I'm struggling to understand the meaning/motivation behind injective objects in (abelian) categories, especially in the context of group cohomology. They seem to be mostly mysterious as one mostly cares about having enough of them. Having enough injectives is great since we can define derived functors and derived functors are useful. But this reason for defining them seems kind of backwards.

I am wondering if there is any "intrinsic" interest / reason in defining and studying injectives objects, which would naturally lead to looking at injective resolutions and defining derived functors.

  • 1
    $\begingroup$ How much have you looked into the standard example of injective modules over a ring? Do you find "meaning/motivation" in those? $\endgroup$ Jan 23 at 21:39
  • $\begingroup$ @TorstenSchoeneberg Well I'm aware of the baer criterion and seen some examples but explicit examples don't really answer my question, because we only care about the property of being injectives, not what the objects explicitly look like. $\endgroup$
    – Ben S.
    Jan 23 at 21:44
  • $\begingroup$ @TorstenSchoeneberg Maybe to rephrase my question, is there any reason, why you would first look at these "injective objects" which allow you to lift maps, if your goal is to define something like derived functors and do stuff with exact sequences, except for the reason that "it just works". $\endgroup$
    – Ben S.
    Jan 23 at 21:50

3 Answers 3


Imagine that you have an additive left-exact functor $F$ (on an unspecified abelian category, which we assume is nice enough for the following informal reasoning to make some pretense of sense – typically modules) and that you are told by an oracle that $R^1F$ exists.

Past the euphoria, you realize with some dismay that this told you little useful – because of its “definition” (or rather, lack thereof, since we’ll pretend you don’t know injectives yet), the properties are so hopelessly self-referent that you’ll never get to compute even a single instance of $R^1F$.

Then an idea strikes you. “Okay”, you tell yourself, “this ideal, Platonic version of $R^1F$ exists. But I can’t do anything with it. So instead I’ll try to make my own version of this functor.” [which we’ll call $DF$]

What was the purpose of $(R^1F)A$ anyway? It was to quantify the cokernel of $FB \rightarrow FC$ where $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is exact: every such cokernel injects into $(R^1F)A$.

“So,” you decide, “the useful part of $(R^1F)A$, which we’ll call $DF(A)$, could be the subobject of $(R^1F)A$ generated by all these cokernels [yes, there’s an issue here. Yes, it’s irrelevant and we’ll pretend it doesn’t exist]. In a fair world, it should be exactly like $(R^1F)A$.”

So, when would $DF(A)$ vanish? “When,” you reason, “every exact sequence as above remains exact after applying $F$. If that’s even possible,” you hastily add, frightened at how bold your assumption was, “because it’s only a sufficient condition, the actual criterion is probably a lot more intricate.”

But how could you check that unrealistic property? “Well,” you muse, “since we’re already in the realm of wild speculation, perhaps… when every exact sequence as above splits? I know I’m grasping at straws, but what else can I suggest?”

As you may have guessed, the objects $A$ such that any exact $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ splits are the injective objects.

  • 2
    $\begingroup$ That reminds me, I still don't have an answer to my question from 10 years ago... $\endgroup$
    – Zhen Lin
    Jan 23 at 23:46
  • $\begingroup$ @Zhen Lin: That’s an interesting question too… the “obvious” approach would be to find interesting $\delta$-functors to which one could apply the universality property to – but the only non-trivial I know [unless you count such inapplicable lemmas as “there is no nonzero morphism from $T^n$, $n >1$, to any exact functor”] would be the derived functor (which basically solves the problem, except we’re trying to avoid them!). $\endgroup$
    – Aphelli
    Jan 24 at 0:33
  • $\begingroup$ Thank you, I really like your answer, this has helped me alot. $\endgroup$
    – Ben S.
    Jan 25 at 18:29

I don’t know about you, but for me the purely algebraic motivations work best for motivating constructions in homological algebra.

(Short) exact sequences are without doubt a really useful tool in commutative algebra and I find it really natural to ask, what happens to them after applying canonical functors such as $-\otimes_R M$ or $\operatorname{Hom}_R(M,-)$ or $\operatorname{Hom}(-,M)$. All three of them preserve biproducts and thus split exact sequences. One might hope that such additive functors preserve exact sequences in general, but unfortunately as it turns out they just preserve the property of having successive morphisms compose to zero. (This is the best reason to consider the category of chain complexes I know of, but I guess this was not the question).

All of the three functors above do not preserve short exact sequences. But they are so important that one might wish to try to find objects $M$, for which they do preserve short exact sequences. These are precisely the flat, projective and injective objects respectively. Studying them in detail one notes that they may be characterized in terms of lifting properties.

Now suppose you have some object $M_0$ for which one of the functors is not exact. Maybe you can somehow replace $M_0$ with some $N_0$ for which it is exact? We cannot expect $M_0$ and $N_0$ to be isomorphic (a functor isomorphic to an exact functor is itself exact) so we will get some object $M_1$ measuring their difference. This may result in another non-exact functor, but we might try to replace it again by some $N_1$, which gives an exact functor. This is the idea of resolving an object and the lifting properties come in really handy to make it precise: we just need to have enough flat/projective/injective objects…


This is going to be a long story, but let me try to explain how we are inevitably led to injective resolutions and derived functors if we want to study chain complexes up to "homology invariance" (= quasi-isomorphism).

Let $\mathcal{A}$ be an abelian category. Let $\textbf{Ch} (\mathcal{A})$ be the category of chain complexes in $\mathcal{A}$. I will use homological grading and write $\partial$ for the differential operator. Since $\mathcal{A}$ is abelian, basic homological algebra can be applied to chain complexes in $\mathcal{A}$. Suppose we are interested in studying chain complexes in $\mathcal{A}$ but we want to consider chain complexes the same if they have the same homology. This leads us to define:

Definition. A quasi-isomorphism of chain complexes in $\mathcal{A}$ is a morphism $f : A \to B$ in $\textbf{Ch} (\mathcal{A})$ such that, for every integer $n$, the induced morphism $\mathrm{H}_n (f) : \mathrm{H}_n (A) \to \mathrm{H}_n (B)$ of homology objects is an isomorphism in $\mathcal{A}$. The derived category $\mathbf{D} (\mathcal{A})$ is the category obtained by freely inverting the quasi-isomorphisms in $\textbf{Ch} (\mathcal{A})$.

That means we have a functor $\gamma : \textbf{Ch} (\mathcal{A}) \to \mathbf{D} (\mathcal{A})$ that sends quasi-isomorphisms in $\textbf{Ch} (\mathcal{A})$ to isomorphisms in $\mathbf{D} (\mathcal{A})$, and every functor $\textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ that sends quasi-isomorphisms in $\textbf{Ch} (\mathcal{A})$ to isomorphisms in $\mathcal{C}$ factors through the functor $\gamma : \textbf{Ch} (\mathcal{A}) \to \mathbf{D} (\mathcal{A})$ in a unique way. For example, the homology functors $\textrm{H}_n : \textbf{Ch} (\mathcal{A}) \to \mathcal{A}$ certainly send quasi-isomorphisms to isomorphisms, so they induce functors $\mathbf{D} (\mathcal{A}) \to \mathcal{A}$.

But what are the morphisms of $\mathbf{D} (\mathcal{A})$? It turns out that every morphism $\gamma P \to \gamma Q$ in $\mathbf{D} (\mathcal{A})$ is of the form $(\gamma j)^{-1} \circ \gamma f$ for some morphism $f : P \to \hat{Q}$ and quasi-isomorphism $j : Q \to \hat{Q}$ in $\textbf{Ch} (\mathcal{A})$. Furthermore, $\mathbf{D} (\mathcal{A})$ has an additive structure and $\gamma : \textbf{Ch} (\mathcal{A}) \to \mathbf{D} (\mathcal{A})$ is additive. So, for each $Q$ in $\textbf{Ch} (\mathcal{A})$, $P \mapsto \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (\gamma P, \gamma Q)$ defines a functor $\textbf{Ch} (\mathcal{A})^\textrm{op} \to \textbf{Ab}$ that sends quasi-isomorphisms to isomorphisms, and $Q \mapsto (P \mapsto \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (\gamma P, \gamma Q))$ defines a functor $\textbf{Ch} (\mathcal{A}) \to [\textbf{Ch} (\mathcal{A})^\textrm{op}, \textbf{Ab}]$ that sends quasi-isomorphisms to isomorphisms. What is its relation to $Q \mapsto (P \mapsto \textrm{Hom}_{\textbf{Ch} (\mathcal{A})} (P, Q))$?

Definition. Let $P$ and $Q$ be chain complexes in $\mathcal{A}$. The chain complex $\textrm{Hom}_\mathcal{A} (P, Q)$ of abelian groups is defined as follows: $$\begin{align} \textrm{Hom}_\mathcal{A} (P, Q)_n & = \prod_m \textrm{Hom}_\mathcal{A} (P_m, Q_{m + n}) \\ (\partial (f))_m & = \partial \circ f_{m+1} - (-1)^n f_m \circ \partial \end{align}$$

Observe that the 0-cycles of $\textrm{Hom}_\mathcal{A} (P, Q)$ are precisely the morphisms $P \to Q$ in $\textbf{Ch} (\mathcal{A})$, whereas the homology group $\mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (P, Q))$ can be identified with the chain homotopy classes of morphisms $P \to Q$. Anyway, varying $P$, we obtain a functor $\textrm{Hom}_\mathcal{A} (-, Q) : \textbf{Ch} (\mathcal{A})^\textrm{op} \to \textbf{Ch} (\textbf{Ab})$. Unfortunately, it does not preserve quasi-isomorphisms in general, so $\mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (P, Q))$ is different from $\textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P, Q)$.

Definition. A K-injective chain complex in $\mathcal{A}$ is a chain complex $Q$ in $\mathcal{A}$ such that $\textrm{Hom}_\mathcal{A} (-, Q) : \textbf{Ch} (\mathcal{A})^\textrm{op} \to \textbf{Ch} (\textbf{Ab})$ preserves quasi-isomorphisms.

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. $\gamma \textrm{Hom}_\mathcal{A} (-, Q) : \textbf{Ch} (\mathcal{A})^\textrm{op} \to \mathbf{D} (\textbf{Ab})$ sends quasi-isomorphisms in $\textbf{Ch} (\mathcal{A})$ to isomorphisms in $\mathbf{D} (\textbf{Ab})$.
  3. $\mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (-, Q)) : \textbf{Ch} (\mathcal{A})^\textrm{op} \to \textbf{Ab}$ sends quasi-isomorphisms in $\textbf{Ch} (\mathcal{A})$ to isomorphisms in $\textbf{Ab}$.

Proof. Mostly straightforward. The trick to get from statement 3 to statement 1 is to observe that $$\mathrm{H}_n (\textrm{Hom}_\mathcal{A} (P, Q)) \cong \mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (P [-n], Q))$$ where $P [-n]$ is the chain complex defined by $P [-n]_m = P_{m - n}$ with differential $P [-n]_m \to P [-n]_{m - 1}$ given by $(-1)^n$ times the differential $P_{m - n} \to P_{m - n - 1}$, and this isomorphism is natural in $P$ (and $Q$ too). ◼

In other words, a K-injective chain complex is a chain complex that "perceives" quasi-isomorphisms as chain homotopy equivalences. Here is an instance of this phenomenon:

Lemma. Let $g : Q \to P$ be a morphism in $\textbf{Ch} (\mathcal{A})$. If $Q$ is K-injective and $g : Q \to P$ is a quasi-isomorphism, then there is a morphism $f : P \to Q$ in $\textbf{Ch} (\mathcal{A})$ such that $f \circ g : Q \to Q$ is chain homotopic to $\textrm{id}_Q$.

Proof. The induced homomorphism $\mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (P, Q)) \to \mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (Q, Q))$ is an isomorphism. In particular, the homology class of $\textrm{id}_Q$ has a preimage, say represented by a 0-cycle $f$ in $\textrm{Hom}_\mathcal{A} (P, Q)$, i.e. a morphism $P \to Q$ in $\textbf{Ch} (\mathcal{A})$. But that means $g \circ f$ is in the same homology class as $\textrm{id}_Q$, and the homology classes of 0-cycles in $\textrm{Hom}_\mathcal{A} (Q, Q)$ are the chain homotopy classes of morphisms $Q \to Q$ in $\textbf{Ch} (\mathcal{A})$, so we are done. ◼

Corollary. If both $P$ and $Q$ are K-injective chain complexes in $\mathcal{A}$, then the following are equivalent:

  1. $f : P \to Q$ is a quasi-isomorphism in $\textbf{Ch} (\mathcal{A})$.
  2. $f : P \to Q$ is a chain homotopy equivalence in $\textbf{Ch} (\mathcal{A})$. ◼

Why does this matter? Well, for a general additive functor $F : \mathcal{A} \to \mathcal{B}$, the induced functor $\textbf{Ch} (F) : \textbf{Ch} (\mathcal{A}) \to \textbf{Ch} (\mathcal{B})$ is not guaranteed to preserve quasi-isomorphisms. Indeed, $\textbf{Ch} (F)$ preserving quasi-isomorphisms is equivalent to $F$ preserving (short) exact sequences. But $\textbf{Ch} (F)$ always preserves chain homotopy equivalences. So if there were a "best approximation" of $\textbf{Ch} (F)$ by a functor that preserves quasi-isomorphisms, and every chain complex in $\mathcal{A}$ is quasi-isomorphic to a K-injective one, then the "best approximation" should be determined by the restriction of $\textbf{Ch} (F)$ to the full subcategory of K-injective chain complexes in $\mathcal{A}$.

Definition. The right derived functor of a functor $G : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ is a functor $\mathbf{R} G : \mathbf{D} (\mathcal{A}) \to \mathcal{C}$ equipped with a natural transformation $\eta : G \Rightarrow (\mathbf{R} G) \gamma$ such that, for every functor $H : \mathcal{C} \to \mathcal{D}$ and every functor $K : \mathbf{D} (\mathcal{A}) \to \mathcal{D}$ and every natural transformation $\phi : H G \Rightarrow K \gamma$, there is a unique natural transformation $\psi : H \mathbf{R} G \Rightarrow K$ such that $\psi \gamma \bullet \eta = \phi$.

(This is stronger than the original definition by Verdier.)

Theorem. Let $G : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ be a functor that sends chain homotopy equivalences in $\textbf{Ch} (\mathcal{A})$ to isomorphisms in $\mathcal{C}$. Assume for every chain complex $P$ in $\mathcal{A}$ we have a quasi-isomorphism $j_P : P \to \hat{P}$ in $\textbf{Ch} (\mathcal{A})$ where $\hat{P}$ is a K-injective chain complex in $\mathcal{A}$. Then the right derived functor $\mathbf{R} G : \textbf{D} (\mathcal{A}) \to \mathcal{C}$ exists and $\eta_Q : G Q \to (\mathbf{R} G) (\gamma Q)$ is an isomorphism for every K-injective chain complex $Q$ in $\mathcal{A}$.

Proof. You can do this directly, but it is helpful to introduce the category $\mathbf{K} (\mathcal{A})$, which is $\textbf{Ch} (\mathcal{A})$ modulo chain homotopy. In $\mathbf{K} (\mathcal{A})$, for every morphism $f : P \to Q$, there is a unique morphism $\hat{f} : \hat{P} \to \hat{Q}$ such that the following diagram commutes: $$\require{AMScd} \begin{CD} P @>{j_P}>> \hat{P} \\ @V{f}VV @VV{\hat{f}}V \\ Q @>>{j_Q}> \hat{Q} \end{CD}$$ (This is not always true in $\textbf{Ch} (\mathcal{A})$!) Thus we may make $P \mapsto \hat{P}$ and $f \mapsto \hat{f}$ into a functor from $\mathbf{K} (\mathcal{A})$ to the full subcategory $\mathbf{K} (\mathcal{A})_\textrm{K-inj}$ of K-injective chain complexes (modulo chain homotopy). Since $j_P : P \to \hat{P}$ is an isomorphism in $\mathbf{K} (\mathcal{A})$ if $P$ is K-injective, this exhibits $\mathbf{K} (\mathcal{A})_\textrm{K-inj}$ as a reflective subcategory of $\mathbf{K} (\mathcal{A})$. Notice that $\hat{f} : \hat{P} \to \hat{Q}$ is an isomorphism in $\textbf{K} (\mathcal{A})_\textrm{K-inj}$ if $f : P \to Q$ is a quasi-isomorphism, so we get a functor $R : \mathbf{D} (\mathcal{A}) \to \textbf{K} (\mathcal{A})_\textrm{K-inj}$ where $R (\gamma P) = \hat{P}$ and $R (\gamma f) = \hat{f}$.

You can also show that $\mathbf{K} (\mathcal{A})$ is also the category obtained by freely inverting the chain homotopy equivalences in $\textbf{Ch} (\mathcal{A})$, so $G : \textbf{Ch} (\mathcal{A}) \to \mathcal{C}$ factors through the quotient $\textbf{Ch} (\mathcal{A}) \to \mathbf{K} (\mathcal{A})$, say as $\bar{G} : \mathbf{K} (\mathcal{A}) \to \mathcal{C}$. Then you can check that taking $(\mathbf{R} G) = \bar{G} R$ and $\eta_P : G P \to (\mathbf{R} G) (\gamma P)$ to be $G j_P : G P \to G \hat{P}$ works. ◼

Hopefully the above convinces you that K-injective chain complexes are important. But what are they, in more elementary terms?

Proposition. Let $Q$ be a chain complex in $\mathcal{A}$ concentrated in degree 0. The following are equivalent:

  1. $Q$ is a K-injective chain complex in $\mathcal{A}$.
  2. $Q_0$ is an injective object in $\mathcal{A}$.

Proof. Essentially, $\textrm{Hom}_\mathcal{A} (-, Q) : \textbf{Ch} (\mathcal{A})^\textrm{op} \to \textbf{Ch} (\textbf{Ab})$ can be identified with $\textrm{Hom}_\mathcal{A} (-, Q_0): \textbf{Ch} (\mathcal{A})^\textrm{op} \to \textbf{Ch} (\textbf{Ab})$. (There is a subtlety about the signs in the differentials.) The latter preserves quasi-isomorphisms if and only if $\textrm{Hom}_\mathcal{A} (-, Q_0) : \mathcal{A}^\textrm{op} \to \textbf{Ab}$ is exact. But that is the definition of injective object, so we are done. ◼

So the simplest examples of K-injective chain complexes are injective objects. More generally:

Proposition. Let $Q$ be a chain complex of injective objects in $\mathcal{A}$. If $Q$ is bounded above (i.e. there is an integer $N$ such that $Q_n = 0$ for all $n \ge N$), then $Q$ is K-injective.

(Proof omitted.)

Unfortunately it is not true that K-injective chain complexes are chain complexes of injectives. For example, for any object $A$ in $\mathcal{A}$, $$\begin{CD} \cdots @>>> 0 @>>> A @>{\textrm{id}}>> A @>>> 0 @>>> \cdots \end{CD}$$ is a K-injective chain complex in $\mathcal{A}$, because the property of being K-injective is chain homotopy equivalence invariant, and $0$ is certainly a K-injective chain complex. Nonetheless, I hope that this convinces you that injectivity is a natural notion in the context of homological algebra and not merely something that happens to work well.


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