Why do universal $\delta$-functors annihilate injectives?

Question

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?

Answer 1

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 all-or-nothing; 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.)

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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 quasi-isomorphisms. (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 quasi-isomorphisms), 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 quasi-isomorphisms. 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 K-injective 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 quasi-isomorphisms.

Example. Any bounded above (homological!) chain complex of injective objects is K-injective.

Proposition. Let $Q$ be a chain complex in $\mathcal{A}$. The following are equivalent.

1. $Q$ is a K-injective chain complex in $\mathcal{A}$.

2. The functor $\textrm{Hom}_{\mathbf{K} (\mathcal{A})} (-, Q) : \mathbf{K} (\mathcal{A})^\textrm{op} \to \textbf{Ab}$ sends quasi-isomorphisms to isomorphisms.

3. 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 K-injective 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 well-defined 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.

Answer 2

Here is a proof, that injectives are acyclic in the case, that $T^\bullet$ is an effaceable $\delta$-functor. That is, for each object $A\in\mathcal{A}$ and each $n\ge 1$ there is a mono $f:A \hookrightarrow B$ such that $T^n(f)=0$. Effaceable $\delta$-functors are universal, whether $\mathcal{A}$ has enough injectives or not.

Assuming $T^\bullet$ is effaceable, let $I$ be an injective object and $n\ge 1$. Choose a mono $f:I\hookrightarrow B$ with $T^n(f)=0$. Because $I$ is injective, the sequence $0\to I\to B\to \text{ker}(f) \to 0$ is split exact. Additivity of $T^n$ implies $0\to T^n I \to T^n B \to T^n\text{ker}(f)\to 0$ to be split exact, so $T^nI\hookrightarrow T^nB$ is still mono. But $T^n(f)=0$, therefore $T^nI=0$.

This proof does not require $\mathcal{A}$ to have enough injectives. However, effaceability seems to me a stronger condition than sheer universality. Do you know a neat counterexample for a universal $\delta$-functor, that is not effaceable?

Answer 3

I'm learning derived functors and encountered this question. I think I have a example of a universal $\delta$-functor not effaceable as asked by @Lucina, though I'm still familiarizing myself with these concepts so it might be incorrect. This still does not answer the original question because in my example the only injective is $0$.

Consider the abelian category of finitely generated abelian groups $\newcommand{\FGAb}{\mathsf{FGAb}} \newcommand{\Ab}{\mathsf{Ab}} \FGAb$. It has no injectives apart from $0$, since to extend along $\mathbb Z \xrightarrow{\times n} \mathbb Z$ we need divisibility by $n$, and no finitely generated abelian group is divisible apart from $0$. Consider the functors $\newcommand{\Ext}{\operatorname{Ext}}\Ext^i(\mathbb Z/p\mathbb Z, -) : \FGAb \to \Ab$. They still form a universal $\delta$-functor after restricting from $\Ab$. Now consider effacing $\mathbb Z$. For a large enough prime $p$, all the finite factors are erased and we are left with $\Ext^i (\mathbb Z/p \mathbb Z, \varphi)$, where $\varphi : \mathbb Z \to \mathbb Z^n$. After a change of basis, we can assume $\varphi$ is factored as $\mathbb Z \xrightarrow{\times k} \mathbb Z \hookrightarrow \mathbb Z^n$, where $k > 0$. If $p$ is larger than $k$, then this map isn't effaced by $\Ext^1 (\mathbb Z/p \mathbb Z, -)$.

Now consider the product of countably many copies of $\Ab$, which is still an abelian category (since it is the $\Ab$-valued sheaf category on the discrete space $\mathbb N$). Combine all the universal functors into one $\FGAb \to \Ab^{\mathbb N}$. Then by our previous discussion $\mathbb Z$ cannot be effaced, since every injective map out of it is not mapped to zero by a factor.

As already discussed, to answer the original question, we need a category with some injectives but not enough, and the $\delta$-functors need to be universal but not effaceable, which is quite stringent.