## 19.12 K-injectives in Grothendieck categories

The material in this section is taken from the paper [serpe] authored by SerpĂ©. This paper generalizes some of the results of by Spaltenstein to general Grothendieck abelian categories. Our Lemma 19.12.3 is only implicit in the paper by SerpĂ©. Our approach is to mimic Grothendieck's proof of Theorem 19.11.7.

Lemma 19.12.1. Let $\mathcal{A}$ be a Grothendieck abelian category with generator $U$. Let $c$ be the function on cardinals defined by $c(\kappa ) = |\bigoplus _{\alpha \in \kappa } U|$. If $\pi : M \to N$ is a surjection then there exists a subobject $M' \subset M$ which surjects onto $N$ with $|M'| \leq c(|N|)$.

Proof. For every proper subobject $N' \subset N$ choose a morphism $\varphi _{N'} : U \to M$ such that $U \to M \to N$ does not factor through $N'$. Set

$M' = \mathop{\mathrm{Im}}\left( \bigoplus \nolimits _{N' \subset N} \varphi _{N'} : \bigoplus \nolimits _{N' \subset N} U \longrightarrow M\right)$

Then $M'$ works. $\square$

Lemma 19.12.2. Let $\mathcal{A}$ be a Grothendieck abelian category. There exists a cardinal $\kappa$ such that given any acyclic complex $M^\bullet$ we have

1. if $M^\bullet$ is nonzero, there is a nonzero subcomplex $N^\bullet$ which is bounded above, acyclic, and $|N^ n| \leq \kappa$,

2. there exists a surjection of complexes

$\bigoplus \nolimits _{i \in I} M_ i^\bullet \longrightarrow M^\bullet$

where $M_ i^\bullet$ is bounded above, acyclic, and $|M_ i^ n| \leq \kappa$.

Proof. Choose a generator $U$ of $\mathcal{A}$. Denote $c$ the function of Lemma 19.12.1. Set $\kappa = \sup \{ c^ n(|U|), n = 1, 2, 3, \ldots \}$. Let $n \in \mathbf{Z}$ and let $\psi : U \to M^ n$ be a morphism. In order to prove (1) and (2) it suffices to prove there exists a subcomplex $N^\bullet \subset M^\bullet$ which is bounded above, acyclic, and $|N^ m| \leq \kappa$, such that $\psi$ factors through $N^ n$. To do this set $N^ n = \mathop{\mathrm{Im}}(\psi )$, $N^{n + 1} = \mathop{\mathrm{Im}}(U \to M^ n \to M^{n + 1})$, and $N^ m = 0$ for $m \geq n + 2$. Suppose we have constructed $N^ m \subset M^ m$ for all $m \geq k$ such that

1. $\text{d}(N^ m) \subset N^{m + 1}$, $m \geq k$,

2. $\mathop{\mathrm{Im}}(N^{m - 1} \to N^ m) = \mathop{\mathrm{Ker}}(N^ m \to N^{m + 1})$ for all $m \geq k + 1$, and

3. $|N^ m| \leq c^{\max \{ n - m, 0\} }(|U|)$

for some $k \leq n$. Because $M^\bullet$ is acyclic, we see that the subobject $\text{d}^{-1}(\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})) \subset M^{k - 1}$ surjects onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$. Thus we can choose $N^{k - 1} \subset M^{k - 1}$ surjecting onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$ with $|N^{k - 1}| \leq c^{n - k + 1}(|U|)$ by Lemma 19.12.1. The proof is finished by induction on $k$. $\square$

Lemma 19.12.3. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $\kappa$ be a cardinal as in Lemma 19.12.2. Suppose that $I^\bullet$ is a complex such that

1. each $I^ j$ is injective, and

2. for every bounded above acyclic complex $M^\bullet$ such that $|M^ n| \leq \kappa$ we have $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$.

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

Proof. Let $M^\bullet$ be an acyclic complex. We are going to construct by induction on the ordinal $\alpha$ an acyclic subcomplex $K_\alpha ^\bullet \subset M^\bullet$ as follows. For $\alpha = 0$ we set $K_0^\bullet = 0$. For $\alpha > 0$ we proceed as follows:

1. If $\alpha = \beta + 1$ and $K_\beta ^\bullet = M^\bullet$ then we choose $K_\alpha ^\bullet = K_\beta ^\bullet$.

2. If $\alpha = \beta + 1$ and $K_\beta ^\bullet \not= M^\bullet$ then $M^\bullet /K_\beta ^\bullet$ is a nonzero acyclic complex. We choose a subcomplex $N_\alpha ^\bullet \subset M^\bullet /K_\beta ^\bullet$ as in Lemma 19.12.2. Finally, we let $K_\alpha ^\bullet \subset M^\bullet$ be the inverse image of $N_\alpha ^\bullet$.

3. If $\alpha$ is a limit ordinal we set $K_\beta ^\bullet = \mathop{\mathrm{colim}}\nolimits K_\alpha ^\bullet$.

It is clear that $M^\bullet = K_\alpha ^\bullet$ for a suitably large ordinal $\alpha$. We will prove that

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K_\alpha ^\bullet , I^\bullet )$

is zero by transfinite induction on $\alpha$. It holds for $\alpha = 0$ since $K_0^\bullet$ is zero. Suppose it holds for $\beta$ and $\alpha = \beta + 1$. In case (1) of the list above the result is clear. In case (2) there is a short exact sequence of complexes

$0 \to K_\beta ^\bullet \to K_\alpha ^\bullet \to N_\alpha ^\bullet \to 0$

Since each component of $I^\bullet$ is injective we see that we obtain an exact sequence

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K_\beta ^\bullet , I^\bullet ) \leftarrow \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K_\alpha ^\bullet , I^\bullet ) \leftarrow \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N_\alpha ^\bullet , I^\bullet )$

By induction the term on the left is zero and by assumption on $I^\bullet$ the term on the right is zero. Thus the middle group is zero too. Finally, suppose that $\alpha$ is a limit ordinal. Then we see that

$\mathop{\mathrm{Hom}}\nolimits ^\bullet (K_\alpha ^\bullet , I^\bullet ) = \mathop{\mathrm{lim}}\nolimits _{\beta < \alpha } \mathop{\mathrm{Hom}}\nolimits ^\bullet (K_\beta ^\bullet , I^\bullet )$

with notation as in More on Algebra, Section 15.71. These complexes compute morphisms in $K(\mathcal{A})$ by More on Algebra, Equation (15.71.0.1). Note that the transition maps in the system are surjective because $I^ j$ is injective for each $j$. Moreover, for a limit ordinal $\alpha$ we have equality of limit and value (see displayed formula above). Thus we may apply Homology, Lemma 12.31.8 to conclude. $\square$

Lemma 19.12.4. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $(K_ i^\bullet )_{i \in I}$ be a set of acyclic complexes. There exists a functor $M^\bullet \mapsto \mathbf{M}^\bullet (M^\bullet )$ and a natural transformation $j_{M^\bullet } : M^\bullet \to \mathbf{M}^\bullet (M^\bullet )$ such

1. $j_{M^\bullet }$ is a (termwise) injective quasi-isomorphism, and

2. for every $i \in I$ and $w : K_ i^\bullet \to M^\bullet$ the morphism $j_{M^\bullet } \circ w$ is homotopic to zero.

Proof. For every $i \in I$ choose a (termwise) injective map of complexes $K_ i^\bullet \to L_ i^\bullet$ which is homotopic to zero with $L_ i^\bullet$ quasi-isomorphic to zero. For example, take $L_ i^\bullet$ to be the cone on the identity of $K_ i^\bullet$. We define $\mathbf{M}^\bullet (M^\bullet )$ by the following pushout diagram

$\xymatrix{ \bigoplus _{i \in I} \bigoplus _{w : K_ i^\bullet \to M^\bullet } K_ i^\bullet \ar[r] \ar[d] & M^\bullet \ar[d] \\ \bigoplus _{i \in I} \bigoplus _{w : K_ i^\bullet \to M^\bullet } L_ i^\bullet \ar[r] & \mathbf{M}^\bullet (M^\bullet ). }$

Then $M^\bullet \to \mathbf{M}^\bullet (M^\bullet )$ is a functor. The right vertical arrow defines the functorial injective map $j_{M^\bullet }$. The cokernel of $j_{M^\bullet }$ is isomorphic to the direct sum of the cokernels of the maps $K_ i^\bullet \to L_ i^\bullet$ hence acyclic. Thus $j_{M^\bullet }$ is a quasi-isomorphism. Part (2) holds by construction. $\square$

Lemma 19.12.5. Let $\mathcal{A}$ be a Grothendieck abelian category. There exists a functor $M^\bullet \mapsto \mathbf{N}^\bullet (M^\bullet )$ and a natural transformation $j_{M^\bullet } : M^\bullet \to \mathbf{N}^\bullet (M^\bullet )$ such

1. $j_{M^\bullet }$ is a (termwise) injective quasi-isomorphism, and

2. for every $n \in \mathbf{Z}$ the map $M^ n \to \mathbf{N}^ n(M^\bullet )$ factors through a subobject $I^ n \subset \mathbf{N}^ n(M^\bullet )$ where $I^ n$ is an injective object of $\mathcal{A}$.

Proof. Choose a functorial injective embeddings $i_ M : M \to I(M)$, see Theorem 19.11.7. For every complex $M^\bullet$ denote $J^\bullet (M^\bullet )$ the complex with terms $J^ n(M^\bullet ) = I(M^ n) \oplus I(M^{n + 1})$ and differential

$d_{J^\bullet (M^\bullet )} = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)$

There exists a canonical injective map of complexes $u_{M^\bullet } : M^\bullet \to J^\bullet (M^\bullet )$ by mapping $M^ n$ to $I(M^ n) \oplus I(M^{n + 1})$ via the maps $i_{M^ n} : M^ n \to I(M^ n)$ and $i_{M^{n + 1}} \circ d : M^ n \to M^{n + 1} \to I(M^{n + 1})$. Hence a short exact sequence of complexes

$0 \to M^\bullet \xrightarrow {u_{M^\bullet }} J^\bullet (M^\bullet ) \xrightarrow {v_{M^\bullet }} Q^\bullet (M^\bullet ) \to 0$

functorial in $M^\bullet$. Set

$\mathbf{N}^\bullet (M^\bullet ) = C(v_{M^\bullet })^\bullet [-1].$

Note that

$\mathbf{N}^ n(M^\bullet ) = Q^{n - 1}(M^\bullet ) \oplus J^ n(M^\bullet )$

with differential

$\left( \begin{matrix} - d^{n - 1}_{Q^\bullet (M^\bullet )} & - v^ n_{M^\bullet } \\ 0 & d^ n_{J^\bullet (M)} \end{matrix} \right)$

Hence we see that there is a map of complexes $j_{M^\bullet } : M^\bullet \to \mathbf{N}^\bullet (M^\bullet )$ induced by $u$. It is injective and factors through an injective subobject by construction. The map $j_{M^\bullet }$ is a quasi-isomorphism as one can prove by looking at the long exact sequence of cohomology associated to the short exact sequences of complexes above. $\square$

Theorem 19.12.6. Let $\mathcal{A}$ be a Grothendieck abelian category. For every complex $M^\bullet$ there exists a quasi-isomorphism $M^\bullet \to I^\bullet$ such that $M^ n \to I^ n$ is injective and $I^ n$ is an injective object of $\mathcal{A}$ for all $n$ and $I^\bullet$ is a K-injective complex. Moreover, the construction is functorial in $M^\bullet$.

Proof. Please compare with the proof of Theorem 19.2.8 and Theorem 19.11.7. Choose a cardinal $\kappa$ as in Lemmas 19.12.2 and 19.12.3. Choose a set $(K_ i^\bullet )_{i \in I}$ of bounded above, acyclic complexes such that every bounded above acyclic complex $K^\bullet$ such that $|K^ n| \leq \kappa$ is isomorphic to $K_ i^\bullet$ for some $i \in I$. This is possible by Lemma 19.11.4. Denote $\mathbf{M}^\bullet (-)$ the functor constructed in Lemma 19.12.4. Denote $\mathbf{N}^\bullet (-)$ the functor constructed in Lemma 19.12.5. Both of these functors come with injective transformations $\text{id} \to \mathbf{M}$ and $\text{id} \to \mathbf{N}$.

Using transfinite recursion we define a sequence of functors $\mathbf{T}_\alpha (-)$ and corresponding transformations $\text{id} \to \mathbf{T}_\alpha$. Namely we set $\mathbf{T}_0(M^\bullet ) = M^\bullet$. If $\mathbf{T}_\alpha$ is given then we set

$\mathbf{T}_{\alpha + 1}(M^\bullet ) = \mathbf{N}^\bullet (\mathbf{M}^\bullet (\mathbf{T}_\alpha (M^\bullet )))$

If $\beta$ is a limit ordinal we set

$\mathbf{T}_\beta (M^\bullet ) = \mathop{\mathrm{colim}}\nolimits _{\alpha < \beta } \mathbf{T}_\alpha (M^\bullet )$

The transition maps of the system are injective quasi-isomorphisms. By AB5 we see that the colimit is still quasi-isomorphic to $M^\bullet$. We claim that $M^\bullet \to \mathbf{T}_\alpha (M^\bullet )$ does the job if the cofinality of $\alpha$ is larger than $\max (\kappa , |U|)$ where $U$ is a generator of $\mathcal{A}$. Namely, it suffices to check conditions (1) and (2) of Lemma 19.12.3.

For (1) we use the criterion of Lemma 19.11.6. Suppose that $M \subset U$ and $\varphi : M \to \mathbf{T}^ n_\alpha (M^\bullet )$ is a morphism for some $n \in \mathbf{Z}$. By Proposition 19.11.5 we see that $\varphi$ factor through $\mathbf{T}^ n_{\alpha '}(M^\bullet )$ for some $\alpha ' < \alpha$. In particular, by the construction of the functor $\mathbf{N}^\bullet (-)$ we see that $\varphi$ factors through an injective object of $\mathcal{A}$ which shows that $\varphi$ lifts to a morphism on $U$.

For (2) let $w : K^\bullet \to \mathbf{T}_\alpha (M^\bullet )$ be a morphism of complexes where $K^\bullet$ is a bounded above acyclic complex such that $|K^ n| \leq \kappa$. Then $K^\bullet \cong K_ i^\bullet$ for some $i \in I$. Moreover, by Proposition 19.11.5 once again we see that $w$ factor through $\mathbf{T}^ n_{\alpha '}(M^\bullet )$ for some $\alpha ' < \alpha$. In particular, by the construction of the functor $\mathbf{M}^\bullet (-)$ we see that $w$ is homotopic to zero. This finishes the proof. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).