I would like to understand the relationship between the derived category definition of a right derived functor $Rf$ (which involves an initial natural transformation $n: Qf \rightarrow (Rf)Q$, 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?
Best Answer
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.