I'm confused by the significance of Cartan-Eilenberg resolutions when constructing hyper-derived functors.
Let $F$ be a right-exact functor, and let $A^\bullet$ be a chain complex. According to this Wikipedia page, the left hyper-derived functor of $F$ on a chain complex $A^\bullet$ can be constructed by taking a Cartan-Eilenberg resolution of projective objects $P^{\bullet,\bullet}$ of $A^\bullet$, applying $F$ to $P^{\bullet,\bullet}$ and take cohomology of the total complex.
On the other hand, according to this Wikipedia page, the left hyper-derived functor of $F$ on $A^\bullet$ is given by $H^i(F(P^\bullet))$, where $P^\bullet \to A^\bullet$ is any quasi-isomorphism.
Question. What is the significance/advantage of using a Cartan-Eilenberg resolution instead of an arbitrary quasi-isomorphism?
Comments, hints, and references are welcome.
Best Answer
To use arbitrary quasi-isomorphisms $P^\bullet \to A^\bullet$, where $P^\bullet$ is a complex of projectives, you first need to show that such quasi-isomorphisms exist. That's what the Cartan-Eilenberg resolution accomplishes in the case when $A^\bullet$ is bounded above: it gives you a double complex of projectives such that its total complex is the complex $P^\bullet$ quasi-isomorphic to $A^\bullet$. Compare it to how we first show that any module over a commutative ring has a free resolution, even though we then use arbitrary projective resolutions in some computations.
The fact that a Cartan-Eilenberg resolution also induces resolutions of the homology of $A^\bullet$ is also sometimes useful, even though it's not needed to make sure hyperhomology is well-defined.