# Derived functors vs universal delta functors

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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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 $F:\mathsf A\rightarrow \mathsf B$ be an additive left exact functor between two abelian categories. Take an injective resolution of an object $A$ in $\mathsf A$:

$$0\rightarrow A \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots$$

Let us call $i: A \rightarrow I^0$ the first morphism. Apply $F$ to this exact sequence:

$$0\rightarrow FA \rightarrow FI^0 \rightarrow FI^1 \rightarrow \cdots$$

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

$$\mathbb RF(A) = [ FI^0 \rightarrow FI^1 \rightarrow FI^2 \rightarrow \cdots ]$$

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

$R^nF(A) = H^n(\mathbb RF(A)) = H^n(FI^)$.

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

$$qF \Rightarrow (\mathbb RF)q$$

which is essentially

$$Fi: FA \rightarrow \mathbb RF(A)$$

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

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

To do this, observe that, in general, given two right-derivable functors between two, say, model categories $$F,G: \mathsf C \rightarrow \mathsf D$$, and a natural transformation between them $t: F \Rightarrow G$, we have a natural transformation between the total right derived functors $\mathbb Rt : \mathbb RF \Rightarrow \mathbb RG$ because of the universal property of the derived functors:

Indeed, if $f : qF \Rightarrow (\mathbb RF)q$ and $g : qG \Rightarrow (\mathbb RG)q$ are the universal morphisms of the derived functors, then we have a natural transformation

$$gt : F \Rightarrow (\mathbb R G)q$$

and, so, because of the universal property of derived functors, a unique natural transformation $\mathbb R t : \mathbb R F \rightarrow \mathbb R G$ such that $(\mathbb R t)qf = g$.

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

$$\mathbb R f^0 : \mathbb R F \Rightarrow \mathbb R T^0.$$

Taking cohomology, for each $n$, we get

$$H^n(\mathbb R f^0) : H^n (\mathbb R F) \Rightarrow H^n (\mathbb R T^0).$$

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

$$R^nf : R^n F \Rightarrow R^nT^0$$

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

$$i^n : R^nT^0 \Rightarrow T^n$$

which extend the identity

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

The composition

$$i^n \circ R^f : R^F \Rightarrow T^n$$

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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