The following is the definition of the right-derived functor from Lang's Algebra -
Let $\mathcal A$ and $\mathcal B$ be abelian categories with enough injectives. Consider a covariant functor $F:\mathcal A\to\mathcal B$ that is left exact and additive. Let $M$ be an object in $\mathcal A$, let $$0\to M\to I^0\to I^1\to I^2\to$$ be an injective resolution, which we abbreviate by, $0\to M\to I_M$, where $I_M$ is the complex $I^0\to I^1\to I^2\to$. We let $I$ be the complex, $$0\to I^0\to I^1\to I^2\to$$
Then we define the right-derived functor $R^nF$ by $$R^nF(M)=H^n(F(I)),$$ in other words, the $n$-th homology of the complex, $$0\to F(I^0)\to F(I^1)\to F(I^2)\to$$
My question :
There is probably something very silly that I am missing but what I don't get is why are $I_M$ and $I$ complexes? Isn't $I_M$ exact and except at the $I^0$ stage $I$ is also exact? If so then $0\to F(I^0)\to F(I^1)\to F(I^2)\to\cdots$ is also exact except at the $F(I^0)$ stage. So then all the homologies for $n>0$ will be trivial. But that isn't true. So what am I missing?
Thank you.
You have to remember carefully what does left exact functor mean. Recall that $F$ is left exact if, for every short exact sequence $0\to M^\prime \to M\to M^{\prime\prime}\to 0$ there is an exact sequence $0\to F(M^\prime)\to F(M)\to F(M^{\prime\prime})$.
First, note that an injective resolution $0\to M \to I_M$ is an exact long sequence. If we "truncate" this sequence and force what remains into a complex we get the desired $0\to I_M$ which is exact except in the zeroth part.
But applying $F$, which is exact only from the left, we get a sequence $0\to F(I_M)$ which is not exact because $F$ doesn't preserve exactness at right side; basically, the only property that remains is that $\rm{im}\subseteq \ker $, which follows by additivity.
The point is that you don't need side-exactness to define derived functors at this stage; the left exactness is necessary only to obtain some more interesting results as $R^0(F)\simeq F$.