Let R be a commutative ring. {Footnote: If, instead, R is a ring with anti-involution (such as a group-ring) all that is below appears to still hold.} Let R-fpmod be the category of finitely presented R-modules and let *F* be the full category of finitely presented covariant abelian-group-valued functors on R-fpmod. A functor is in *F* iff it appears in an exact sequence H^A --> H^B --> T --> O where the first two objects are representable functors. It is an easy exercise to see that *F* is an exact subcategory of the ambient functor category, hence an abelian category. For any f.p.module A we will use A@ to denote the functor that carries B to A@B (using, here, @ for $\otimes$). If we choose a finite-rank free resolution F --> G --> A --> O we obtain exact F@ --> G@ --> A@ --> O. Since F@ = H^F, G@ = H^G and *F* is closed under cokernel formation, we know that A@ is in *F*. (I'll continue to denote isomorphisms here with equality signs.) Let I denote the forgetful functor from R-fpmod to the category of abelian groups. It has two other notations, to wit, H^R and R@. I pointed out in my last note that *F* is the free abelian category generated by R. (That is, given any abelian category with an object and an action of R thereon there is an exact functor from *F* that carries I to the given object, unique up to natural equivalence.) But I'm not going to use that fact here. (I did use it to learn all this stuff, though.). NOTATION: Every group-valued functor from a category of R-modules, commutative R, can be canonically lifted to a module-valued functor. Given two such functors, S and T, we follow the CS tradition of denoting their composition, "first apply S then T" as S;T (hence (S;T)(A) = T(S(A)). For fixed S the functor S;T is exact in the second variable. (That property, together with S:I = S, characterizes this bifunctor; we don't actually need to define it in terms of composition.) We will need: (H^A);(H^B) = H^{A@B} and (A@);(B@) = (A@B)@ DEFINE: [S,T] is the functor such that sends A to the group of natural transformations Hom((H^A;S),T) Note: [S,T] carries right exact sequences in the first variable to left exact sequences, and preserves left-exactness in the second variable. We will need the formula [H^A,T] = (A@);T verifiable by evaluating on an arbitrary B: [H^A,T](B) = Hom((H^B;H^A),T) = Hom(H^{A@B},T) = T(A@B) = ((A@);T)(B) A couple of things implicit in the definition of [S,T] should be explicated. First: LEMMA: If S and T are in *F* then so is S;T. BECAUSE: If T is in *F* let H^A --> H^B --> T --> 0 be exact. We obtain an exact sequence S;H^A --> S;H^B --> S;T --> O Since *F* is closed under cokernel formation it suffices to show that S;H^A is in *F* for any f.p. R-module A. Let F and G be finite-rank free modules and F --> G --> A --> O exact. Then so is O --> S;H^A --> S;H^G --> S;H^F Since *F* is closed under kernel formation it suffices to show that S;H^F is in *F* for any finite-rank free module F. But S;H^F is just a finite direct sum of copies of S and *F* is, of course, closed under direct sums. [] Second explication: LEMMA: If S and T are in *F* then so is [S,T]. BECAUSE: If S is in *F* let H^A --> H^B --> S --> 0 be exact. We obtain an exact sequence O --> [S,T] --> [H^B,T] --> [H^A,T] which we know can be rewritten as O --> [S,T] --> (B@);T --> (A@);T Since *F* is closed under kernel formation and composition we're done. [] DEFINE a *-FUNCTOR, S* as [S,I]. LEMMA: The duality functor is exact BECAUSE: If S --> T --> U is exact then H^A;S --> H^A;T --> H^A;U is exact for every A. We will obtain, therefore, the exactness of U*(A) --> S*(A) --> T*(A) if we know: LEMMA: The object I is injective in *F*. BECAUSE: Being a representable functor, I is, of course, projective in the ambient functor category, hence in *F*. Let O | I | H^A --> H^B --> T --> O be exact (all vertical arrows point down). We seek a retraction for I --> T. Since I is projective we can choose a map I --> H^B to yield a commutative triangle. The fact that I --> T is monic says that O ----> I | | H^A --> H^B is a pullback. It is, therefore, the Yoneda-image of a pushout square: B --> A | | R --> O Let O --> K --> B --> A. It is an exercise in abelian categories that K --> B --> R is epi. (The dual exercise is easy in the category of abelian groups -- which, of course, suffices.) We may choose a retraction R --> K. The map it induces, H^B --> H^K --> I, is such that H^A --> H^B --> H^K --> I is a zero-map and we obtain a factorization H^B --> I = H^B --> T --> I. The map T --> I is what we seek. [] {Footnote: The object I need not be injective in the ambient functor category. In the case R = Z let T be the functor that assigns to a group its torsion subgroup. If I were injective then any endo- transformation on T would extend to an endo-transformation on I. But there are uncountably many endo-transformations on T and only countably many on I (the ring of endo-transformations on T is the "n-adic" completion of the integers -- it may be constructed as the product of all the p-adic completions.)} THEOREM: The *-functor on is a duality on *F*. WE NEED TO SHOW that S** = S. First note that I* = I, hence, of course, I** = I. Second, for any finite-rank free module, F, (H^F)** = H^F, just because the *-functor is additive. Third, for any f.p. module A let F and G be finite rank free modules and F --> G --> A --> O exact. Then O --> H^A --> H^G --> H^F is exact and is carried to exact O --> (H^A)** --> H^G --> H^F, hence (H^A)** = H^A. Finally, for any T in *F*, choose H^A --> H^B --> T --> O exact. The **-functor carries it to exact H^A --> H^B --> T** --> O Hence T** = T. [] Note that (H^A)* = [H^A,I] = (A@);I = A@. Hence (A@)* = H^A. SLOGAN: H^A and A@ are dual. It follows that A@ is injective in *F*. (It's a good exercise to find a direct proof.) And, of course, every object has an injective resolution (if H^A --> H^B --> T*--> O is exact then so is O --> T --> B@ --> A@). Note that an f.p, functor is projective iff it is representable and, dually, it is injective iff it is of the form A@. {Footnote: The category of f.p. abelian-group-valued sheaves on R-fpmod, that is, the category of f.p. abelian-group-valued _contravariant_ functors on R-fpmod need not have injective resolutions. In the case that R = Z any f.p. functor will carry Z to an f.p. abelian group. If E is injective in the category of f.p functors then for any n > 0 the map n:Z --> Z induces monic H_Z --> H_Z hence epic (H_Z,E) --> (H_Z,E). The latter is just multiplication by n on E(Z), hence E(Z) is a divisible group. But the only divisible finitely generated abelian group is O, thus H_Z can not have an injective extension.} DEFINE S@T = [S,T*]*. It follows immediately that this bi-functor on *F* is right-exact in both variables and that it is a commutative with I as unit. Note: H^A @ H^B = H^{A@B} (which together with the right-exactness characterizes @). THEOREM *F* is *-autonomous. BECAUSE: The only thing left to prove is Hom(S@T,U) = Hom(S,[T,U]) Both sides of this equation carry right-exact sequences in the first two positions to left-exact sequences in the category of abelian groups. It therefore suffices to verify the isomorphism when S = H^A and T = H^B: Hom(H^A @ H^B, U) = Hom(H^{A@B},U) = U(A@B) Hom(H^A,[H^B,U]) = Hom(H^A,[H^B,U]) = ]H^B,U](A) = U(A@B) [] COR: [S,T] = [T*,S*] LEMMA: For S and T in *F*, (S;T)* = S*;T*. BECAUSE: Both (S;T)* and S*;T* are exact in T. Because every object in *F* is obtainable using finite limits and co-limits starting with the object I, it therefore suffices to establish that (S;I)* = S*;I* and that's immediate. [] COR: (H^A)@S = [H^A,S*]* = ((A@);S*)* = H^A;S An interesting adjointness arises: PROPOSITION: For S and T in *F* Hom( H^A;S , T ) = Hom( S , A@;T ) BECAUSE: H^A;S = S@H^A and A@;T = [H^A,T]. [] {Footnote: When the compositions are reversed we obtain the routine adjointness: Hom( S;A@ , T ) = Hom( S , T;H^A ).} COR: [S,T](A) = Hom(S, A@;T) S*(A) = Hom(S,A@). {Footnote: The last equation is the one I used at Ottawa to compute S*. If all one knows is that there is a duality on *F* such that (H^A)* = A@ then we may infer S*(A) = Hom(H^A,S*) = Hom(S**,(H^A)*) = Hom(S,A@).} Recall that a coherent ring is one such that all finitely generated ideals are finitely presented. It follows that f.p. modules are closed under finite limits (hence form an exact subcategory). {Footnote: Clearly Noetherian implies coherent. For a quick separating example take the group ring of the rationals (usually called "polynomials with rational exponents"). All finitely generated ideals are principal. The important non-Noetherian examples arise in algebraic geometry.} LEMMA: For f.p. modules over a coherent ring Ext^n(A,-) and Tor_n(A,-) are dual. BECAUSE: Choose a free resolution ...--> F_{n+1} --> F_n -->....--> F_2 --> F_1 --> A --> 0 The Ext^n(A,--) functors are obtainable as the homology of the sequence: 0 --> H^A --> H^{F_1} --> H^{F_2} --> ... The Tor_n(A,--) functors are obtainable as the homology of the sequence: ... --> (F_2)@ --> (F_1)@ --> A@ --> O. These complexes are dual. [] It has not escaped my notice that this *-autonomous structure suggests a possible alternate approach to classical homological algebra. Just one example: a "connecting homomorphism" between half-exact f.p. functors S and T may be identified as an exact sequence of the form O --> S --> E --> P --> T --> O where E is injective and P is projective and this says that the duality works well with the notion of satellites and derived functors.