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Here is an excerpt from some notes I stumbled upon online:

In fact, the elementary homological algebra proof that right derived functors’ definitions do not depend upon the choice of injective resolution shows that, if $$ 0 \longrightarrow \mathcal{S} \longrightarrow \mathcal{I}_0 \xrightarrow{\enspace f_0 \enspace} \mathcal{I}_1 \xrightarrow{\enspace f_1 \enspace} \dotsb $$ is injective and $$ 0 \longrightarrow \mathcal{S} \longrightarrow \mathcal{A}_0 \xrightarrow{\enspace g_0 \enspace} \mathcal{A}_1 \xrightarrow{\enspace g_1 \enspace} \dotsb $$ is mereley $F$-acyclic, then we have a natural chain homotopy from $$ 0 \longrightarrow F \mathcal{S} \longrightarrow F \mathcal{I}_0 \xrightarrow{\enspace F f_0 \enspace} F \mathcal{I}_1 \xrightarrow{\enspace F f_1 \enspace} \dotsb $$ to $$ 0 \longrightarrow F \mathcal{S} \longrightarrow F \mathcal{A}_0 \xrightarrow{\enspace F g_0 \enspace} F \mathcal{A}_1 \xrightarrow{\enspace F g_1 \enspace} \dotsb $$ Therefore, we have a natural $$ \operatorname{R}^n F (\mathcal{S}) \cong \ker F g_n / {\operatorname{im} F g_{n-1}}. $$ That is, if there is at least one injective resolution of an object $\mathcal{S}$, then the derived functors $\operatorname{R}^n F$ of $F$ can be computed via any $F$-acyclic resolution.

(Original image of this text here.)


From what I understand, the "elementary proof" is just the fundamental lemma of homological algebra which says the homotopy type of chain maps out of projective resolutions is determined by maps between the objects being resolved.

It seems that the author says that every $F$-acyclic resolution is homotopic to some injective/projective resolution, but I don't think this follows from the fundamental lemma. What am I missing?

Is there a way to show one can compute derived functors using projective resolutions without dimension shifting?

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  • $\begingroup$ Paul Garrett is active on mathoverflow.net. Best you repost the question there. $\endgroup$ – tj_ Jul 18 '15 at 11:48
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    $\begingroup$ He's not saying that every $F$-acyclic resolution is chain homotopy equivalent to some injective resolution (which is not true), but that after applying $F$ the $F$-acyclic resolution and injective resolution become chain homotopy equivalent. $\endgroup$ – Jeremy Rickard Jul 18 '15 at 12:46
  • $\begingroup$ @JeremyRickard why is this true, and how does the conclusion that one can compute derived functors with acyclics follow? $\endgroup$ – user153312 Jul 18 '15 at 13:55
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I'm a bit puzzled by the use of the term "chain homotopy", as a chain homotopy is something you have between chain maps, not between complexes. It's certainly not true in general that the complexes $F\mathcal{A}_*$ and $F\mathcal{I}_*$ are chain homotopy equivalent, but what is true is that they have the same cohomology, and so either can be used to calculate the derived functors of $F$.

To sketch how to see this:

First, the identity map of $S$ can be lifted to a cochain map $\varphi:\mathcal{A}_*\to\mathcal{I}_*$ that is an isomorphism on cohomology (the cohomology being only in degree zero).

A cochain map is an isomorphism on cohomology if and only if its mapping cone is acyclic, so the mapping cone of $\varphi$ is acyclic. It's also a bounded below cochain complex of $F$-acyclic objects, which implies that it remains exact upon applying $F$, and so $F(\varphi)$ is also an isomorphism on cohomology.

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  • $\begingroup$ Don't one need F to be additive in order to conclude the mapping cone has F-acyclic objects ? $\endgroup$ – tj_ Jul 19 '15 at 10:14
  • $\begingroup$ @tj_ I was assuming $F$ was meant to be additive. That's the usual context of derived functors. $\endgroup$ – Jeremy Rickard Jul 19 '15 at 10:21

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