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Given abelian categories $\mathcal A, \mathcal B$ and a left exact functor $F: \mathcal A \to \mathcal B$, we can define the notion of a right derived functor (or universal $\delta$-functor) by its universal property. Roughly speaking, we have functors $R^iF$ for any $i$ and short exact sequences give rise to long exact sequences in a functorial way. Furthermore the $R^iF$ are universal with this property, hence they are unique if they exist.

The following is well known:

If $\mathcal A$ has enough injectives, one can construct right derived functors via injective resolutions, hence they always exist and as an immediate consequence we have $R^iF(I)=0 ~\forall i > 0$ for any injective object $I$.

If the category does not have enough injectives (there might be some - but not enough - non-zero injective objects though), we still can ask the following question:

If the right derived functors of $F$ exist, does $R^iF(I)=0 ~ \forall i >0$ for any injective object $I$ still hold?

Morally the question is the following: In the "usual case" with enough injectives, is the construction of the right derived functors crucial to the acyclity of injective objects or is it a formal consequence of the universal property, regardless of the existence?

Note. The question was already posed here: Why do universal $\delta$-functors annihilate injectives?, but unfortunately nobody came up with an answer.

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MooS
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  • Injectives are somewhat crucial in the proof of universality of the $\delta$-functor in that injective resolutions are functorial. Concretely, if $H$ is a function which assigns to any object some choice of injective resolution, $H$ has a uniquely induced functoriality. This is not true for merely $F$-acyclic objects for any particular functor $F$. – Mellon Nov 23 '16 at 10:46
  • I do not see how this approaches the question in any sense. You basically say that we need injectives to make sure our construction is sufficiently functorial. But my question targets at the situation, where we do not even have any construction. We are just given an universal $\delta$-functor and ask whether injective objects are acyclic. – MooS Nov 23 '16 at 10:54
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    @Bubbles Since you mention it, strictly speaking, universality is proved in Tohoku for effacable functors: the condition is that for any $A$ and $n > 0$ there exists a monomorphism $A \rightarrowtail B$ such that $T^n (B) = 0$. The situation with enough injectives is a particular case. –  Nov 23 '16 at 17:51
  • You might get more traction by posting this question on MathOverflow; I bet it would be welcome there. – Ben Blum-Smith Feb 24 '17 at 15:11
  • Do you know any examples of non-effaceable universal $\delta$-functors? The reason I ask is that an effaceable $\delta$-functor $(T^i)$ satisfies $T^i(I) = 0$ for all $i > 0$ and injective $I$. This is pretty easy to prove; see, e.g., p. 7 in these notes by Kedlaya. Maybe a non-effaceable universal $\delta$-functor could give a counterexample? – Takumi Murayama Mar 20 '17 at 03:31
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    Yes, for effaceable functors it is clear, since any injective object is a direct summand of an acyclic object in this case. Unfortunately I don't know an universal non-effaceable delta functor. Maybe this is worth another question. – MooS Mar 20 '17 at 08:14

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