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.


1 Answer 1


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.

  • $\begingroup$ Thank you. Do you know any example or reference where the fact that a C-E resolution induces resolutions of the homology of the complex? $\endgroup$
    – Svinto
    Commented Dec 8, 2021 at 11:47
  • $\begingroup$ @Svinto That's a part of the definition of the Cartan-Eilenberg resolution. What you want is the existence, which is shown, for example, in Charles Weibel's "An introduction to homological algebra", chapter 5.7, or here: stacks.math.columbia.edu/tag/015G $\endgroup$ Commented Dec 8, 2021 at 11:57
  • $\begingroup$ My comment was more of a follow-up question. It would be interesting to see an example on how/why these "extra properties" are useful sometimes. I'll look around for a bit, and maybe I will post it as a new question if I can't find anything. $\endgroup$
    – Svinto
    Commented Dec 8, 2021 at 13:08
  • $\begingroup$ @Svinto the extra properties are very, very useful. Without them you cannot conclude one of the most important properties of the Cartan-Eilenberg resolution, which is that (a) maps $f:A\to B$ of complexes always lift to maps $P\to Q$ of the Cartan-Eilenberg double complexes (b) such lifts are unique up to double-complex homotopy. $\endgroup$
    – FShrike
    Commented Dec 21, 2023 at 21:31

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