19.12 K-injectives in Grothendieck categories

The material in this section is taken from the paper [serpe] authored by Serpé. 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 Serpé. 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 $|N'| \leq c(|N|)$.

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

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

Then $N'$ 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

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

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 surjective 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. $\square$

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