Right-deriving a component of bifunctor through F-exact resolutions

Question

This is an exercise in Aluffi's Chapter 0 (Exercise IX.8.14).

We are given a bifunctor $\mathcal{F}: \text{A}^{op}\times \text{B} \to \text{C}$. Denote $\mathcal{F}^A(-)=\mathcal{F}(A,-),\ \mathcal{F}_B(-)=\mathcal{F}(-,B)$. Assume that both these functors are left-exact, $\text{B}$ has enough injectives, while in $\text{A}$ every object has an $\mathcal{F}$-exact resolution, i.e. a resolution by objects $E_i$ such that the functors $\mathcal{F}^{E_i}$ are exact. We can then define 'right-derived' functors $\text{T}^i\mathcal{F}_B(A)=\text{R}^i\mathcal{F}^A(B)$, where $\text{R}^i\mathcal{F}^A(-)$ is the classic right-derived functor of $\mathcal{F}^A(-)$ constructed using the injective resolutions in $\text{B}$.

The task is then to show that $\text{T}^i\mathcal{F}_B(A)$ form a $\delta$-functor, that is give a natural long exact sequence

$$
0\longrightarrow \mathcal{F}_B(A_3)\longrightarrow \mathcal{F}_B(A_2)\longrightarrow \mathcal{F}_B(A_1)\longrightarrow \text{T}^1\mathcal{F}_B(A_3)\longrightarrow \text{T}^1\mathcal{F}_B(A_2)\longrightarrow \text{T}^1\mathcal{F}_B(A_1)\longrightarrow \cdots
$$
for every short exact sequence
$$
0\longrightarrow A_3\longrightarrow A_2\longrightarrow A_1\longrightarrow 0
$$

This exercise is in the chapter on double complexes, so I tried looking at double complexes with elements $\mathcal{F}(E_i,Q_j)$, where $E_i$ and $Q_j$ are the objects in the $\mathcal{F}$-exact and injective resolutions of some 2 objects respectively, but I have no idea how to join the three $\mathcal{F}$-exact resolutions of the three $A_k$. The fact that this can be done for projective resolutions is what allows us to define the right-derived contravariant functor in the first place, but I can't see how the fact that $\mathcal{F}$ is a bifunctor allows us to do something similar here.

Answer 1

I started with the same approach and found the same stumbling block on being unable to make compatible $\mathcal F$-exact resolutions.

Instead, it was able to go through with injective resolutions of $B$. However, it requires the additional assumption that for $Q$ injective, $\mathcal F_Q$ is exact. This (by definition) the case for Hom, but I don't believe its generally true. This material is covered in Tohoku[0] on page 25, and that assumption is made explicitly:

Let $C_1, C_2, C_0$ be three abelian categories. Let $\mathcal T(A, B)$ be an additive bifunctor $C_1 \times C_2 \rightarrow C_0$ which, to fix the ideas, we will assume to be contravariant in $A$ and covariant in $B$. Suppose that every object of $C_2$ is isomorphic to a subobject of an injective object. Then we can construct the right partial derived functors of $\mathcal T$ with respect to the second variable $B$. $$ R^i_2 \mathcal T(A, B) = H^i (\mathcal T(A, C(B))) $$ where $C(B)$ is the complex defined by a right resolution of B by injective objects. Of course, the $R^i_ 2 \mathcal T$ are bifunctors. We now suppose that for any injective object $B$ in $C_2$, the functor defined on C1 by $A \mapsto \mathcal T(A, B)$ is exact. We will show that for any $B ∈ C_2$ the sequence $(R^i_ 2 \mathcal T(A, B))$ can be considered a cohomological functor in $A$. Let $C(B)$ be the complex defined by a right resolution of $B$ by injective objects. For any exact sequence $0 \rightarrow A' \rightarrow A \rightarrow A'' \rightarrow 0$ in $C_1$, the sequence of homomorphisms of complexes $0 \rightarrow \mathcal T(A', C(B)) \rightarrow \mathcal T(A, C(B)) \rightarrow \mathcal T(A'', C(B)) \rightarrow $ is exact (according to the assumption on $\mathcal T$, the terms of $C(B)$ are injective), and it thus defines an exact cohomology sequence, that is $$ \cdot \cdot \cdot \rightarrow R^i_2 \mathcal T(A'', B) \rightarrow R^i_2 \mathcal T(A, B) \rightarrow R^i_2 \mathcal T(A', B) \xrightarrow{\partial} R^{i+1}_2 \mathcal T(A'', B) \rightarrow \cdot \cdot \cdot $$

[0] https://www.math.mcgill.ca/barr/papers/gk.pdf