Definition 13.31.1. Let $\mathcal{A}$ be an abelian category. A complex $I^\bullet $ is K-injective if for every acyclic complex $M^\bullet $ we have $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$.
13.31 K-injective complexes
The following types of complexes can be used to compute right derived functors on the unbounded derived category.
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 13.31.2. Let $\mathcal{A}$ be an abelian category. Let $I^\bullet $ be a complex. The following are equivalent
$I^\bullet $ is K-injective,
for every quasi-isomorphism $M^\bullet \to N^\bullet $ the map
is bijective, and
for every complex $N^\bullet $ the map
is an isomorphism.
\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet ) \]
\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(N^\bullet , I^\bullet ) \]
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 13.31.3. 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 13.31.4. Let $\mathcal{A}$ be an abelian category. A bounded below complex of injectives is K-injective.
Lemma 13.31.5. 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. Observe that the complex
\[ C : \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I^{b - 1}) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I^ b) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I^{b + 1}) \]
has cohomology $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$ in the middle. Similarly, the complex
\[ C_ t : \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I_ t^{b - 1}) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I_ t^ b) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I_ t^{b + 1}) \]
computes $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet )$. Next, observe that we have
\[ C = \prod \nolimits _{t \in T} C_ t \]
as complexes of abelian groups by our choice of $I$. Taking products is an exact functor on the category of abelian groups. Hence if $K^\bullet $ is acyclic, then $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet ) = 0$, hence $C_ t$ is acyclic, hence $C$ is acyclic, hence we get $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ) = 0$. Thus we find that $I^\bullet $ is K-injective. Having said this, we can use Lemma 13.31.2 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 13.31.6. 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 13.14.4 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 13.31.2. Thus $I^\bullet \to I^\bullet $ is a final object of $I^\bullet /\text{Qis}(\mathcal{A})$ and we win. $\square$
Lemma 13.31.7. 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 and $RF(I^\bullet ) = F(I^\bullet )$ for K-injective complexes $I^\bullet $.
\[ RF : D(\mathcal{A}) \longrightarrow \mathcal{D}' \]
Proof. To see this we apply Lemma 13.14.15 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 13.31.2. $\square$
The following lemma can be generalized to limits over bigger ordinals.
Lemma 13.31.8. Let $\mathcal{A}$ be an abelian category. Let be an inverse system of complexes. Assume
each $I_ n^\bullet $ is $K$-injective,
each map $I_{n + 1}^ m \to I_ n^ m$ is a split surjection,
the limits $I^ m = \mathop{\mathrm{lim}}\nolimits I_ n^ m$ exist.
\[ \ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet \]
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 12.31.4. $\square$
It appears that a combination of Lemmas 13.29.3, 13.31.4, and 13.31.8 produces “enough K-injectives” for any abelian category with enough injectives and countable products. Actually, this may not work! See Lemma 13.34.4 for an explanation.
Lemma 13.31.9. 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
$u$ is right adjoint to $v$, and
$v$ 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. $\square$
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