
## 13.29 K-injective complexes

The following types of complexes can be used to compute right derived functors on the unbounded derived category.

Definition 13.29.1. Let $\mathcal{A}$ be an abelian category. A complex $I^\bullet$ is K-injective if for every acyclic complex $M^\bullet$ we have $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$.

In the situation of the definition we have in fact $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet [i], I^\bullet ) = 0$ for all $i$ as the translate of an acyclic complex is acyclic.

Lemma 13.29.2. Let $\mathcal{A}$ be an abelian category. Let $I^\bullet$ be a complex. The following are equivalent

1. $I^\bullet$ is K-injective,

2. for every quasi-isomorphism $M^\bullet \to N^\bullet$ the map

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet )$

is bijective, and

3. for every complex $N^\bullet$ the map

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(N^\bullet , I^\bullet )$

is an isomorphism.

Proof. Assume (1). Then (2) holds because the functor $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}( - , I^\bullet )$ is cohomological and the cone on a quasi-isomorphism is acyclic.

Assume (2). A morphism $N^\bullet \to I^\bullet$ in $D(\mathcal{A})$ is of the form $fs^{-1} : N^\bullet \to I^\bullet$ where $s : M^\bullet \to N^\bullet$ is a quasi-isomorphism and $f : M^\bullet \to I^\bullet$ is a map. By (2) this corresponds to a unique morphism $N^\bullet \to I^\bullet$ in $K(\mathcal{A})$, i.e., (3) holds.

Assume (3). If $M^\bullet$ is acyclic then $M^\bullet$ is isomorphic to the zero complex in $D(\mathcal{A})$ hence $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$, whence $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$ by (3), i.e., (1) holds. $\square$

Lemma 13.29.3. Let $\mathcal{A}$ be an abelian category. Let $(K, L, M, f, g, h)$ be a distinguished triangle of $K(\mathcal{A})$. If two out of $K$, $L$, $M$ are K-injective complexes, then the third is too.

Proof. Follows from the definition, Lemma 13.4.2, and the fact that $K(\mathcal{A})$ is a triangulated category (Proposition 13.10.3). $\square$

Lemma 13.29.4. Let $\mathcal{A}$ be an abelian category. A bounded below complex of injectives is K-injective.

Lemma 13.29.5. Let $\mathcal{A}$ be an abelian category. Let $T$ be a set and for each $t \in T$ let $I_ t^\bullet$ be a K-injective complex. If $I^ n = \prod _ t I_ t^ n$ exists for all $n$, then $I^\bullet$ is a K-injective complex. Moreover, $I^\bullet$ represents the product of the objects $I_ t^\bullet$ in $D(\mathcal{A})$.

Proof. Let $K^\bullet$ be an complex. Then we have

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ) = \prod \nolimits _{t \in T} \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet )$

Since taking products is an exact functor on the category of abelian groups we see that if $K^\bullet$ is acyclic, then $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$ is acyclic because this is true for each of the complexes $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet )$. Having said this, we can use Lemma 13.29.2 to conclude that

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet , I^\bullet ) = \prod \nolimits _{t \in T} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet , I_ t^\bullet )$

and indeed $I^\bullet$ represents the product in the derived category. $\square$

Lemma 13.29.6. Let $\mathcal{A}$ be an abelian category. Let $F : K(\mathcal{A}) \to \mathcal{D}'$ be an exact functor of triangulated categories. Then $RF$ is defined at every complex in $K(\mathcal{A})$ which is quasi-isomorphic to a K-injective complex. In fact, every K-injective complex computes $RF$.

Proof. By Lemma 13.15.4 it suffices to show that $RF$ is defined at a K-injective complex, i.e., it suffices to show a K-injective complex $I^\bullet$ computes $RF$. Any quasi-isomorphism $I^\bullet \to N^\bullet$ is a homotopy equivalence as it has an inverse by Lemma 13.29.2. Thus $I^\bullet \to I^\bullet$ is a final object of $I^\bullet /\text{Qis}(\mathcal{A})$ and we win. $\square$

Lemma 13.29.7. Let $\mathcal{A}$ be an abelian category. Assume every complex has a quasi-isomorphism towards a K-injective complex. Then any exact functor $F : K(\mathcal{A}) \to \mathcal{D}'$ of triangulated categories has a right derived functor

$RF : D(\mathcal{A}) \longrightarrow \mathcal{D}'$

and $RF(I^\bullet ) = F(I^\bullet )$ for K-injective complexes $I^\bullet$.

Proof. To see this we apply Lemma 13.15.15 with $\mathcal{I}$ the collection of K-injective complexes. Since (1) holds by assumption, it suffices to prove that if $I^\bullet \to J^\bullet$ is a quasi-isomorphism of K-injective complexes, then $F(I^\bullet ) \to F(J^\bullet )$ is an isomorphism. This is clear because $I^\bullet \to J^\bullet$ is a homotopy equivalence, i.e., an isomorphism in $K(\mathcal{A})$, by Lemma 13.29.2. $\square$

The following lemma can be generalized to limits over bigger ordinals.

Lemma 13.29.8. Let $\mathcal{A}$ be an abelian category. Let

$\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$

be an inverse system of K-injective complexes. Assume

1. each $I_ n^\bullet$ is $K$-injective,

2. each map $I_{n + 1}^ m \to I_ n^ m$ is a split surjection,

3. the limits $I^ m = \mathop{\mathrm{lim}}\nolimits I_ n^ m$ exist.

Then the complex $I^\bullet$ is K-injective.

Proof. We urge the reader to skip the proof of this lemma. Let $M^\bullet$ be an acyclic complex. Let us abbreviate $H_ n(a, b) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M^ a, I_ n^ b)$. With this notation $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet )$ is the cohomology of the complex

$\prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m - 2) \to \prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m - 1) \to \prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m) \to \prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m + 1)$

in the third spot from the left. We may exchange the order of $\prod$ and $\mathop{\mathrm{lim}}\nolimits$ and each of the complexes

$\prod _ m H_ n(m, m - 2) \to \prod _ m H_ n(m, m - 1) \to \prod _ m H_ n(m, m) \to \prod _ m H_ n(m, m + 1)$

is exact by assumption (1). By assumption (2) the maps in the systems

$\ldots \to \prod _ m H_3(m, m - 2) \to \prod _ m H_2(m, m - 2) \to \prod _ m H_1(m, m - 2)$

are surjective. Thus the lemma follows from Homology, Lemma 12.28.4. $\square$

It appears that a combination of Lemmas 13.28.3, 13.29.4, and 13.29.8 produces “enough K-injectives” for any abelian category with enough injectives and countable products. Actually, this may not work! See Lemma 13.32.4 for an explanation.

Lemma 13.29.9. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $u : \mathcal{A} \to \mathcal{B}$ and $v : \mathcal{B} \to \mathcal{A}$ be additive functors. Assume

1. $u$ is right adjoint to $v$, and

2. $v$ is exact.

Then $u$ transforms K-injective complexes into K-injective complexes.

Proof. Let $I^\bullet$ be a K-injective complex of $\mathcal{A}$. Let $M^\bullet$ be a acyclic complex of $\mathcal{B}$. As $v$ is exact we see that $v(M^\bullet )$ is an acyclic complex. By adjointness we get

$0 = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(v(M^\bullet ), I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{B})}(M^\bullet , u(I^\bullet ))$

hence the lemma follows. $\square$

Comment #2301 by on

Maybe in the Definition of K-injective it was meant that if for every acyclic object the Hom is acyclic (not zero).

Comment #2302 by on

The thing that is asked to be zero in the definition is not the Hom complex, but the group of morphisms in the homotopy category. Moreover the Hom complex is not defined until later. It first shows up in the chapter on differential graded algebra when we define the differential graded category of complexes, see Example 22.19.6.

Comment #3067 by ykm on

Minor typo: Lemma 13.29.2 in proof of 3->1, the two instances of $N^\dot$ should be $M^\dot$

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