Let $G$ be a group, $M$ be $\Bbb{Z}[G]-$ module (where $\Bbb{Z}[G]$ is the group ring). We can define the functor $M\to M^G$ where $M^G$ is the subgroup of $G-$ invariant.
It's the same as the hom functor $$\text{Hom}_{\Bbb{Z}[G]}(\Bbb{Z}, -)(M) = M^G$$
since it's left exact, and R-module has enough projective, we can (right) derive it. and $\Bbb{Z}$ has the canonical free resolution shown for example in wiki
$${\displaystyle \cdots \to F_{n}\to F_{n-1}\to \cdots \to F_{0}\to \mathbb {Z} \to 0.}$$
We can use the resolution to compute the k-th derived functor $R^k \text{Hom}_{\Bbb{Z}[G]}(\Bbb{Z},-)(M)$ above.
However I was a bit confused here since when computing the cohomology of free resolution we use another functor $$\text{Hom}_{\Bbb{Z}[G]}(-,M)$$ not the functor we want to derive $\text{Hom}_{\Bbb{Z}[G]}(\Bbb{Z},-)$. I'm not sure if my question makes sense?
For example when we compute the sheaf cohomology we derive the global section functor $$\Gamma(X,-)$$
And we construct a injective resolution of $\Bbb{R} \to \mathcal{A}^0 \to ….$ (e.g. the deRham resolution) then we compute the cohomology using the same functor $\Gamma(X,-)$.
Best Answer
A priori, the derived functor $R^k\mathrm{Hom}_{\Bbb Z[G]}(\Bbb Z,-)(M)$ is computed by taking an injective resolution of $M$. However, there is a fundamental result in homological algebra, which is called "balancing the ext functor", which states the following:
Let $R$ be a ring and let $M$ and $N$ be the left $R$-modules, then for each integer $k\geq 0$, we have a natural isomorphism $$R^k\mathrm{Hom}_R(M,-)(N)\cong R^k\mathrm{Hom}_R(-,N)(M)$$
Because $\mathrm{Hom}_R(-,N)$ is contravariant, the right derived functor is computed by taking a projective resolution of $M$, that's what is done in group cohomology if we use the canonical free resolution.