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I am working through Weibel's "An Introduction to Homological Algebra". After reading the first 3 chapters, to me it seems that there is a much easier way to think about derived functors as follows.

  • Given an abelian category $A$, homology functors $H_n$ are functors
    from the category of chain complexes bounded from below $H_n: \operatorname{Ch}^+(A)\to A$, defined naturally as $H_n=\operatorname{ker}_{n-1}/\operatorname{im}_n$ (forgive my abuse of notation here - I hope it is clear what I mean).
  • A category with enough projectives has a natural map(not necessarily functorial) $S: A\to \operatorname{Ch}^+(A)$, associating to each object its projective resolution. Considering $\operatorname{Ch}^+(A)$ as a kind of "fiber bundle" over $A$, with a map $\pi: \operatorname{Ch}(A)\to A$, associating to every chain complex its $0^{\text{th}}$ object, the functor $S$ can be thought of as a natural "section" of $\operatorname{Ch}(A)\xrightarrow{\pi}A$.
  • Given a right exact functor $F: A\to A$, we define its left derived functor as just the composition $L_nF:=H_n\circ \tilde{F}\circ S: A\xrightarrow{S} \operatorname{Ch}^+(A)\xrightarrow{\tilde{F}}\operatorname{Ch}^+(A)\xrightarrow{H_n} A$, where $\tilde{F}:\operatorname{Ch}^+(A)\to \operatorname{Ch}^+(A)$ here is the functor induced by $F: A\to A$.

Is there a way to make the above heuristics more precise? In particular,

Question 1: Can $\operatorname{Ch}^+(A)$ be thought of as a "fiber bundle" over $A$, and can the map that associates to every object in $A$ its projective resolution be thought of as a section $S:A\to \operatorname{Ch}^+(A)$ of $\operatorname{Ch}^+(A)\xrightarrow{\pi} A$ in this sense?

Question 2: Are there other useful "sections" $A\to \operatorname{Ch}^+(A)$, besides projective (injective) resolutions? Can they be used to define something similar to derived functors?

Question 3: Are there other useful "fiber bundles" besides $\operatorname{Ch}^+(A)$, or is there a deep reason we are only concerned with the category of chain complexes?

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    The first issue is that $S$ isn’t a functor! Because projective resolutions aren’t canonically defined (and lifting of morphisms only work up to homotopy!), you will run into trouble with composition of morphisms. This is why, I think, one wants to consider at least the homotopy category. Note also that the point of projective resolutions is to have complexes with only projectives, so if you keep your initial object in the resolution, then you lose some good things (and the definition of $L_n$ isn’t the same for $n\geq 1$ and $n=0$). So the “proper” $\pi$ would rather be $H_0$. – Aphelli Jun 13 '23 at 21:42
  • You are right about $S$ - the homotopy category is probably more appropriate, thank you! For the second note, the definition of $L_n$ I am familiar with is the same for for $n=0$ and $n\geq 1$ - it is simply $L_nF(O)=H_n(F(P))$, where $O\in A$ is an object, and $P:=...P_1\to P_0\to O\to 0$ its projective resolution. Then $L_0=F$ by default, since $F$ is right exact. I suppose this definition only works for $\operatorname{Ch}^+(A)$ of complexes bounded from below. For arbitrary complexes, $H_0$ is probably more appropriate $\pi$, you are right. I will edit the question accordingly. –  Jun 13 '23 at 21:52
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    As you observe, there is nothing in your setup that forces you to pick the projective resolution or take homology here. So it is too simple. You might like to have a look at this answer, which explains the relationship between derived functors and resolutions but takes for granted the meaningfulness of the derived category. – Zhen Lin Jun 13 '23 at 22:31

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