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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 $R^nG$)?

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    $\begingroup$ 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. $\endgroup$
    – Wojowu
    Commented Oct 24, 2020 at 9:58

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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 $n\geq 2$.

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