# Is every middle exact functor a derived functor?

Assume for the sake of simplicity we are working with categories of modules over some ring. Call a functor $$F$$ middle exact if for an exact sequence $$0 \to A \to B \to C \to 0$$, we have that $$FA \to FB \to FC$$ is exact.

We know that for any right (resp. left) exact functor $$F$$, $$L_nF$$ (resp. $$R^nF$$) are middle exact, since they fit into a long exact sequence $$... \to L_n(A) \to L_n(B) \to L_n(C) \to ...$$

1. Is it then true that any middle exact functor $$F$$ comes from this construction, i.e. $$F = L_nG$$ or $$R^nG$$ for some $$G$$?

2. Is there a way to compute $$G$$ and $$n$$, given that we know $$F=L_nG$$ (or $$R^nG$$)?

• Counterexamples should come from combining left and right derived functors, e.g. take some right exact $F$, left exact $G$ and consider functor given by $H(A)=L_1F(A)\oplus R_1G(A)$. For a suitable $F,G$ this will neither vanish on all projective nor all injective objects, so cannot be left or right derived. I find the second question more interesting. The answer won't be unique (e.g. if $F=0$) but maybe it can be made to work in some cases. Oct 24, 2020 at 9:58

It is not possible in general to infer $$n$$ and $$G$$ from a middle-exact functor $$F$$. Consider the module category of some self-injective, finite-dimensional algebra over a field, say a group algebra of a finite group. Then every left-derived functor vanishes on projectives and therefore factors through the stable module category. But the stable module category has a shift, the Heller operator $$\Omega$$. It satisfies $$L_{n+1}G(M) = L_nG(\Omega M)$$. It is possible to lift $$\Omega$$ to a functor on the module category (by picking a surjection $$P_M \twoheadrightarrow M$$ with a projective $$P_M$$ for every $$M$$ that is functorially in $$M$$) so that $$L_{n+1}(G) = L_n(G\circ\Omega)$$. In other words, you cannot distinguish $$(G,n+1)$$ and $$(G\circ\Omega,n)$$ if you only know one of the left derived functors of $$G$$.
As Wojowu already pointed out in the comments, even if you fix $$n$$, you cannot figure out $$G$$, because there are non-zero functors with $$L_n F = 0$$. For example on the arrow category over some abelian category the functor $$\operatorname{coker}$$ is right exact with $$L_1$$ equal to $$\ker$$ and all higher $$L_n$$ equal to zero. Therefore you cannot distinguish $$(G,n)$$ and $$(G\oplus\operatorname{coker},n)$$ for any $$n\geq 2$$.