ag.algebraic geometry - Derived functors vs universal delta functors

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

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

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

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

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.