Timeline for Injective resolution for right derived functor
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6 events
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| Aug 1, 2022 at 8:13 | comment | added | Denis Nardin | @PrimeRibeyeDeal Let $K(A)$ be the homotopy category of complexes (i.e. complexes and homotopy classes of maps). Then it is easy to see that any additive functor $F:A→B$ induces a functor $F:K(A)→K(B)$. The left derived functor $LF$ is just the right Kan extension of the composition $K(A)→K(B)→D(B)$ along the projection $K(A)→D(A)$. Similarly the right derived functor is the left Kan extension. This is Deligne's approach to derived functors (stacks.math.columbia.edu/tag/05S7) | |
| Jul 30, 2022 at 4:31 | comment | added | PrimeRibeyeDeal | For the less initiated, could you say briefly the connection to Kan extensions? | |
| Feb 23, 2015 at 17:56 | comment | added | Ingo Blechschmidt | For unbounded complexes, the claim is false. Consider the acyclic complex $\cdots \stackrel{2}{\to} \mathbb{Z}/4 \stackrel{2}{\to} \cdots$ of $\mathbb{Z}/4$-modules. This is a complex of projective modules (so, in the dual category, it's a complex of injective objects). But tensoring it with $\mathbb{Z}/2$, one obtains a complex which has cohomology in every degree. (This counterexample is in Gelfand/Manin.) | |
| Feb 23, 2015 at 17:53 | comment | added | Ingo Blechschmidt | If $I^\bullet$ is an acyclic complex of injective objects, bounded below, it is contractible, i.e. homotopy equivalent to the zero complex. Since any additive functor (exact or not) has to preserve homotopy equivalences, it follows that in such a case $I^\bullet \otimes M$ is acyclic as well. | |
| Mar 7, 2014 at 23:07 | comment | added | Li Yutong | Thank you for your answer! I see where I was wrong. However, for the statement $I^{\bullet}$ is acyclic then $I^{\bullet} \otimes M$ is also acyclic, you can prove by using "Residues and Duality" Chapt I, Lemma 4.5 Page 41. | |
| Mar 7, 2014 at 19:52 | history | answered | Denis Nardin | CC BY-SA 3.0 |