Chapter 8 identifies the objects of our theory with module categories. This chapter identifies the permitted functors with bimodules and their natural transformations with bimodule morphisms. Rings remain unital and need not be commutative; modules remain left modules.
9.1. Which functors?
Definition 9.1. The permitted functors
Fix rings \(R,S\). A functor \(F:R\text{-Mod}\to S\text{-Mod}\) is additive if its maps on Hom groups preserve addition. It is right exact if it preserves cokernels. Equivalently, whenever
\[X\xrightarrow uY\xrightarrow qZ\longrightarrow0\]is exact, so is the image diagram under \(F\). For additive functors these formulations agree: in modules such an exact diagram means precisely that \(q\) is a cokernel of \(u\), as proved in Chapter 5. We require also preservation of every set-indexed coproduct, including the empty coproduct.
Write \(\mathscr F(R,S)\) for the permitted functors and their natural transformations, and \(\mathscr B(S,R)\) for \(S\)-\(R\) bimodules and their bimodule morphisms. The operations on transformations are componentwise identities and composition.
There is a size convention in \(\mathscr F\): its objects may themselves be class functors, so this notation is a scheme of categorical assertions for fixed functors, rather than a claim that proper classes are elements of a set. Theorem 9.5 will encode each transformation by the set-sized map \(\eta_R\). Thus the Hom collections in this scheme are sets. No category of all categories or of all classes is being assumed.
Worked example for Definition 9.1
Counterexample: Dropping right exactness destroys reconstruction from F(R)
Counterexample: An infinite product functor misses the coproduct condition
These are exactly the additive functors preserving all small colimits. To see this, a colimit of a small diagram \(D\) is the cokernel of the map
\[\bigoplus_{\alpha:i\to j}D(i)\longrightarrow\bigoplus_iD(i),\qquad \iota_\alpha x\longmapsto\iota_i x-\iota_j D(\alpha)(x).\]A map from the second sum annihilates this map exactly when its components form a cocone; the cokernel universal property is therefore the colimit universal property. Additivity, coproduct preservation, and cokernel preservation carry this construction to the same construction for \(FD\). Conversely a cokernel is the coequalizer of \(u,0\), and a coproduct is a colimit, so colimit preservation implies both requirements. This formula also proves module categories have all small colimits.
Alphabeta Math: the cocontinuity criterion
For every \(S\)-\(R\) bimodule \(B\), Chapter 6 proves that \(T_B(M)=B\otimes_R M\), \(T_B(u)=1_B\otimes u\), is additive, right exact, and coproduct-preserving. The task is the converse.
9.2. Extracting the bimodule
Lemma 9.2. The two actions on F(R)
For an additive functor \(F\), the left \(S\)-module \(B=F({}_R R)\) becomes an \(S\)-\(R\) bimodule by
\[b r=F(\rho_r)(b),\qquad \rho_r:R\to R,\quad \rho_r(x)=xr.\]Here \(\rho_r\) is a left \(R\)-module morphism, even when \(R\) is noncommutative.
Proof.
Distributivity and associativity in \(R\) give \(\rho_r(x+y)=\rho_r(x)+\rho_r(y)\) and \(\rho_r(ax)=a\rho_r(x)\), so \(\rho_r\) is indeed \(R\)-linear. Also
\[\rho_{r+r'}=\rho_r+\rho_{r'},\qquad \rho_1=1_R,\qquad \rho_{rr'}=\rho_{r'}\circ\rho_r.\]Additivity and functoriality give \(b(r+r')=br+br'\), \(b1=b\), and \((br)r'=b(rr')\). Since \(F(\rho_r)\) is an \(S\)-linear module map, \((b+b')r=br+b'r\) and \((sb)r=s(br)\). These are every right-module axiom and the commuting-action condition.
□9.3. A comparison defined before presentations
Lemma 9.3. The canonical natural transformation
For \(m\in M\), let \(\lambda_m:R\to M\) be the left-module map \(\lambda_m(x)=xm\). There is an \(S\)-linear natural transformation
\[\tau^F_M:B\otimes_R M\longrightarrow F(M),\qquad \tau^F_M(b\otimes m)=F(\lambda_m)(b).\]Counterexample: An infinite product functor misses the coproduct condition
Proof.
For fixed \(m\), \(F(\lambda_m)\) is an additive \(S\)-linear map, giving additivity in \(b\) and the equation \(\tau^F_M(sb\otimes m)=s\tau^F_M(b\otimes m)\). For fixed \(b\), the identity \(\lambda_{m+m'}=\lambda_m+\lambda_{m'}\) and additivity of \(F\) give additivity in \(m\). The balancing relation follows from
\[\lambda_{rm}=\lambda_m\circ\rho_r,\] \[F(\lambda_m)(br)=F(\lambda_m)F(\rho_r)(b)=F(\lambda_{rm})(b).\]The tensor universal property therefore gives a unique additive map \(\tau^F_M\), and the displayed \(S\)-linearity on generators makes it \(S\)-linear on all tensors.
For \(u:M\to N\), \(u\lambda_m=\lambda_{u(m)}\). Consequently
\[F(u)\tau^F_M(b\otimes m)=F(\lambda_{u(m)})(b)=\tau^F_N(b\otimes u(m)).\]Pure tensors generate, so this proves \(F(u)\tau^F_M=\tau^F_N(1_B\otimes u)\). This is naturality; no presentation has entered the definition.
□Theorem 9.4. Eilenberg–Watts
Every additive, right-exact, coproduct-preserving functor \(F:R\text{-Mod}\to S\text{-Mod}\) is naturally isomorphic to \(F(R)\otimes_R-\), through \(\tau^F\). Every \(S\)-\(R\) bimodule gives such a functor.
Worked example for Theorem 9.4
Counterexample: Dropping right exactness destroys reconstruction from F(R)
Counterexample: An infinite product functor misses the coproduct condition
Proof.
The regular module. For \(r\in R\), \(\lambda_r=\rho_r\), so \(\tau^F_R(b\otimes r)=br\). This is the unit isomorphism of Chapter 6, with inverse \(b\mapsto b\otimes1\).
Free modules. Let \(Q=R^{(I)}\) with injections \(\iota_i:R\to Q\). Both \(T_B(Q)\), with the maps \(1_B\otimes\iota_i\), and \(F(Q)\), with the maps \(F(\iota_i)\), are coproducts of copies of \(B\), after the preceding unit identification. Naturality says \(\tau^F_Q\) agrees with that identification on each copy. The coproduct universal property supplies its inverse and verifies both composites are identities. This includes \(I=\varnothing\).
All modules. Chapter 5 gives a free presentation
\[R^{(J)}\xrightarrow dR^{(I)}\xrightarrow qM\longrightarrow0.\]Right exactness of both functors makes \(T_B(q)\) and \(F(q)\) the cokernels of \(T_B(d)\) and \(F(d)\), respectively. Naturality gives a commuting square between the two maps \(d\), whose vertical maps are the free-module isomorphisms just proved. A commuting square of isomorphisms induces an isomorphism of cokernels: use its inverses to construct the reverse map, then cancel the epic cokernel maps to check the two composites. The induced map here is \(\tau^F_M\), by naturality at \(q\) and uniqueness of cokernel factorization. Therefore every component is invertible. The componentwise inverses are natural: rearrange \(F(u)\tau^F_M=\tau^F_NT_B(u)\) to get the inverse naturality equation. The converse was proved in Chapter 6.
□9.4. Transformations are bimodule maps
Theorem 9.5. Evaluation at R classifies every transformation
For permitted functors \(F,G\), the assignment \(\eta\mapsto\eta_R\) is a bijection
\[\operatorname{Nat}(F,G)\cong\operatorname{Hom}_{S\text{-}R}(F(R),G(R)).\]In particular, for bimodules \(B,D\), every transformation \(T_B\Rightarrow T_D\) is uniquely \(\beta\otimes1\) for an \(S\)-\(R\) bimodule morphism \(\beta:B\to D\).
Proof.
If \(\eta:F\Rightarrow G\) is natural, then \(\eta_R\) is \(S\)-linear because it is a morphism in \(S\text{-Mod}\). Naturality at \(\rho_r\) gives
\[\eta_R(br)=\eta_RF(\rho_r)(b)=G(\rho_r)\eta_R(b)=\eta_R(b)r.\]Thus \(\beta=\eta_R\) is a bimodule morphism. Naturality at \(\lambda_m\) gives
\[\eta_M\tau^F_M(b\otimes m)=\eta_MF(\lambda_m)(b)=G(\lambda_m)\beta(b)=\tau^G_M(\beta(b)\otimes m).\]Pure tensors generate and \(\tau^F_M\) is invertible, so necessarily
\[\eta_M=\tau^G_M\circ(\beta\otimes1_M)\circ(\tau^F_M)^{-1}.\]Conversely, for any bimodule morphism \(\beta:F(R)\to G(R)\), Chapter 6 gives an \(S\)-linear map \(\beta\otimes1_M\) for each \(M\). On pure tensors it commutes with \(1\otimes u\); hence these maps form a natural transformation. Conjugating by the two natural isomorphisms \(\tau\) gives a natural \(\eta:F\Rightarrow G\) through the displayed formula. At \(R\), the unit identifications \(b\otimes1\mapsto b\) show \(\eta_R=\beta\). This proves both surjectivity and injectivity.
For \(F=T_B\), evaluation at \(R\) identifies \(T_B(R)=B\otimes_RR\) with \(B\) as a bimodule, because the right action is induced by \(1_B\otimes\rho_r\). Indeed \(b\otimes x\) maps to \(bx\), and \(b\otimes xr\) maps to \((bx)r\). Under this identification naturality at \(\lambda_m\) reads directly \(\eta_M(b\otimes m)=\beta(b)\otimes m\). Identity transformations correspond to identity maps; vertical composites correspond to composites by taking their components at \(R\). An invertible transformation therefore corresponds exactly to an invertible bimodule map, with the inverse recovered in the same way.
□Corollary 9.6. The equivalence of Hom categories
The assignment
\[\mathscr B(S,R)\longrightarrow\mathscr F(R,S),\qquad B\longmapsto T_B,\qquad \beta\longmapsto\beta\otimes1,\]is fully faithful and essentially surjective. It has the explicit inverse assignment \(F\mapsto F(R)\), \(\eta\mapsto\eta_R\), with comparisons \(T_{F(R)}\cong F\) and \(T_B(R)\cong B\). Thus it is an equivalence in the componentwise size convention of Definition 9.1.
Proof.
Functoriality of the tensor assignment follows from \((\gamma\beta)\otimes1=(\gamma\otimes1)(\beta\otimes1)\) on pure generators. Functoriality of evaluation is componentwise. Full faithfulness is Theorem 9.5, and essential surjectivity is Theorem 9.4. The comparison \(\tau^F\) is natural also in \(F\) by the equation in Theorem 9.5. The comparison \(T_B(R)\to B\) is natural in \(B\), since it sends \(\beta(b)\otimes r\) to \(\beta(b)r=\beta(br)\). These are the two inverse-functor comparisons.
□Example 9.7. Reduction modulo 2 and its transformations
Take \(R=\mathbb Z\), \(S=\mathbb F_2\), and \(B=\mathbb F_2\), with \(b n\) given by reducing \(n\) modulo \(2\). Then
\[\mathbb F_2\otimes_{\mathbb Z}M\cong M/2M.\]Proposition 6.8 proves this natural, \(\mathbb F_2\)-linear isomorphism, including the quotient action and both inverse maps. Thus this example satisfies every hypothesis of Theorem 9.4 by Theorems 6.6 and 6.7.
A bimodule endomorphism of \(\mathbb F_2\) is multiplication by its value at \(1\), hence either zero or identity; either map commutes with integers. Theorem 9.5 says these are also the only natural endomorphisms of the functor \(M\mapsto M/2M\).
Sources and further reading
- Alphabeta Math, the Eilenberg–Watts theorem for arbitrary unital rings, including the arbitrary-ring theorem, classification of transformations, and equivalence of Hom categories.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Reconstructing reduction modulo three
Applies to: Definition 9.1, Lemma 9.2, Lemma 9.3, Theorem 9.4.
Take \(R=\mathbb Z,S=\mathbb F_3\), \(F(M)=M/3M\). It is additive on maps. Direct sums commute with this quotient by finite support, including the empty sum. A cokernel \(N/\operatorname{im}f\) becomes \(N/(\operatorname{im}f+3N)\), which is also the cokernel after reduction, so \(F\) is right exact. Thus it is a permitted functor. Its value \(B=F(\mathbb Z)=\mathbb F_3\) has right action \(\bar b\cdot n=\overline{bn}\), obtained by reducing \(\rho_n\).
For \(m\in M\), reducing \(\lambda_m(n)=nm\) gives \(\tau_M(\bar b\otimes m)=bm+3M\). Its inverse is \(m+3M\mapsto\bar1\otimes m\); changing \(m\) by \(3a\) changes this tensor by zero. Balancing checks both composites. For a homomorphism \(u:M\to N\), both naturality paths give \(bu(m)+3N\). At \(M=\mathbb Z/6\), both sides are \(\mathbb F_3\), with \(\bar1\otimes\bar2\) mapped to \(\bar2+3M\). This verifies the canonical rather than an unspecified Eilenberg–Watts isomorphism.
Worked example 2. A transformation between two different quotients
Applies to: Theorem 9.5, Corollary 9.6.
View \(F(M)=M/6M\) and \(G(M)=M/3M\) as functors to abelian groups; the quotient argument above makes both permitted. A \(\mathbb Z\)-\(\mathbb Z\) bimodule map \(\mathbb Z/6\to\mathbb Z/3\) is determined by the image \(a\in\mathbb Z/3\) of \(\bar1\). All three choices are allowed because \(6a=0\). They give exactly the transformations \(\eta^a_M(m+6M)=am+3M\). These are well defined since \(6M\subset3M\), and natural since homomorphisms preserve integer multiplication. Any natural transformation has this form by testing \(\mathbb Z\to M,1\mapsto m\). Thus evaluation at the regular module is a bijection on transformations. The identifications \((\mathbb Z/n)\otimes M\cong M/nM\) and \((\mathbb Z/n)\otimes\mathbb Z\cong\mathbb Z/n\) recover both objects and these arrows under the two inverse assignments, illustrating the Hom-category equivalence.
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. Dropping right exactness destroys reconstruction from F(R)
Related statements: Definition 9.1, Theorem 9.4.
Claim refuted. Additivity and preservation of coproducts alone force a functor to be tensoring with its value on the ring.
Proof.
Let \(R=S=\mathbb Z\) and \(F(M)=\operatorname{Hom}_{\mathbb Z}(\mathbb Z/2,M)\), with postcomposition on arrows. This is the subgroup of elements \(m\in M\) with \(2m=0\), via evaluation at \(\bar1\). Addition of maps shows \(F\) is additive. An element of a direct sum is killed by 2 exactly when each of its finitely many nonzero entries is, so \(F\) preserves arbitrary coproducts, including zero. Yet \(F(\mathbb Z)=0\) and \(F(\mathbb Z/2)\cong\mathbb Z/2\). Thus \(F(\mathbb Z)\otimes M=0\) cannot recover \(F\).
The failed hypothesis is right exactness: the cokernel presentation \(\mathbb Z\xrightarrow2\mathbb Z\to\mathbb Z/2\to0\) is sent to \(0\to0\to\mathbb Z/2\to0\), whose penultimate arrow is not onto. Every construction and failure is within the stated module categories.
Counterexample 2. An infinite product functor misses the coproduct condition
Related statements: Definition 9.1, Lemma 9.3, Theorem 9.4.
Claim refuted. An additive exact functor is automatically a tensor functor, without the coproduct hypothesis.
Proof.
On abelian groups let \(F(M)=\prod_{n\ge1}M\), with maps applied coordinatewise. It is additive. It is exact: a coordinatewise kernel is a product of kernels, and a tuple in the product of images has a tuple of preimages, choosing one per coordinate. In particular a product of surjections is onto, and image equals kernel in a product of short exact sequences.
Take \(M_i=\mathbb Z\) for every positive integer \(i\). In \(F(\bigoplus_iM_i)\), the tuple whose \(n\)-th row is the basis vector \(e_n\) is legitimate: every row has finite support. In the image of \(\bigoplus_iF(M_i)\), however, only finitely many columns can be nonzero, because an input has only finitely many summand components. The diagonal tuple has infinitely many nonzero columns, so the comparison is not surjective. The canonical tensor map \(F(\mathbb Z)\otimes\bigoplus_i\mathbb Z\to F(\bigoplus_i\mathbb Z)\) also misses it: a tensor is a finite sum of \(b_j\otimes m_j\), and the finite union of the supports of the \(m_j\) contains every nonzero column of its image. Thus the natural transformation of Lemma 9.3 need not be invertible when coproduct preservation is removed.
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