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(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.
