Can we also define $R^iF(i=0,1,\cdots)$?

Question

Let $F:\mathcal C\to \mathcal D$ be an additive covariant functor between Abelian categories. Assume $\mathcal C$ has enough injective objects. We can also define $R^iF(i=0,1,\cdots)$, can't we?

Answer 1

Yes, we can. There are two reasons people usually assume left exact though. The first is that it is convenient to have $R^0 F(A)=F(A)$, which uses the left exactness of $F$.

The other reason is that you can also define the left derived functor of $F$, and you would like it to be zero above degree zero so that the right derived functors capture all the cohomological data. This is a weaker reason than the first, but still nice to have.

That being said, left and right derived functors of arbitrary functors can be very useful.