). Thus every kernel arrow is monic. Dually every cokernel arrow is epic.

For parallel \(f,g:X\to Y\), the equation \(ft=gt\) is equivalent to \((f-g)t=0\). Therefore their equalizer is \(\ker(f-g)\). The equation \(tf=tg\) is likewise equivalent to \(t(f-g)=0\), giving the cokernel description of their coequalizer.

The complete general additive zero-kernel and zero-cokernel arguments are retained in the first paragraph of the proof of Trace coreflections, Proposition 4.1. They give \[ \begin{gathered} f\text{ monic}\ \Longleftrightarrow\ \ker f\simeq0,\\ f\text{ epic}\ \Longleftrightarrow\ \operatorname{coker}f\simeq0. \end{gathered} \tag{1.3} \] Existence of the indicated universal object is understood. In particular, the kernel of a kernel arrow is zero; its universal arrow is \(0\to K\), since a monomorphism kills only zero arrows.

If \(f=0\), the identity of \(X\) is its kernel: every map to \(X\) factors uniquely through that identity. Conversely, if a kernel arrow \(k\) is invertible, \(fk=0\) implies \(f=0\). Thus \(\ker f\to X\) is invertible exactly when \(f=0\). The dual says that \(Y\to\operatorname{coker}f\) is invertible exactly when \(f=0\). In particular, the cokernel of \(0\to X\) and the kernel of \(Y\to0\) have their identity structural maps.

2. Finite diagrams use differences

Suppose now that every arrow has a kernel and a cokernel. For \(g_0:X_0\to Y\) and \(g_1:X_1\to Y\), the pullback is the kernel of the signed map \[ \begin{gathered} X_0\oplus X_1\longrightarrow Y,\\ (g_0,-g_1)=g_0p_0-g_1p_1. \end{gathered} \tag{2.1} \] For \(f_0:X\to Y_0\) and \(f_1:X\to Y_1\), the pushout is the cokernel of \[ i_0f_0-i_1f_1:X\longrightarrow Y_0\oplus Y_1. \tag{2.2} \] Here \(p_\nu,i_\nu\) are the biproduct projections and injections. The full signed-square proof retained in Exact squares, §1, identifies a kernel factor with a pair satisfying the pullback equation, and a map out of the cokernel with a pair satisfying the pushout equation. These steps use only the biproduct and kernel or cokernel universal properties. They therefore prove (2.1)–(2.2) in the present additive category, even if it is not abelian. The later exact-obstruction statements of that lesson keep their abelian hypotheses.

It is convenient to write the pushout \(Y_0\amalg_XY_1\) as \(Y_0\oplus_XY_1\). This notation means a pushout; it does not mean an ordinary biproduct without the relation imposed by \(X\).

More generally every finite diagram has a limit and a colimit. Here finite means that the indexing category has finitely many objects and arrows. For \(D:I\to\mathcal C\), set \[ \begin{gathered} V=\bigoplus_{i\in\operatorname{Ob}I}D_i,\\ W=\bigoplus_{a:i\to j}D_j,\\ p_a\delta=D(a)p_i-p_j, \\ L=\ker\delta. \end{gathered} \tag{2.3} \] The projection \(p_a\) is from \(W\); \(p_i,p_j\) are from \(V\). Finite products specify the unique \(\delta:V\to W\) with these components. A map \(x:T\to V\) is exactly a tuple \((x_i:T\to D_i)_i\). The equation \(\delta x=0\) is exactly \(D(a)x_i=x_j\) for every arrow, namely the cone equations. The kernel property gives a unique factor \(T\to L\). Composing its kernel arrow with the \(p_i\) gives the required cone projections. The same equivalence proves their universal property and uniqueness for every \(T\), so \(L\) is a limit.

For empty \(I\), both sums are zero and the construction gives the terminal zero object. Apply the proved limit construction to \(D^{op}:I^{op}\to\mathcal C^{op}\). The opposite has finite biproducts and kernels because the original has cokernels. Reversing the resulting universal cone gives a colimit of \(D\), including the initial zero object for the empty diagram. No infinite object products or sums are asserted by this finite argument.

3. Retain the specified coimage and image

The complete general additive proof in Trace coreflections, §3, identifies the kernel pair of \(f:X\to Y\) with \(X\oplus\ker f\) and its self-pushout with \(Y\oplus\operatorname{coker}f\), with their specified projections and injections. It proves \[ \begin{gathered} \operatorname{Coim}f=\operatorname{coker}(\ker f\to X),\\ \operatorname{Im}f=\ker(Y\to\operatorname{coker}f). \end{gathered} \tag{3.1} \] This identifies the intrinsic kernel-pair and self-pushout definitions with the additive ones, through the full universal arguments already in that lesson. It applies to every additive category with kernels and cokernels; no generator or ambient abelian category is used in that general paragraph.

Section 2 supplies the finite limits and colimits required by the complete construction in Coimages, §1. Retain its specified factorization \[ \begin{gathered} X\xrightarrow{q_f}\operatorname{Coim}f \xrightarrow{u_f}\operatorname{Im}f,\\ \operatorname{Im}f\xrightarrow{i_f}Y,\\ f=i_fu_fq_f. \end{gathered} \tag{3.2} \] There \(q_f\) is epic, \(i_f\) monic, and the compatible \(u_f\) is unique. The comparison in (3.1) is this same arrow by the retained general proof in Trace coreflections. Thus an arbitrary isomorphism between the two objects is insufficient: the arrow attached to \(f\) must be invertible. Exercise 3 proves the precise compatibility of these comparisons with a commutative square of arrows.

4. The extra abelian axiom

An additive category is abelian when every arrow admits a kernel and a cokernel and every specified comparison \(u_f\) in (3.2) is invertible. Equivalently, every arrow is strict in the retained intrinsic convention. Section 2 then supplies all finite limits and colimits.

In such a category a map that is both monic and epic is invertible. Indeed, its kernel and cokernel are zero by (1.3). Equation (3.1) identifies its coimage quotient with the cokernel of \(0\to X\), hence with \(1_X\), and its image inclusion with the kernel of \(Y\to0\), hence with \(1_Y\). Consequently all three maps in (3.2) are invertible. Their composite is \(f\), so \(f\) is invertible, with the reverse composite of the three inverses as inverse.

The converse to this last implication is false. The complete example in Trace coreflections, §4, is an additive category with kernels and cokernels in which every map that is both monic and epic is invertible, while a particular coimage–image comparison is not. Exercise 4 extracts its exact obstruction. Thus the abelian axiom tests every arrow's specified comparison, not just the cancellation properties of an arrow that happens to be both monic and epic.

5. Module kernels and a missing finite kernel

Let \(R\) be any unital ring and use left modules. For \(f:M\to N\), put \[ K=f^{-1}(0),\qquad Q=N/f(M). \tag{5.1} \] The subset \(K\) is closed under addition, negatives and left scalar multiplication because \(f\) is linear, so is a submodule. A linear \(t:L\to M\) with \(ft=0\) lands in \(K\); the same element function, with codomain \(K\), is its unique linear factor through inclusion. Thus inclusion is the kernel. Likewise \(f(M)\) is a submodule, and \(N/f(M)\) has the induced module operations. A linear \(t:N\to L\) with \(tf=0\) is constant on each coset of \(f(M)\), and therefore defines a unique linear map \(Q\to L\), \([n]\mapsto t(n)\). This is exactly the cokernel property.

Module maps form abelian groups, composition is bilinear, and zero and finite direct sums give the additive structure by their componentwise universal properties. The full computation in Coimages, §5, proves that every module arrow's intrinsic comparison is the compatible linear bijection \(M/\ker f\to f(M)\), with its linear inverse. Retaining that proof and the kernel/cokernel constructions above proves that the category of all left \(R\)-modules is abelian, with no commutativity or finiteness restriction on \(R\).

For use with the retained trace construction, this category is also cocomplete. The direct sum of any small family consists of tuples with finite support. Its inclusions are linear, and any family of maps from the summands determines the unique linear map given by a finite sum on such a tuple. This proves the coproduct property. For a small diagram take the direct sum of its objects and quotient by the submodule generated by all differences between an element in its source summand and its image in the target summand, for every diagram arrow. A map out of the direct sum kills these differences exactly when its component maps form a compatible cocone. Quotient factorization then proves the colimit universal property. The empty diagram gives zero. This verifies the needed ambient colimits rather than inferring them from abelianity alone.

Now let \(I\) be a left ideal that is not finitely generated. Both \(R\) and \(R/I\) are finitely generated modules. Nevertheless the quotient \[ \pi:R\longrightarrow R/I \tag{5.2} \] has no kernel in the full category of finitely generated left modules.

Proof. Suppose \(\kappa:K\to R\) were such a kernel, with \(K\) finitely generated. The equation \(\pi\kappa=0\) gives \(\kappa(K)\subseteq I\). If \(\kappa(z)=0\), the linear map \(R\to K\), \(r\mapsto rz\), and the zero map are both factors of the zero map \(R\to R\). The kernel's uniqueness forces them to agree, hence \(z=0\). Thus \(\kappa\) is injective as a module map, even though its universal property was only tested in the smaller category.

For each \(x\in I\), the linear map \(R\to R\), \(r\mapsto rx\), is killed by \(\pi\). Since \(R\) is one of the allowed finitely generated test objects, it factors through \(\kappa\). Evaluating that factor at \(1\) shows \(x\in\kappa(K)\). Therefore \(\kappa(K)=I\). Images of a finite generating family of \(K\) would then generate \(I\), a contradiction. This proves the claim. Conversely, if \(I\) is finitely generated, its ordinary inclusion is an object and kernel arrow of this full subcategory, by restriction of the kernel property proved above. \(\square\)

The obstruction is therefore the absence of the required universal object, rather than merely the failure of a proposed inclusion to satisfy a formula. Exercise 2 gives an explicit ring and ideal.

6. Four graded exercises with full solutions

Exercise 1 (introductory: signs over the integers). Compute the pullback of \(\mathbb Z\xrightarrow{2}\mathbb Z\xleftarrow{3}\mathbb Z\) and the pushout of \(\mathbb Z\xleftarrow{2}\mathbb Z\xrightarrow{3}\mathbb Z\) in abelian groups. Specify both structural maps and verify the universal properties, including their signs.

Solution. The pullback's elements are pairs \((a,b)\) with \(2a=3b\). Since 2 and 3 are coprime, every pair is uniquely \((3t,2t)\). Thus its object is \(\mathbb Z\), with projections multiplication by 3 and 2. For maps \(u,v:T\to\mathbb Z\) satisfying \(2u=3v\), the factor is \(h=u-v\): the equations give \(3h=u\) and \(2h=v\). It is unique since \(3-2=1\).

The pushout is \(\mathbb Z^2/\mathbb Z(2,-3)\). The homomorphism \((a,b)\mapsto3a+2b\) is surjective since \(3-2=1\). Its kernel consists precisely of \((2t,-3t)\), by the same coprimality argument. It therefore identifies the quotient with \(\mathbb Z\), whose structural maps from the two summands are multiplication by 3 and 2. For \(u,v:\mathbb Z\to T\) with \(2u=3v\), the unique factor \(w:\mathbb Z\to T\) is \(u-v\). Again \(3w=u\), \(2w=v\), and uniqueness follows by subtracting these two equations. Hence \[ \begin{gathered} P\to\mathbb Z^2:\quad t\mapsto(3t,2t),\\ \mathbb Z^2\to Q:\quad(a,b)\mapsto3a+2b. \end{gathered} \tag{6.1} \] The pullback equation uses \((2,-3)\) as a row, while the pushout kills \((2,-3)\) as a column. The universal computations explain both occurrences of the minus sign.

Exercise 2 (intermediate: an explicit absent kernel). For a field \(k\), let \(R=k[x_1,x_2,\ldots]\) and \(I=(x_1,x_2,\ldots)\). Prove \(I\) is not finitely generated, and conclude that \(R\to k\) has no kernel in finitely generated \(R\)-modules, although its source and target are objects of that category.

Solution. The ideal \(I\) consists exactly of the polynomials with zero constant term: every nonconstant monomial is divisible by one of its variables, and every polynomial has only finitely many monomials. Suppose a finite list of polynomials \(f_1,\ldots,f_m\) generated \(I\). There is a finite set \(S\) of variables occurring in those polynomials. Choose \(x_j\) outside \(S\). The ring map \(R\to k[x_j]\) that keeps \(x_j\) and sends every other variable to zero sends each \(f_i\) to zero, because its constant term is zero and all its variables lie in \(S\). It would therefore send their generated ideal to zero. But it sends \(x_j\in I\) to the nonzero polynomial \(x_j\), a contradiction.

Evaluation at zero identifies \(R/I\) with \(k\). The free module \(R\) and its cyclic quotient \(k\) are finitely generated. Section 5's complete rank-one-probe argument now applies to this specific non-finitely-generated ideal and proves the absence of any kernel object in the smaller category. In the category of all modules the kernel is the existing module \(I\). No assertion that the finitely generated category is abelian is being assumed.

Exercise 3 (hard: the comparison respects a square). In an additive category with kernels and cokernels, let \(f:X\to Y\), \(f':X'\to Y'\), \(a:X\to X'\), and \(b:Y\to Y'\) satisfy \(bf=f'a\). Prove that the square induces canonical maps \(c:\operatorname{Coim}f\to\operatorname{Coim}f'\) and \(d:\operatorname{Im}f\to\operatorname{Im}f'\), and that \[ du_f=u_{f'}c. \tag{6.2} \] Check identity squares and composition of squares as well.

Solution. Let \(k_f:K_f\to X\) and \(r_f:Y\to Q_f\) be the kernel and cokernel of \(f\), and use primed versions for \(f'\). Since \(f'a k_f=bfk_f=0\), there is a unique \(\alpha:K_f\to K_{f'}\) with \(k_{f'}\alpha=a k_f\). Hence \(q_{f'}a k_f=0\), and the cokernel \(q_f\) gives a unique \(c\) with \(cq_f=q_{f'}a\).

Since \(r_{f'}bf=r_{f'}f'a=0\), there is a unique \(\beta:Q_f\to Q_{f'}\) with \(\beta r_f=r_{f'}b\). Consequently \(r_{f'}b i_f=\beta r_fi_f=0\), and the kernel \(i_{f'}\) gives a unique \(d\) with \(i_{f'}d=bi_f\). The choices are determined by the universal structural maps, so are canonical up to their unique compatible identifications.

Now \[ \begin{aligned} i_{f'}d u_fq_f&=bf,\\ i_{f'}u_{f'}c q_f&=f'a=bf. \end{aligned} \tag{6.3} \] Cancel the epic \(q_f\) and then the monic \(i_{f'}\) to get (6.2). For an identity square, the identity maps satisfy the defining equations for \(c,d\), so uniqueness gives those identities. For two composable squares, the composites of their \(c\)-maps satisfy the equation for the composite source map after \(q_f\); uniqueness identifies them with the composite square's coimage map. The same argument after \(i_{f''}\) handles the image maps. Thus all comparisons commute compatibly with identities and composition, without any abelian axiom.

Exercise 4 (advanced: cancellation leaves a defect). Use the full trace construction and example in Trace coreflections, §§1–4. For \(A=k[x,y]\), \(\mathfrak a=(x,y)\), let \(\mathcal C_0\) consist of the \(\mathfrak a\)-generated modules. Compute the canonical comparison for \(v:\mathfrak a\to\mathfrak a/(Ax)\), including the kernel of that comparison. Explain why \(\mathcal C_0\) passes the mono-and-epi invertibility test but fails the abelian axiom.

Solution. The retained trace theorem gives intrinsic kernel \(T_{\mathfrak a}(\ker_Av)\) and ambient cokernel. Here \(\ker_Av=Ax\). Multiplication by \(x\) identifies \(A\) with \(Ax\), since the polynomial ring is a domain. The retained full ideal calculation gives \(T_{\mathfrak a}A=\mathfrak a\), so naturality under this isomorphism gives \(T_{\mathfrak a}(Ax)=x\mathfrak a\). The quotient is generated, and \(v\) is epic, so its intrinsic cokernel is zero. Therefore its intrinsic image is the whole target. Equation (3.1) gives \[ \begin{gathered} u_v:\mathfrak a/(x\mathfrak a) \longrightarrow\mathfrak a/(Ax),\\ \ker_Au_v=Ax/(x\mathfrak a) \simeq A/\mathfrak a\simeq k. \end{gathered} \tag{6.4} \] The isomorphism sends the class of \(a\in A\) to the class of \(xa\). Its kernel is exactly \(\mathfrak a\), by cancellation of \(x\), and its image is all of \(Ax/(x\mathfrak a)\). The retained ideal presentation shows \(k\) is itself \(\mathfrak a\)-generated. Hence the intrinsic kernel of \(u_v\) is also this nonzero \(k\), by the trace theorem. The comparison is not invertible.

The complete proof of Trace coreflections, Proposition 4.1, shows that mono and epi in \(\mathcal C_0\) agree with injective and surjective maps in all \(A\)-modules. A map with both properties has its module-linear inverse in the full subcategory, so passes the invertibility test. This test nevertheless leaves the nonzero defect in (6.4) untouched. The actual abelian axiom forbids that defect for every arrow. The ambient module category and its needed small colimits satisfy the hypotheses by Section 5; the earlier full trace and ideal proofs supply the intrinsic calculations.

7. References