Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Suppose $\mathcal{A}$ has enough injectives, and consider a universal (cohomological) $\delta$functor $T^\bullet$ from $\mathcal{A}$ to $\mathcal{B}$. By the theory of derived functors, we know that that $T^n (A) = 0$ for all injective objects $A$ and all $n \ge 1$ and that $T^n$ is effaceable for $n \ge 1$. But can this be shown directly without invoking derived functors?
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1If $T^*$ is universal you can show that $T^i$ must be the $i$th right derived functor of $T^0$ and so you get higher vanishing on injectives. This is because the $R^* T^0$ is also a universal delta functor, and hence (by universality) must be the same as $T^*$. In fact, universal delta functors are just a different formalization of derived functors, so I don't know why you'd want to avoid derived functors. – Parsa Apr 02 '12 at 10:27

7@Parsa: Universal $\delta$functors make sense in the absence of enough injectives, but derived functors require some form of "enough acyclics" assumption. I am basically wondering if it is true that injectives are acyclic even when there are not enough injectives. – Zhen Lin Apr 02 '12 at 10:33

2Why do you want to show this without using derived functors? – Jimmy R Jul 23 '13 at 21:18

Here’s an idea that might work (if the details work out): define $D^0=T^0$, and, if $n \geq 1$, $D^nA$ as the image in $T^nA$ of the cokernel of $T^{n1}I \rightarrow T^{n1}C$ (with $0 \rightarrow A \rightarrow I \rightarrow C \rightarrow 0$ exact and $I$ injective). I think it might be welldefined, functorial, even cohomological. It follows that there’s a natural map $U:T^{\cdot} \Rightarrow D^{\cdot}$ whose composition with $in: D^{\cdot} \Rightarrow T^{\cdot}$ is an isomorphism. So $in$ isomorphism, thus (exactness) $T^nA \rightarrow T^nI$ is null, which concludes. – Aphelli Jan 24 '22 at 01:09

Hmmm. That’s basically the Cartan–Eilenberg satellite construction. I think I knew about that when I asked the question though, so I’m not sure why I specified the assumption of enough injectives. As I said in the comments, what I really want to know is whether injectives are acyclic even without assuming that there are enough injectives. – Zhen Lin Jan 24 '22 at 02:26
1 Answers
In the 10 years since I asked this question, I learned some other theories of derived functors, and the fact that there has been no answer has been forthcoming leads me to believe that the question may be unanswerable because the definition of (universal) $\delta$functor is simply too weak. Perhaps Grothendieck himself realised this as he and Verdier developed the theory of (triangulated categories and) derived categories. So let me answer a modified question.
The main source of long exact sequences in abelian categories – as far as I know – is short exact sequences of chain complexes in abelian categories, or more generally triangles in triangulated categories. However, it seems that not every long exact sequence arises this way: Neeman [Long exact sequences coming from triangles] showed that there is a necessary condition involving $\textrm{Ext}^3$ and Šťovíček [A characterization of long exact sequences coming from the snake lemma] showed that Neeman's condition is not sufficient. So if what we really want is a long exact sequence coming from a triangle we should just ask for a triangle.
Another problem with the $\delta$functor formalism is that it is very hard to express how the universal $\delta$functor of a composite functor is related to the composites of the corresponding universal $\delta$functors. (cough Grothendieck spectral sequence cough) It is easier to express in the derived category formalism – simply from the definition, there is a canonical natural transformation from the right derived functor of a composite to the composite of the right derived functors – but I feel the original definition is deficient – in practice, when right derived functors are constructed by injective resolutions, the comparison is an isomorphism, but as far as I know this cannot be shown from the definition alone.
The above I already more or less understood 6 or 7 years ago, but recently I happened to read some remarks on MathOverflow and in the Stacks project that made me realise there is a way to modify the definition of derived functor so that:
 the existence of a derived functor does not have to be allornothing; it can be defined one object at a time; and
 it is a theorem that evaluating at an injective resolution computes the right derived functor.
As such, with this definition, it is a theorem that injective objects are acyclic for left exact functors – whether or not there are enough injectives!
(I think someone – maybe t.b. – tried to explain these ideas to me many years ago, but I did not understand derived categories at all back then. Even now I only have a faint understanding of derived categories, and I still do not understand triangulated categories.)
Now with the preface out of the way we can discuss the mathematical details.
Given an abelian category $\mathcal{A}$, let $\textbf{Ch} (\mathcal{A})$ be the category of (unbounded) chain complexes, let $\mathbf{K} (\mathcal{A})$ be $\textbf{Ch} (\mathcal{A})$ modulo chain homotopy, and let $\mathbf{D} (\mathcal{A})$ be $\textbf{Ch} (\mathcal{A})$ or $\mathbf{K} (\mathcal{A})$ localised with respect to quasiisomorphisms. (It is a simple exercise to show that both definitions are equivalent.)
Given an additive functor $F : \mathcal{A} \to \mathcal{B}$ where $\mathcal{A}$ and $\mathcal{B}$ are abelian categories, we automatically get induced functors $\textbf{Ch} (F) : \textbf{Ch} (\mathcal{A}) \to \textbf{Ch} (\mathcal{B})$ and $\mathbf{K} (F) : \mathbf{K} (\mathcal{A}) \to \mathbf{K} (\mathcal{B})$, and if $F : \mathcal{A} \to \mathcal{B}$ is exact (or, equivalently, $\textbf{Ch} (F) : \textbf{Ch} (\mathcal{A}) \to \textbf{Ch} (\mathcal{B})$ preserves quasiisomorphisms), then we get an induced functor $\mathbf{D} (F) : \mathbf{D} (\mathcal{A}) \to \mathbf{D} (\mathcal{B})$. The derived functors of $F$, or rather $\textbf{Ch} (F)$, are essentially the best approximations of $\textbf{Ch} (F)$ by functors that preserve quasiisomorphisms. However, strict functoriality is difficult to achieve so we often work at the level of $\mathbf{K}$ instead of $\textbf{Ch}$.
Definition. Let $F : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ be a functor and let $P$ be a chain complex in $\mathcal{A}$. The value of the right derived functor of $F$ at $P$ is an object $(\mathbf{R} F) P$ in $\mathcal{C}$ together with a map assigning to each object $P'$ and morphism $p : P' \to P$ in $\mathbf{D} (\mathcal{A})$ a morphism $\eta (p) : F P' \to (\mathbf{R} F) P$ in $\mathcal{C}$, such that the following conditions are satisfied:
 For every morphism $p' : P'' \to P'$ in $\textbf{Ch} (\mathcal{A})$: $$\eta (p \circ p') = \eta (p) \circ F p'$$ In other words, $\eta$ is map $$\textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P', P) \to \textrm{Hom}_\mathcal{C} (F P', (\mathbf{R} F) P)$$ that is natural (in the technical sense!) as $P'$ varies in $\textbf{Ch} (\mathcal{A})$.
 For every object $C$ in $\mathcal{C}$ and every map $\psi : \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P', P) \to \textrm{Hom}_\mathcal{C} (F P', C)$ natural as $P'$ varies in $\textbf{Ch} (\mathcal{A})$, there is a unique morphism $c : (\mathbf{R} F) P \to D$ such that $$\psi (p) = c \circ \eta (p)$$ for all objects $P'$ in $\textbf{Ch} (\mathcal{A})$ and all morphisms $p : P' \to P$ in $\mathbf{D} (\mathcal{A})$.
(Mutatis mutandis for $\mathbf{K} (\mathcal{A})$ instead of $\textbf{Ch} (\mathcal{A})$. It is clear that if $F : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ factors through the quotient $\textbf{Ch} (\mathcal{A}) \to \mathbf{K} (\mathcal{A})$ then it does not matter we consider it as a functor $\textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ or a functor $\mathbf{K} (\mathcal{A}) \to \mathcal{C}$.)
Experts will recognise this as an instance of a pointwise left Kan extension. If $\mathcal{C}$ has colimits of diagrams of shape $\textbf{Ch} (\mathcal{A}) \times_{\mathbf{D} (\mathcal{A})} \mathbf{D} (\mathcal{A})_{/ P}$ then the value of the right derived functor of $F$ at $P$ exists, for every $F : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$. (Mutatis mutandis for $\mathbf{K} (\mathcal{A})$.) This happens if e.g. $\mathcal{A}$ is small and $\mathcal{C}$ is cocomplete – perhaps this is what Grothendieck was alluding to in his remark after théorème 2.2.2 in his Tôhoku paper. Anyway, back to injectives.
Definition. A Kinjective chain complex in $\mathcal{A}$ is a chain complex $Q$ in $\mathcal{A}$ such that the functor $\textrm{Hom}_\mathcal{A} (, Q) : \textbf{Ch} (\mathcal{A})^\textrm{op} \to \textbf{Ch} (\textbf{Ab})$ preserves quasiisomorphisms.
Example. Any bounded above (homological!) chain complex of injective objects is Kinjective.
Proposition. Let $Q$ be a chain complex in $\mathcal{A}$. The following are equivalent.
$Q$ is a Kinjective chain complex in $\mathcal{A}$.
The functor $\textrm{Hom}_{\mathbf{K} (\mathcal{A})} (, Q) : \mathbf{K} (\mathcal{A})^\textrm{op} \to \textbf{Ab}$ sends quasiisomorphisms to isomorphisms.
For every chain complex $P$, the map $\textrm{Hom}_{\mathbf{K} (\mathcal{A})} (P, Q) \to \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P, Q)$ defined by the localisation functor $\mathbf{K} (\mathcal{A}) \to \mathbf{D} (\mathcal{A})$ is an isomorphism.
Proof. That 1 implies 2 is clear, because $\textrm{Hom}_{\mathbf{K} (\mathcal{A})} (P, Q) \cong \mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (P, Q))$. It is straightforward to show that 2 implies $\textrm{Hom}_{\mathbf{K} (\mathcal{A})} (P, Q) \to \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P, Q)$ is surjective, but it seems that verifying injectivity requires the use of the Gabriel–Zisman calculus of fractions. That 3 implies 2 is obvious, and 2 implies 1 by a degree shift argument. ◼
Theorem. Let $Q$ be a Kinjective chain complex in $\mathcal{A}$ and let $F : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ be a functor. If $F$ sends chain homotopy equivalences of chain complexes in $\mathcal{A}$ to isomorphisms in $\mathcal{C}$, then $F Q$, together with the map assigning to each $P'$ and $p : P' \to Q$ in $\mathbf{D} (\mathcal{A})$ the morphism $F p : F P' \to F Q$ in $\mathcal{C}$, is the value of the right derived functor of $F$ at $Q$.
Proof. The hypothesis implies $F : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ factors as the quotient $\textbf{Ch} (\mathcal{A}) \to \mathbf{K} (\mathcal{A})$ followed by a (uniquely determined) functor $\mathbf{K} (\mathcal{A}) \to \mathcal{C}$. On the other hand, the proposition says that every morphism $p : P' \to Q$ in $\mathbf{D} (\mathcal{A})$ lifts uniquely to $\mathbf{K} (\mathcal{A})$, so (abusing notation somewhat) we do have a welldefined morphism $F p : F P' \to Q$ in $\mathcal{C}$. The rest of the verification is essentially the Yoneda lemma. ◼
Corollary. If $B$ is an object in $\mathcal{A}$ and we have an exact sequence in $\mathcal{A}$ of the form below, $$0 \longrightarrow B \overset{j}{\longrightarrow} Q_0 \longrightarrow Q_{1} \longrightarrow Q_{2} \longrightarrow \cdots$$ where each $Q_n$ is an injective object in $\mathcal{A}$, then (with $F$ as in the theorem, and assuming $Q_n = 0$ for $n > 0$) $F Q$, together with the map assigning to each $P'$ and $p : P' \to B [0]$ the morphism $F (j \circ p) : F P' \to F Q$, is the value of the right derived functor of $F$ at $B [0]$. ◼
Example. Let $\mathcal{B}$ be an abelian category and let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor. Then the induced functor (abusing notation!) $F : \textbf{Ch} (\mathcal{A}) \to \textbf{Ch} (\mathcal{B})$ preserves chain homotopy equivalences, hence the composite $\textbf{Ch} (\mathcal{A}) \to \mathbf{D} (\mathcal{B})$ sends chain homotopy equivalences to isomorphisms. Thus, putting $\mathcal{C} = \mathbf{D} (\mathcal{B})$, we may apply the theorem and its corollary. The classical derived functors are recovered by taking homology of the modern derived functors (when there are enough injectives), so we have proved (for our definition of derived functors, without assuming that there are enough injectives) that $$(\mathrm{R}^n F) I \overset{\textrm{def}}{=} \mathrm{H}_{n} ((\mathbf{R} F) I) \cong \begin{cases} I & \textrm{if } n = 0 \\ 0 & \textrm{if } n \ne 0 \end{cases}$$ for any injective object $I$ in $\mathcal{A}$, as desired.
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