Right derived functor

In mathematics, certain functors may be derived to obtain other functors closely related to the original ones. This operation, while fairly abstract, unifies a number of constructions throughout mathematics.

It was noted in various quite different settings that a short exact sequence often gives rise to a "long exact sequence". The concept of derived functors explains and clarifies many of these observations.

Suppose we are given a covariant left exact functor F : A → B between two abelian categories A and B. If 0 → A → B → C → 0 is a short exact sequence in A, then applying F yields the exact sequence 0 → F(A) → F(B) → F(C) and one could ask how to continue this sequence to the right to form a long exact sequence. Strictly speaking, this question is ill-posed, since there are always numerous different ways to continue a given exact sequence to the right. But it turns out that (if A is "nice" enough) there is one canonical way of doing so, given by the right derived functors of F. For every i≥1, there is a functor RⁱF: A → B, and the above sequence continues like so: 0 → F(A) → F(B) → F(C) → R¹F(A) → R¹F(B) → R¹F(C) → R²F(A) → R²F(B) → ... . From this we see that F is an exact functor if and only if R¹F = 0; so in a sense the right derived functors of F measure "how far" F is from being exact.

If the object A in the above short exact sequence is injective, then the sequence splits. Applying any additive functor to a split sequence results in a split sequence, so in particular R¹F(A) = 0. Right derived functors (for i>0) are zero on injectives: this is the motivation for the construction given below.

The crucial assumption we need to make about our abelian category A is that it has enough injectives, meaning that for every object A in A there exists a monomorphism A → I where I is an injective object in A.

...
Wikipedia