Given an exact (additive) functor $F$, i.e. an additive functor preserving exact sequences, it is not hard to show that all derived functors of $F$ vanish.

At the same time given a right exact functor (a similar argument holds for the left exact case) one can show that for every short exact sequence $$0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0$$ there exists an induced long exact sequence of the form $$\cdots \longrightarrow L_1F(B)\longrightarrow L_1F(C)\longrightarrow F(A) \longrightarrow F(B)\longrightarrow F(C) \longrightarrow 0$$ and hence if the first derived functor $L_1F$ vanishes, the functor is exact.

This seems to imply that the vanishing of the first derived functor is a sufficient condition for the vanishing of all higher derived functors. Is this true? This feels like a very strong result/constraint, so I get the feeling that I am missing something.


Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to higher derived functors on other objects.

The following illustration may make this feel less surprising. Let $A$ be any object and take a short exact sequence $$0\to B\to P \to A\to 0$$ where $P$ is projective. There is then an induced long exact sequence $$\dots\to L_{n+1}F(P)\to L_{n+1}F(A)\to L_nF(B)\to L_nF(P)\to\cdots$$ When $n\geq 1$, $L_nF(P)$ and $L_{n+1}F(P)$ are trivial since $P$ is projective, and so the map $L_{n+1}F(A)\to L_nF(B)$ is an isomorphism. So, for instance, the vanishing of $L_1F(B)$ is equivalent to the vanishing of $L_2F(A)$. Iterating this construction, we can similarly find an object $C$ such that the vanishing of $L_1F(C)$ is equivalent to the vanishing of $L_3F(A)$, and so on. So if we know $L_1F$ vanishes on all objects, that actually tells us $L_nF$ vanishes on $A$ for all $n\geq 1$.

  • $\begingroup$ Thanks for the quick and very informative answer. This construction indeed gives a lot of insight. $\endgroup$ – NDewolf Jun 17 at 17:03
  • $\begingroup$ This trick is known as "dimension-shifting". $\endgroup$ – Lord Shark the Unknown Jun 17 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.