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Let $F:\mathcal A\to \mathcal B$ be a functor between abelian categories. Suppose $F$ is, say, left exact (plus additive and covariant). We have built its right derived functors $R^iF$.

I see no reason why $R^iF$ should be left or right exact (of course $R^0F\cong F$ is left exact, but apart from this I see nothing else; or are $R^iF$ necessarily left exact?). My question is:

What does it mean, for $F$, to have exact derived functors? What are examples of this behavior?

I put the "algebraic geometry" tag because I have in mind functors like $F=\Gamma=H^0(X,-)$ where $X$ is a scheme, or (more generally) $F=f_\ast$ for $f:X\to Y$ a morphism of schemes.

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3 Answers 3

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Total derived functor is an exact functor between triangulated categories (i.e. respects distinguished triangles and translation functor). Exactness of each individual $R^i F$ perhaps is not the best way to think about the situation.

From technical point of view when you try to calculate $R^iF$ you need a resolution, but you can choose any resolution because they are all quasi-isomorphic, this leads to a notion of derived category i.e. complexes of objects of a given abelian category up to quasi-isomorphisms. The definition has some technical difficulties and not very straightforward but you can find it in some textbooks on homological algebra. I like Weibel"s An Introduction to Homological Algebra. Derived category is an additive category but in general not abelian, it has structure of a triangulated category where notion of exactness is replaced by distinguished triangles and exact functors are functors between triangulated categories that respects distinguished triangles (and also translations).

Exact functor between abelian categories immediately gives you an exact functor between derived categories, but if you start with a non exact functor you can derive it and get an exact (total) derided functor between derived categories. Now remember that objects of derived category were complexes, in particular value of the total derived functor is a complex and you can take cohomology of that comlex and that is $R^iF$.

To sum up, there is a total derived functor between derived categories and it is exact, and (partial) derived functors $R^iF$ are cohomology if that functor.

Because we are in the algebraic geometry section I ought to mention one more reference where you can find discussion of this machinery with application to algebraic geometry: Hartshorne R. Residues and duality.

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  • $\begingroup$ Thanks. Could you please expand your answer a bit? I still do not know how to "think about the situation". It seems that the situation I describe has no meaning at all, is it the case? $\endgroup$
    – Brenin
    Sep 22, 2013 at 8:26
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    $\begingroup$ Exactness of the total derived functor $RF$ is nothing else than the well-known long exact cohomology sequence. If $0 \to A \to B \to C \to 0$ is a short exact sequence, then we get a long exact sequence $\dotsc \to R^n F (A) \to R^n F (B) \to R^n F (C) \to R^{n+1} F (A) \to \dotsc$, in the derived category this becomes an exact triangle $R F (A) \to R F (B) \to R F (C) \to R F (A) [1]$. $\endgroup$
    – Martin Brandenburg
    Sep 22, 2013 at 11:34
  • $\begingroup$ So, again we get nothing more about $F$ apparently. So the exactness of the derived functors seems to be a meaningless condition. $\endgroup$
    – Brenin
    Sep 22, 2013 at 17:13
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    $\begingroup$ It is by no means meaningless. But I admit that it doesn't really address your original question. $\endgroup$
    – Martin Brandenburg
    Sep 22, 2013 at 18:58
  • $\begingroup$ @Martin Brandenburg And exact functors between derived categories respect isomorphisms in the derived category, right? Meaning, if $M\cong N$ are isomorphic in the derived category $\bf{D}$ and $F: {\bf D}\rightarrow{\bf D}'$ is an exact functor from the derived category ${\bf D}$ to the derived category ${\bf D}'$, then $F(M)\cong F(N)$ in ${\bf D}'$. Is that right? $\endgroup$
    – Flavius Aetius
    May 13, 2022 at 13:04
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The functor $R^i F$ can be right exact can happen when the higher cohomology vanishes, but left exactness is a strange thing to ask for. Suppose the domain of $F$ is an abelian category with enough injective objects, and consider a short exact sequence $$0 \longrightarrow A \longrightarrow I \longrightarrow C \longrightarrow 0$$ where $I$ is injective. Then we get a long exact sequence in cohomology: $$0 \longrightarrow R^0 F A \longrightarrow R^0 F I \longrightarrow R^0 F C \longrightarrow R^1 F A \longrightarrow R^1 F I \longrightarrow \cdots$$ However, $R^i F I = 0$ for $i > 0$, so if $R^1 F$ is left exact, then we would be forced to conclude that $R^1 F A = 0$, and hence that $R^0 F$ is exact. More generally, if $R^{n + 1} F$ is left exact, then $R^n F$ has to be right exact and $R^{n + 1} F = 0$.

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I think right derived functors are often not left or right exact. But the following special case is quite nice: If $R^{n+1} F = 0$, then $R^n F$ is right exact. This comes from the long exact cohomology sequence induced by a short exact sequence. For example, if $X$ is an $n$-dimensional noetherian space, then the top sheaf cohomology $H^n(X,-)$ is right exact. If $X$ is a smooth curve and $A \in \mathrm{Coh}(X)$, then $\mathrm{Ext}^1(A,-)$ is right exact.

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