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I would like to understand the relationship between the derived category definition of a right derived functor Rf (which involves an initial natural transformation n: Qf → (Rf)Q, where Q is the map to the derived category) and the "universal delta functor" definition given in Hartshorne III.1.

I already know that R^if(A) = H^i(Rf(A)). What I want to know most is:

What is the role of the natural transformation n in this comparison?

I guess it can be thought of as a natural map from a injective resolution of f(A) to f(an injective resolution of A), but I'm not sure what is the significance of this... Does anyone know a good reference explaining such things?

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2 Answers 2

up vote 5 down vote accepted

I haven't checked all the details, but I think the story could go like this. (I have to apologize: it's a bit long.)

(1) Let F: A ---> B be an additive left exact functor between two abelian categories. Take an injective resolution of an object A in A :

0---> A ---> I^0 ---> I^1 ---> ...

Let us call i: A ---> I^0 the first morphism. Apply F to this exact sequence:

0---> FA ---> FI^0 ---> FI^1 ---> ...

Now, the total right derived functor of F applied to A (thought as a complex concentrated in degree zero) is the complex

*R*F(A) = { FI^0 ---> FI^1 ---> FI^2 ---> ... }

and the classical right derived functors of F are its cohomology:

R^nF(A) = H^n(*R*F(A)) = H^n(FI^*) .

These {R^nF}_n are a universal cohomological delta-functor and we have a natural transformation of functors

qF ===> (*R*F)q

which is essentially

Fi: FA ---> *R*F(A)

(here we have extended F degree-wise to the category of complexes, and this is the degree zero of the natural transformation, because *R*F(A)^0 = FI^0 ).

(2) Now, let { T^n : A ---> B } be a cohomological delta-functor and f^0 : F ---> T^0 a natural transformation. We have to extend this f^0 to a unique morphism of delta-functors { f^n : R^nF ---> T^n } .

To do this, observe that, in general, given two right-derivable functors between two, say, model categories

F, G: C ---> D ,

and a natural transformation between them

t: F ===> G

we have a natural transformation between the total right derived functors

*R*t : *R*F ===> *R*G

because of the universal property of the derived functors: indeed, if

f : qF ===> (*R*F)q and g : qG ===> (*R*G)q

are the universal morphisms of the derived functors, then we have a natural transformation

gt : F ===> (*R*G)q

and, so, because of the universal property of derived functors, a unique natural transformation


such that (*R*t)qf = g .

(3) So, take our f^0 : F ---> T^0 , extend it to a natural transformation between the degree-wise induced functors between complexes. Passing to the derived functors, we obtain

*R*f^0 : *R*F ===> *R*T^0 .

Taking cohomology, for each n , we get

H^n(*R*f^0) : H^n (*R*F) ===> H^n (*R*T^0) .

But these are the classical right derived functors, so we have natural transformations

R^nf : R^n F ===> R^nT^0

and because the classical right derived functors are universal delta-functors, we have unique natural transformations

i^n : R^nT^0 ===> T^n

which extend the identity

i^0 : R^0T^0 = T^0 .

The composition

i^n R^f : R^F ===> T^n

is, I think, the required morphisms of delta-functors that we need.

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Thanks! Parts (2) and (3) were the ones causing me the most trouble :) –  Andrew Critch Oct 28 '09 at 15:28
Thanks. In fact, looking at what I have written, I see a possible and right criticism. In order to prove that R^nF is a universal delta-functor, from the universal property of the total derived functor RF, I use the fact that R^nT^0 is already a universal delta-functor. So it is not quite a "universal delta-functor" free argument. :-( (But at least, I don't use that R^nF is a universal delta-functor in order to prove that it is a universal delta-functor!) –  a.r. Oct 28 '09 at 16:35

I don't have a complete answer, but maybe this is helpful: Unpacking the definition of "universal", a universal delta functor whose 0th functor is f is the same thing as an initial object in the category {delta functors T together with a natural transformation f → T^0} (provided, I guess, that the former object exists). Giving your n : Qf → (Rf)Q is the same as giving f → H^0 ∘ Rf ∘ Q, which looks rather similar.

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