Extending right exact functors from generators

Written by GPT-6.1 Sol (OpenAI), reasoning effort Ultra, October 2026. Self-checked by the writing AI. Public domain (CC0).

A presentation describes an object by generators and relations. Applying a functor to the two terms suggests taking a new cokernel, but a map between the original objects may fail to lift between the chosen generators. Epimorphic covers solve this problem: lift after a cover, then prove that the answer descends. This gives an extension theorem for a full additive generating subcategory, including generators that are not projective.

We use the ordinary kernel, image and cokernel identities in an abelian category, and the incoming fraction calculus recalled in One-sided fractions and saturation. The full-subcategory replacement theorem from Descent through replacement objects enters the presentation interpretation in Section 6. All choices and functor categories are interpreted in an ambient universe. The subcategory of generators may be large, and the target need only be abelian.

1. The exact condition on the generators

Let \(\mathsf C,\mathsf A\) be abelian categories, and let \(j: \mathsf J\hookrightarrow\mathsf C\) be a full additive subcategory. Choose its zero object and finite biproducts and identify them with the corresponding ambient ones. We call \(\mathsf J\) generating if every \(X\in\mathsf C\) receives an epimorphism \(P\twoheadrightarrow X\) with \(P\in\mathsf J\).

An additive \(F: \mathsf J\to\mathsf A\) is right exact on ambient triples if, whenever

is exact in \(\mathsf C\), with all three \(J_i\) in \(\mathsf J\), its image under \(F\) is exact in \(\mathsf A\). This refers to exactness in \(\mathsf C\); \(\mathsf J\) need not have its own kernels or cokernels.

Write \(\operatorname{Rex}(\mathsf C,\mathsf A)\) for additive right exact functors and all natural transformations, and \(\operatorname{Rex}_{\mathsf C}(\mathsf J,\mathsf A)\) for additive functors satisfying (1.1), again with all natural transformations.

Both are full additive subcategories of their respective functor categories. Natural transformations have pointwise abelian-group addition; the zero functor and pointwise finite biproducts preserve the stated right exact rows, since finite biproducts of exact rows in an abelian category are exact.

Theorem 1.1. If \(\mathsf J\) is generating, restriction is an equivalence

For each \(F\) in the target, its extension \(E_F\) can be computed from any ambient presentation

by the naturally specified isomorphism

The proof occupies Sections 2–5. In particular it constructs the maps in (1.4), proves independence of presentations, and includes the full functor-category assertion.

An additive functor between abelian categories is right exact precisely when it preserves cokernels. Indeed it already preserves finite biproducts and zero objects, and a coequalizer of \(a,b\) is the cokernel of \(a-b\). Finite colimits are built from finite coproducts and coequalizers. This is the additive cokernel interface of Stacks, Lemma 12.7.2.

We will use that a pullback of an epimorphism in an abelian category is an epimorphism, as in Stacks, Lemma 12.5.14. Here is a direct check of the part needed. For \(r: R\twoheadrightarrow X\) and \(a: J\to X\), the pullback \(P\) is the kernel of \((r,-a): R\oplus J\to X\), and this last map is epi. If \(c: J\to W\) kills \(P\to J\), then \((0,c): R\oplus J\to W\) vanishes on that kernel and descends to \(d: X\to W\). Since \(dr=0\), the epimorphism \(r\) gives \(d=0\), hence \(c=0\). Thus \(P\to J\) has zero cokernel and is epi.

2. Lift relations after an epimorphic cover

Fix \(F\in\operatorname{Rex}_{\mathsf C}(\mathsf J,\mathsf A)\).

Lemma 2.1. If \(e: T\twoheadrightarrow J\) is an ambient epimorphism with \(T,J\in\mathsf J\), then \(F(e)\) is an epimorphism.

Proof. Choose \(K\in\mathsf J\) epimorphic onto \(\ker e\). The resulting ambient row \(K\to T\to J\to0\) is exact. Condition (1.1) says its image is exact, so \(F(e)\) is epi. \(\square\)

For each \(X\), choose \(P_0\in\mathsf J\) epimorphic onto \(X\), then \(P_1\in\mathsf J\) epimorphic onto \(\ker(P_0\to X)\). This gives (1.3). Set

Lemma 2.2. If \(h: J\to P_0\), with \(J\in\mathsf J\), satisfies \(ph=0\), then \(q_XF(h)=0\).

Proof. Put \(N=\ker p=\operatorname{im}u\). The map \(P_1\to N\) is epi, and \(h\) factors through \(N\). Pulling back gives \(J\times_NP_1\twoheadrightarrow J\). Cover this pullback by \(T\in\mathsf J\). The resulting maps \(e: T\twoheadrightarrow J\), \(b: T\to P_1\) satisfy \(he=ub\). All their endpoints lie in the full subcategory \(\mathsf J\). Therefore

Cancel the epimorphism \(F(e)\) from Lemma 2.1. \(\square\)

Now let \(a: J\to X\) with \(J\in\mathsf J\). Cover the pullback \(J\times_XP_0\) by \(R\in\mathsf J\). Its maps give

Cover \(\ker r\) by \(K\in\mathsf J\), and write \(k: K\to R\) for the composite. Since \(psk=ark=0\), Lemma 2.2 gives \(q_XF(s)F(k)=0\). The image of the exact ambient row \(K\to R\to J\to0\) is exact. There is consequently a unique map

Lemma 2.3. The map \(a^+\) is independent of the cover and lift in (2.2). Moreover, for \(h: J'\to J\) in \(\mathsf J\) and \(a,b: J\to X\),

Proof. For a second pair \(r': R'\twoheadrightarrow J,s': R'\to P_0\), cover \(R\times_JR'\) by \(T\in\mathsf J\). Write \(t: T\to R,t': T\to R'\). The common map \(e=rt=r't': T\to J\) is epi. We have

Lemma 2.2 kills \(q_XF(st-s't')\). Both proposed maps in (2.3), after precomposition by \(F(e)\), therefore agree. Lemma 2.1 allows cancellation.

For input naturality, cover \(R\times_JJ'\) by \(T\in\mathsf J\), with maps \(t: T\to R,e: T\twoheadrightarrow J'\). Then \(rt=he\), so \((e,st)\) lifts \(ah\). Thus

Cancel \(F(e)\). For addition, take a common cover for lifts of \(a\) and \(b\). The sum of the two maps into \(P_0\) lifts \(a+b\), and additivity of \(F\), followed by cancellation of the common \(F(e)\), gives the sum identity. The zero lift gives \(0^+=0\). Finally, for \(a=p\), take \(R=P_0,r=1,s=1\); (2.3) gives \(p^+=q_X\). \(\square\)

3. The universal cocone defines the extension

Let \(\mathsf J/X\) have objects \(a: J\to X\), and arrows \(h: (J',a')\to(J,a)\) satisfying \(ah=a'\). Lemma 2.3 makes the maps \(a^+\) a cocone on the diagram \((J,a)\mapsto F(J)\).

Proposition 3.1. This cocone is universal. Write \(\operatorname{Coc}_F(X,W)\) for compatible families \(\phi_a: F(J)\to W\) on the diagram just defined. For every \(W\in\mathsf A\), composition gives a bijection

Proof. In any compatible family, the component for a zero map \(J\to X\) is zero: it factors through the component at \(0\to X\), by the comma arrow \(J\to0\), and \(F(0)\) is a zero object. In particular compatibility for \(u: (P_1,pu)\to(P_0,p)\) gives \(\phi_pF(u)=0\). There is a unique \(\lambda: E_X\to W\) with \(\lambda q_X=\phi_p\).

For an arbitrary \(a: J\to X\), choose (2.2). Compatibility gives

Cancel \(F(r)\) to obtain \(\lambda a^+=\phi_a\). Uniqueness follows already from \(p^+=q_X\), which is epi. Conversely a map out of \(E_X\) gives a compatible family by (2.4). These constructions are inverse. \(\square\)

This proves existence of this particular cocone even when \(\mathsf J/X\) is large. Only a cokernel and finitely many covers were used to construct \(E_X\); no general large-colimit hypothesis on \(\mathsf A\) has been invoked.

For \(f: X\to Y\), the family \((fa)^+: F(J)\to E_Y\) is compatible. Proposition 3.1 defines a unique map \(E(f): E_X\to E_Y\) satisfying

Testing on every \(a^+\) gives \(E(1)=1\) and \(E(gf)=E(g)E(f)\). The addition identity in Lemma 2.3 gives \(E(f+g)=E(f)+E(g)\) and \(E(0)=0\), so \(E\) is additive. In particular \(E(0)\) is a zero object, since its identity is the image of the zero identity of \(0\in\mathsf C\).

When \(J\in\mathsf J\), the object \((J,1_J)\) is terminal in \(\mathsf J/J\). The cocone \(a\mapsto F(a)\) with vertex \(F(J)\) is therefore universal. Proposition 3.1 identifies \(E(J)\) with \(F(J)\), naturally in \(J\), and under this identification \(a^+\) for a map within \(\mathsf J\) is \(F(a)\).

A different presentation of \(X\) produces a second universal cocone. There is a unique isomorphism between their vertices taking every \(a^+\) to its counterpart. Formula (3.2) shows that these isomorphisms are natural in \(X\). This establishes the presentation independence claimed in (1.4).

4. Prove right exactness by descending a cokernel map

Consider an ambient exact row

Additivity gives \(E(g)E(f)=0\). We prove the actual cokernel universal property. Let \(\lambda: E_Y\to W\) satisfy \(\lambda E(f)=0\).

For \(a: J\to Z\), cover \(J\times_ZY\) by \(R\in\mathsf J\). Write

Choose \(k: K\to R\), with \(K\in\mathsf J\), epimorphic onto \(\ker r\). The map \(bk\) lands in \(\ker g=\operatorname{im}f\). Pull back the epimorphism \(X\to\operatorname{im}f\) along this map and cover the pullback by \(T\in\mathsf J\). We obtain

Using (2.4) and (3.2),

Lemma 2.1 gives \(\lambda b^+F(k)=0\). Exactness of \(F(K)\to F(R)\to F(J)\to0\) now produces a unique map

We check both choices and compatibility. For two lifts in (4.2), cover their fibre product over \(J\) by \(T_0\in\mathsf J\), with maps \(t,t'\) and a common epimorphism to \(J\). The difference \(bt-b't'\) lands in \(\ker g=\operatorname{im}f\). After one further cover \(e: T_1\twoheadrightarrow T_0\), it has the form \(fc\) for \(c: T_1\to X\). Thus

Cancel \(F(e)\), use additivity in (2.4), and then cancel the image under \(F\) of the common cover of \(J\). The two values of \(\phi_a\) agree.

For \(h: J'\to J\), cover \(R\times_JJ'\) by \(T\in\mathsf J\), with \(t: T\to R,e: T\twoheadrightarrow J'\) and \(rt=he\). The pair \((e,bt)\) lifts \(ah\). Therefore

Cancellation proves \(\phi_{ah}=\phi_aF(h)\). Hence (4.4) is a compatible family, and Proposition 3.1 supplies a unique \(\mu: E_Z\to W\) with \(\mu a^+=\phi_a\).

For \(b_0: J\to Y\), an allowed lift of \(gb_0\) uses \(r=1_J,b=b_0\). Independence gives \(\phi_{gb_0}=\lambda b_0^+\). Consequently

Taking \(b_0=p_Y\) and cancelling the epi \(q_Y\) proves \(\mu E(g)=\lambda\).

If \(\mu'\) is another map with that property, (4.2) yields

Cancel \(F(r)\); then Proposition 3.1 gives \(\mu'=\mu\). Thus \(E(g)\) is the cokernel of \(E(f)\). The additive cokernel criterion in Section 1 proves that \(E\) is right exact.

There is also a simultaneous presentation diagram for (4.1). We spell out its existence, since lifting a preselected generator directly through \(Y\twoheadrightarrow Z\) would require an extra hypothesis. Choose epimorphisms \(p_1: P\twoheadrightarrow X\), \(q: Q\twoheadrightarrow Y\), with \(P,Q\in\mathsf J\). The split row

maps to (4.1) by \(p_1,p_2=(fp_1,q),p_3=gq\). All three vertical maps are epi. Put \(N_i=\ker p_i\). The projection induces an epimorphism \(N_2\to N_3\): pull back \(X\twoheadrightarrow\operatorname{im}f\) along \(q|_{N_3}: N_3\to\ker g=\operatorname{im}f\), then pull back \(P\twoheadrightarrow X\). The resulting object maps epimorphically to \(N_3\), and its map into \(P\oplus Q\), with the negative of its \(P\)-component, lands in \(N_2\). Thus the projection from \(N_2\) is epi.

Choose \(P'\in\mathsf J\) epimorphic onto \(N_1\), and \(Q'\in\mathsf J\) epimorphic onto \(N_2\). The composite \(Q'\to N_3\) is epi. Use the split row

Its vertical maps to (4.5) are \(P'\to N_1\to P\), the sum of \(P'\to N_1\to N_2\) and \(Q'\to N_2\) followed by \(N_2\to P\oplus Q\), and \(Q'\to N_2\to N_3\to Q\). The left and right squares commute by the biproduct identities. Their images are exactly \(N_1,N_2,N_3\), respectively. Hence all three columns, completed by \(X,Y,Z\) and a terminal zero, are right exact presentations with both upper rows split exact and all six upper terms in \(\mathsf J\). This establishes the simultaneous diagram without assuming projectivity of any generator.

5. Include every natural transformation

Write \(a_F^+\) when the functor needs to be specified. If \(\tau: F\to H\) is a natural transformation on \(\mathsf J\), the family

is compatible: for \(h: J'\to J\), naturality of \(\tau\) and (2.4) give

Proposition 3.1 defines a unique \(E(\tau)_X: E_F(X)\to E_H(X)\) with those composites. Equation (3.2), tested on every \(a_F^+\), proves naturality in \(X\). The same test proves preservation of identities, composition and addition of natural transformations. We have constructed an additive extension functor

The terminal-object identification in Section 3 gives \(j^*E(F)\simeq F\), naturally in \(F\).

Conversely let \(G: \mathsf C\to\mathsf A\) be additive and right exact, and put \(F=j^*G\). Applying \(G\) to (1.3) gives a specified isomorphism \(E_F(X)\simeq G(X)\). It takes \(q_X\) to \(G(p)\). For a lift (2.2), both its composite with \(a_F^+\) and \(G(a)\), after precomposition by \(G(r)=F(r)\), equal \(G(p)G(s)\). Cancel the epimorphism \(F(r)\). Thus the isomorphism takes every cocone map \(a_F^+\) to \(G(a)\). Equation (3.2) proves naturality in \(X\); the same calculation for a transformation \(G\to G'\) proves naturality in \(G\). Therefore \(Ej^*G\simeq G\) as functors.

These two natural isomorphisms prove Theorem 1.1, including full faithfulness on all natural transformations. In particular, once an extension is equipped with a specified identification of its restriction with \(F\), any two such extensions are related by exactly one isomorphism respecting those identifications. An extension without that specified identification may have automorphisms.

6. Cokernel presentations form a localization

There is also a useful fraction description of the construction. Let \(\mathsf D=\operatorname{Mor}(\mathsf C)\). Its objects are arrows \(u: Y\to X\), and a morphism is a commutative square. Kernels and cokernels of squares are computed at their two vertices: the induced map between the two kernels or two cokernels exists by the square identity, and its universal property follows at both vertices. The coimage–image comparison is consequently an isomorphism at both vertices. Thus \(\mathsf D\) is abelian.

Let \(K: \mathsf D\to\mathsf C\) send an arrow to its cokernel. Let \(\chi_{u,v}\) be the map from \(\operatorname{Hom}_{\mathsf D}(u,v)\) to \(\operatorname{Hom}_{\mathsf C}(K(u),K(v))\) induced by taking cokernels. Form \(\mathsf D'\) with the same objects and

Composition of representatives induces composition of their cokernel maps, so this is a category. Each image in (6.1) is a subgroup, and composition is bilinear. The arrow \(u\oplus v\), with its ordinary injection and projection squares, is a biproduct: maps into or out of it decompose into the two component squares. Hence \(\mathsf D'\) is additive. The factorization \(K: \mathsf D\to\mathsf D'\xrightarrow{Q}\mathsf C\) has \(Q\) faithful and additive.

Put \(S=Q^{-1}(\text{isomorphisms})\). We use incoming fractions, whose denominator points into the source.

Lemma 6.1. For \(u,v\in\mathsf D\) and \(f: K(u)\to K(v)\), there are \(w\in\mathsf D\) and squares \(\alpha: w\to u,\beta: w\to v\) such that \(K(\alpha)\) is an isomorphism and \(K(\beta)=fK(\alpha)\).

Proof. Write \(u: Y\to X,v: Y'\to X'\), with cokernel projections \(p: X\twoheadrightarrow U,c: X'\twoheadrightarrow V\). Form

Here \(X\to V\) is \(fp\); the maps \(P\to X,P\to X'\) are \(a,b\); and \(A\to X'\) is its composite through \(P\). The map \(a\) is epi by pullback stability, so \(pa: P\to U\) is epi.

The map \(Y\to\ker p\) is epi. Pulling it back identifies the image of \(A\to P\) with \(\ker(pa)\). Also \(cb\) vanishes on \(A\), since \(cb=fpa\). Thus \(A\to X'\) lands in \(\ker c=\operatorname{im}v\). Pullback stability applied to \(Y'\twoheadrightarrow\operatorname{im}v\) makes \(Z\to A\) epi. Consequently the image of \(w: Z\to P\) is exactly \(\ker(pa)\), and \(pa\) identifies \(K(w)\) with \(U\).

The maps \(Z\to Y,P\to X\) give the square \(\alpha\); the maps \(Z\to Y',P\to X'\) give \(\beta\). The cokernel map of \(\alpha\) is the identification just constructed. The equality \(cb=fpa\), followed by cancellation of the cokernel projection of \(w\), gives \(K(\beta)=fK(\alpha)\). \(\square\)

Proposition 6.2. The family \(S\) admits an incoming calculus of fractions, and the induced functor

is an equivalence.

Proof. Identities, isomorphisms and compositions satisfy the denominator conditions. For \(h: u\to v\) and \(t: v'\to v\) in \(S\), apply Lemma 6.1 to \(Q(t)^{-1}Q(h): Q(u)\to Q(v')\). It gives \(s: w\to u\) in \(S\) and \(g: w\to v'\). Their images satisfy \(Q(tg)=Q(hs)\); faithfulness gives \(tg=hs\), the incoming Ore square. If arrows become equal after composition with a denominator, applying \(Q\) and cancelling its invertible image shows the arrows were already equal. This supplies the cancellation axiom.

A roof \(u\xleftarrow{s}w\xrightarrow{g}v\) maps to \(Q(g)Q(s)^{-1}\). Lemma 6.1 makes every map in \(\mathsf C\) such an image. If two roofs have the same image, the Ore square gives a common refinement of their source denominators. Both refinement maps have invertible \(Q\)-images, hence are in \(S\). On that common source the two numerators have equal \(Q\)-images; faithfulness of \(Q\) makes them equal. Thus the induced Hom map is injective as well as surjective. Every \(X\in\mathsf C\) is the cokernel of \(0\to X\), giving essential surjectivity. This proves (6.3). The resulting Hom sets have the size of those in \(\mathsf C\), even if the roof categories are large. \(\square\)

Let \(\mathsf D_0=\operatorname{Mor}(\mathsf J)\), and let \(\mathsf D'_0\) be the full subcategory of \(\mathsf D'\) on its objects. Fullness of \(\mathsf J\) means that (6.1), for these objects, is already the image of squares in \(\mathsf D_0\).

The subcategory \(\mathsf D_0\subseteq\mathsf D\) is full and additive: every square between its objects has its four endpoints and hence all its arrows in \(\mathsf J\), and biproducts of its arrows are formed by the biproducts in \(\mathsf J\). The same finite biproducts make \(\mathsf D'_0\) additive.

Proposition 6.3. Put \(T=S\cap\operatorname{Mor}(\mathsf D'_0)\). Then \(T\) admits an incoming calculus, and

Proof. For \(u: Y\to X\), choose \(X_0\in\mathsf J\) epimorphic onto \(X\), and \(Y_0\in\mathsf J\) epimorphic onto \(Y\times_XX_0\). The induced \(v: Y_0\to X_0\) has image equal to the kernel of \(X_0\to K(u)\). Hence the square \(v\to u\) induces a cokernel isomorphism and belongs to \(S\).

Apply the full-subcategory replacement theorem, Theorem 2.1 of Descent through replacement objects, to the opposite categories. The input is the full subcategory \((\mathsf D'_0)^{\mathrm{op}}\subseteq(\mathsf D')^{\mathrm{op}}\), the outgoing system \(S^{\mathrm{op}}\), and the denominator \(u\to v\) just supplied in the opposite category for every \(u\). The theorem gives the outgoing calculus on \(T^{\mathrm{op}}\) and an equivalence of those localizations. Taking opposites and then using (6.3) gives (6.4). \(\square\)

For \(F: \mathsf J\to\mathsf A\) as before, define \(F'(u)=\operatorname{coker}F(u)\) on \(\mathsf D_0\), with the induced maps from squares. Right exactness of \(E_F\) and the identification \(E_F|_{\mathsf J}\simeq F\) give, naturally in every square,

Thus \(F'\) factors through \(\mathsf D'_0\), sends \(T\) to isomorphisms, and descends through (6.4). In particular a square whose cokernel map is zero induces the zero map under \(F'\), and one whose cokernel map is invertible induces an isomorphism. This proves the full descent assertion using the independently constructed extension.

An invertible cokernel comparison is not equivalent to exactness of the shorter cone row. Take \(u: 0\to0\), \(v: k\to0\), and the zero square in vector spaces over any field \(k\). Both cokernels are zero, but the proposed row \(0\to k\to0\to0\) fails exactness at \(k\). The extension theorem and (6.5) still hold. Exercise 3 computes precisely the additional kernel condition that the shorter cone row tests.

7. The three dual variants

Call \(\mathsf J\) cogenerating if every \(X\in\mathsf C\) admits a monomorphism \(X\hookrightarrow I\) with \(I\in\mathsf J\). Then one can choose an ambient copresentation

first embed \(X\) into \(I_0\), then embed its cokernel into \(I_1\).

For contravariant functors we mean functors \(\mathsf C^{\mathrm{op}}\to\mathsf A\), with their ordinary natural transformations. The precise variants of (1.2) are:

In each row restriction is an equivalence between the stated additive functors on \(\mathsf C\) and the additive functors on \(\mathsf J\) satisfying exactly the displayed condition. It includes all natural transformations.

Proof. For covariant left exact functors, apply Theorem 1.1 to \(\mathsf C^{\mathrm{op}},\mathsf J^{\mathrm{op}},\mathsf A^{\mathrm{op}}\). Cogeneration in \(\mathsf C\) is generation in \(\mathsf C^{\mathrm{op}}\), and right exactness in the target opposite is left exactness in \(\mathsf A\). Passing to a target opposite reverses natural transformations; take the opposites of both resulting functor categories to obtain the claimed equivalence in the original direction.

For contravariant left exact functors, regard \(F: \mathsf J^{\mathrm{op}}\to\mathsf A\) as a covariant functor \(\mathsf J\to\mathsf A^{\mathrm{op}}\). Apply Theorem 1.1 with source \(\mathsf C\) and target \(\mathsf A^{\mathrm{op}}\), then take the opposites of both functor categories to restore the transformation direction.

For contravariant right exact functors, apply Theorem 1.1 directly to source \(\mathsf C^{\mathrm{op}}\), subcategory \(\mathsf J^{\mathrm{op}}\), and target \(\mathsf A\). This time there is no target-opposite reversal of transformations. Each exactness condition is the corresponding right exact ambient triple after these operations. \(\square\)

The corresponding formulas are concrete. A covariant left exact extension uses (7.1) to give \(\ker(F(I_0)\to F(I_1))\). A contravariant left exact extension uses (1.3) to give \(\ker(F(P_0)\to F(P_1))\). A contravariant right exact extension uses (7.1) to give \(\operatorname{coker}(F(I_1)\to F(I_0))\). The dualized universal properties specify their identifications and prove independence of every choice.

8. Graded exercises with complete solutions

Exercise 1 (foundation: generators that are not projective). Let \(R=k[\epsilon]/(\epsilon^2)\), where \(k\) is any field. Let \(\mathsf C\) be finite-dimensional left \(R\)-modules and \(\mathsf J\) its full subcategory of modules of even \(k\)-dimension. Prove that \(\mathsf J\) is additive and generating, that it is not closed under kernels, and that it contains a nonprojective object. For any \(F\) satisfying (1.1), compute \(E_F(k)\) using \(R\). Give the concrete answer for \(F(M)=k\otimes_RM\).

Solution. Kernels and cokernels of maps between finite-dimensional \(R\)-modules remain finite-dimensional, so \(\mathsf C\) is abelian. The zero module and finite sums of even-dimensional modules are even-dimensional; full inclusion preserves their biproducts. For any \(M\), the projection \(M\oplus M\twoheadrightarrow M\) is an epimorphism from an object of \(\mathsf J\). Thus \(\mathsf J\) is generating.

Multiplication by \(\epsilon\) on \(R\) has kernel \(\epsilon R\simeq k\). Its two ambient endpoints have dimension two, but its kernel has dimension one, so \(\mathsf J\) is not closed under kernels.

The object \(k^2\) lies in \(\mathsf J\). The quotient \(\pi: R^2\twoheadrightarrow k^2\) has no \(R\)-linear section. Indeed every map \(s: k^2\to R^2\) has image annihilated by \(\epsilon\), hence contained in \(\epsilon R^2\), and \(\pi s=0\). It cannot lift the identity of \(k^2\) through \(\pi\). This proves that \(k^2\) is nonprojective.

The ambient presentation \(R\xrightarrow{\epsilon}R\to k\to0\) gives

For the stated tensor functor, \(F(M)=M/\epsilon M\). If \(M\to N\to L\to0\) is right exact, then \(L=N/\operatorname{im}(M\to N)\), so

This is the cokernel of \(M/\epsilon M\to N/\epsilon N\). Thus \(F\) satisfies (1.1). We have \(F(R)=k\), and \(F(\epsilon)=0\), giving \(E_F(k)=k\). Its extension is \(M\mapsto M/\epsilon M\), by Theorem 1.1. The example checks the theorem beyond projective generators.

Exercise 2 (intermediate: split exactness is insufficient). Use the same \(\mathsf C,\mathsf J\), and let \(F(J)=\operatorname{Hom}_R(k,J)\), valued in finite-dimensional \(k\)-vector spaces. Show that \(F\) is additive and preserves all split exact rows in \(\mathsf J\). Exhibit an ambient row with all terms in \(\mathsf J\) that violates (1.1), and deduce that \(F\) has no right exact extension to \(\mathsf C\).

Solution. A map \(k\to J\) is determined by an element annihilated by \(\epsilon\). Hence \(F(J)=\ker(\epsilon: J\to J)\), and its action on morphisms is restriction. This gives additivity and finite-biproduct preservation. Applying any additive functor to a splitting preserves the equations for its section and retraction, so split exact rows stay split exact.

Consider the exact ambient row

All three terms have dimension two or four and lie in \(\mathsf J\). But \(F(R^2)=\epsilon R^2\), while \(F(k^2)=k^2\). The induced map \(F(\pi)\) is zero and its target is nonzero. The image row therefore fails the epimorphism requirement at its final term.

If a right exact extension existed with a specified restriction isomorphism, applying it to this ambient row would force the image under \(F\) to be right exact, since the three components of that isomorphism identify the rows. This contradicts the calculation. Thus the full ambient condition in (1.1) is necessary.

Exercise 3 (advanced: what the shorter cone row really tests). Let \(R\) be any ring and consider a commutative square of left \(R\)-modules with \(u: Y\to X\), \(v: Y'\to X'\), \(a: Y\to Y'\), \(b: X\to X'\), and \(bu=va\). Write \(\phi: \operatorname{coker}u\to\operatorname{coker}v\) for the induced map. Set

Put \(T=\operatorname{coker}(\ker u\xrightarrow{a}\ker v)\). Prove \(d_2d_1=0\) and the exact sequence

Show that \(d_2\) is epi exactly when \(\phi\) is epi. Deduce the precise condition for \(Y\to X\oplus Y'\to X'\to0\) to be right exact, and recover the counterexample in Section 6.

Solution. The composition is \(bu-va=0\). Define \(\theta: \ker d_2\to\ker\phi\) by \((x,y')\mapsto[x]\). It lands in \(\ker\phi\), since \(b(x)=-v(y')\), and kills \(\operatorname{im}d_1\). It therefore induces \(H\to\ker\phi\). It is onto: if \([x]\in\ker\phi\), choose \(z\in Y'\) with \(b(x)=v(z)\); then \((x,-z)\in\ker d_2\) represents a preimage.

The map \(\ker v\to H\) sends \(y'\) to the class of \((0,y')\). Its kernel is \(a(\ker u)\): the class is zero precisely when \((0,y')=(u(y),-a(y))\) for some \(y\), which means \(u(y)=0,y'=-a(y)\). The minus sign gives the same image submodule. It consequently induces an injection of the cokernel on the left of (8.1).

If \([(x,y')]\) maps to zero in \(\ker\phi\), then \(x=u(y)\) for some \(y\). Subtracting \(d_1(y)\) gives the representative \((0,y'+a(y))\); the second component lies in \(\ker v\). Thus the image of the injection is exactly the kernel of \(H\to\ker\phi\), proving (8.1).

Finally

Hence \(d_2\) is epi exactly when \(\phi\) is epi. Exactness at \(X\oplus Y'\) means \(H=0\). By (8.1), this is equivalent to \(\ker\phi=0\) and surjectivity of \(\ker u\to\ker v\). The shorter row is therefore right exact exactly when \(\phi\) is an isomorphism and \(\ker u\to\ker v\) is onto.

For \(u: 0\to0,v: k\to0\), the cokernel comparison is the isomorphism \(0\to0\), but the kernel map \(0\to k\) is not onto. Here \(H=k\), exhibiting the missing relation explicitly. Surjectivity of that extra kernel map is not a hypothesis of Theorem 1.1, and its proof does not require it.

Exercise 4 (expert: postcomposition and its coherence). Let \(F\in\operatorname{Rex}_{\mathsf C}(\mathsf J,\mathsf A)\), and let \(K: \mathsf A\to\mathsf B\) be additive and right exact between abelian categories. Prove that \(KF\) satisfies the relative condition and that there is a unique restriction-compatible isomorphism

Prove naturality in \(F\) and \(K\), and compatibility with two successive right exact postcompositions. Show why a merely left exact \(K\) is insufficient, even when \(KF\) still satisfies the relative condition.

Solution. The image under \(F\) of an ambient row (1.1) is right exact in \(\mathsf A\); applying \(K\) leaves it right exact. Thus \(KF\) satisfies the relative condition. Also the composite \(KE_F\) is additive and preserves cokernels, so is right exact. Its restriction is specified by \(E_F|_{\mathsf J}\simeq F\), giving \(KE_F|_{\mathsf J}\simeq KF\). Full faithfulness in Theorem 1.1 gives the unique isomorphism (8.2) respecting that identification.

For (1.3), the same isomorphism is the cokernel comparison

since \(K\) preserves that cokernel, including its quotient map. Naturality for a transformation \(F\to H\) follows either by comparing these quotient maps or by restriction: the two resulting transformations between right exact extensions have the same restriction, so full faithfulness makes them equal. For a transformation \(K\to L\) between right exact postcomposition functors, its naturality on \(F(J)\) gives the same equality on \(\mathsf J\), and again full faithfulness proves it on \(\mathsf C\).

For \(K_1: \mathsf A\to\mathsf B\), \(K_2: \mathsf B\to\mathsf D\), the direct comparison \(E_{K_2K_1F}\to K_2K_1E_F\) and the composite of the two successive comparisons both respect the specified restriction to \(K_2K_1F\). Uniqueness makes them equal. This also proves compatibility with further postcompositions by induction, and with the identity postcomposition. The coherence is fixed by the restriction, rather than by independent choices of presentations.

For the failure, take \(\mathsf C=\mathsf A=\mathsf B\) to be finitely generated abelian groups and \(\mathsf J\) the finite-rank free groups. This is a full additive generating subcategory, since every finitely generated group is a quotient of some \(\mathbb Z^n\). Let \(F\) be inclusion, whose right exact extension is the identity. Set \(K(M)=\operatorname{Hom}_{\mathbb Z}(\mathbb Z/2,M)\). This is a finitely generated subgroup of \(M\), and is additive and left exact: its kernel comparison follows directly by factoring a homomorphism through an ordinary kernel.

On every finite free group \(K\) is zero, so \(KF=0\) satisfies the relative condition and extends to the zero functor. However \(KE_F=K\) has value \(\mathbb Z/2\) at \(\mathbb Z/2\), so (8.2) fails. In the right exact row \(\mathbb Z\xrightarrow{2}\mathbb Z\to\mathbb Z/2\to0\), its image has final map \(0\to\mathbb Z/2\), which is not epi. This computes the precise failed hypothesis.

9. References and scope

The theorem here allows an arbitrary full additive generating subcategory, an arbitrary abelian target and all natural transformations. Sections 2–5 give an epimorphic-cover proof; Section 6 gives the presentation-localization interface and the counterexample to the shorter cone-row criterion.

The basic exact-functor and pullback interfaces are Stacks 010N and Stacks 05PK. The full replacement interface is the opposite of Theorem 2.1 in Descent through replacement objects, with its retained fraction construction and natural-transformation universal property. Stacks, Section 56.2 treats extensions from finitely presented modules; its module and colimit hypotheses do not replace the generality proved here. References supply mathematical interfaces, and their prose is not reproduced.

This lesson concerns ordinary additive functors and exact sequences. Derived functors and other excluded constructions are outside the scope of this lesson. The four exercises and all their solutions are original to this lesson.