The Stacks project

19.13 Additional remarks on Grothendieck abelian categories

In this section we put some results on Grothendieck abelian categories which are folklore.

Lemma 19.13.1. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $F : \mathcal{A}^{opp} \to \textit{Sets}$ be a functor. Then $F$ is representable if and only if $F$ commutes with colimits, i.e.,

\[ F(\mathop{\mathrm{colim}}\nolimits _ i N_ i) = \mathop{\mathrm{lim}}\nolimits F(N_ i) \]

for any diagram $\mathcal{I} \to \mathcal{A}$, $i \in \mathcal{I}$.

Proof. If $F$ is representable, then it commutes with colimits by definition of colimits.

Assume that $F$ commutes with colimits. Then $F(M \oplus N) = F(M) \times F(N)$ and we can use this to define a group structure on $F(M)$. Hence we get $F : \mathcal{A} \to \textit{Ab}$ which is additive and right exact, i.e., transforms a short exact sequence $0 \to K \to L \to M \to 0$ into an exact sequence $F(K) \leftarrow F(L) \leftarrow F(M) \leftarrow 0$ (compare with Homology, Section 12.7).

Let $U$ be a generator for $\mathcal{A}$. Set $A = \bigoplus _{s \in F(U)} U$. Let $s_{univ} = (s)_{s \in F(U)} \in F(A) = \prod _{s \in F(U)} F(U)$. Let $A' \subset A$ be the largest subobject such that $s_{univ}$ restricts to zero on $A'$. This exists because $\mathcal{A}$ is a Grothendieck category and because $F$ commutes with colimits. Because $F$ commutes with colimits there exists a unique element $\overline{s}_{univ} \in F(A/A')$ which maps to $s_{univ}$ in $F(A)$. We claim that $A/A'$ represents $F$, in other words, the Yoneda map

\[ \overline{s}_{univ} : h_{A/A'} \longrightarrow F \]

is an isomorphism. Let $M \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $s \in F(M)$. Consider the surjection

\[ c_ M : A_ M = \bigoplus \nolimits _{\varphi \in \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, M)} U \longrightarrow M. \]

This gives $F(c_ M)(s) = (s_\varphi ) \in \prod _\varphi F(U)$. Consider the map

\[ \psi : A_ M = \bigoplus \nolimits _{\varphi \in \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, M)} U \longrightarrow \bigoplus \nolimits _{s \in F(U)} U = A \]

which maps the summand corresponding to $\varphi $ to the summand corresponding to $s_\varphi $ by the identity map on $U$. Then $s_{univ}$ maps to $(s_\varphi )_\varphi $ by construction. in other words the right square in the diagram

\[ \xymatrix{ A' \ar[r] & A \ar@{..>}[r]_{s_{univ}} & F \\ K \ar[r] \ar[u]^{?} & A_ M \ar[u]^\psi \ar[r] & M \ar@{..>}[u]_ s } \]

commutes. Let $K = \mathop{\mathrm{Ker}}(A_ M \to M)$. Since $s$ restricts to zero on $K$ we see that $\psi (K) \subset A'$ by definition of $A'$. Hence there is an induced morphism $M \to A/A'$. This construction gives an inverse to the map $h_{A/A'}(M) \to F(M)$ (details omitted). $\square$

Proof. Let $M_ i$, $i \in I$ be a family of objects of $\mathcal{A}$ indexed by a set $I$. The functor $F = \prod _{i \in I} h_{M_ i}$ commutes with colimits. Hence Lemma 19.13.1 applies. $\square$

Remark 19.13.3. In the chapter on derived categories we consistently work with “small” abelian categories (as is the convention in the Stacks project). For a “big” abelian category $\mathcal{A}$ it isn't clear that the derived category $D(\mathcal{A})$ exists because it isn't clear that morphisms in the derived category are sets. In general this isn't true, see Examples, Lemma 104.54.1. However, if $\mathcal{A}$ is a Grothendieck abelian category, and given $K^\bullet , L^\bullet $ in $K(\mathcal{A})$, then by Theorem 19.12.6 there exists a quasi-isomorphism $L^\bullet \to I^\bullet $ to a K-injective complex $I^\bullet $ and Derived Categories, Lemma 13.29.2 shows that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet , L^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ) \]

which is a set. Some examples of Grothendieck abelian categories are the category of modules over a ring, or more generally the category of sheaves of modules on a ringed site.

Lemma 19.13.4. Let $\mathcal{A}$ be a Grothendieck abelian category. Then

  1. $D(\mathcal{A})$ has both direct sums and products,

  2. direct sums are obtained by taking termwise direct sums of any complexes,

  3. products are obtained by taking termwise products of K-injective complexes.

Proof. Let $K^\bullet _ i$, $i \in I$ be a family of objects of $D(\mathcal{A})$ indexed by a set $I$. We claim that the termwise direct sum $\bigoplus _{i \in I} K^\bullet _ i$ is a direct sum in $D(\mathcal{A})$. Namely, let $I^\bullet $ be a K-injective complex. Then we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(\bigoplus \nolimits _{i \in I} K^\bullet _ i, I^\bullet ) & = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(\bigoplus \nolimits _{i \in I} K^\bullet _ i, I^\bullet ) \\ & = \prod \nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet _ i, I^\bullet ) \\ & = \prod \nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet _ i, I^\bullet ) \end{align*}

as desired. This is sufficient since any complex can be represented by a K-injective complex by Theorem 19.12.6. To construct the product, choose a K-injective resolution $K_ i^\bullet \to I_ i^\bullet $ for each $i$. Then we claim that $\prod _{i \in I} I_ i^\bullet $ is a product in $D(\mathcal{A})$. This follows from Derived Categories, Lemma 13.29.5. $\square$

Remark 19.13.5. Let $R$ be a ring. Suppose that $M_ n$, $n \in \mathbf{Z}$ are $R$-modules. Denote $E_ n = M_ n[-n] \in D(R)$. We claim that $E = \bigoplus M_ n[-n]$ is both the direct sum and the product of the objects $E_ n$ in $D(R)$. To see that it is the direct sum, take a look at the proof of Lemma 19.13.4. To see that it is the direct product, take injective resolutions $M_ n \to I_ n^\bullet $. By the proof of Lemma 19.13.4 we have

\[ \prod E_ n = \prod I_ n^\bullet [-n] \]

in $D(R)$. Since products in $\text{Mod}_ R$ are exact, we see that $\prod I_ n^\bullet $ is quasi-isomorphic to $E$. This works more generally in $D(\mathcal{A})$ where $\mathcal{A}$ is a Grothendieck abelian category with Ab4*.

Lemma 19.13.6. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor of abelian categories. Assume

  1. $\mathcal{A}$ is a Grothendieck abelian category,

  2. $\mathcal{B}$ has exact countable products, and

  3. $F$ commutes with countable products.

Then $RF : D(\mathcal{A}) \to D(\mathcal{B})$ commutes with derived limits.

Proof. Observe that $RF$ exists as $\mathcal{A}$ has enough K-injectives (Theorem 19.12.6 and Derived Categories, Lemma 13.29.6). The statement means that if $K = R\mathop{\mathrm{lim}}\nolimits K_ n$, then $RF(K) = R\mathop{\mathrm{lim}}\nolimits RF(K_ n)$. See Derived Categories, Definition 13.32.1 for notation. Since $RF$ is an exact functor of triangulated categories it suffices to see that $RF$ commutes with countable products of objects of $D(\mathcal{A})$. In the proof of Lemma 19.13.4 we have seen that products in $D(\mathcal{A})$ are computed by taking products of K-injective complexes and moreover that a product of K-injective complexes is K-injective. Moreover, in Derived Categories, Lemma 13.32.2 we have seen that products in $D(\mathcal{B})$ are computed by taking termwise products. Since $RF$ is computed by applying $F$ to a K-injective representative and since we've assumed $F$ commutes with countable products, the lemma follows. $\square$

The following lemma is some kind of generalization of the existence of Cartan-Eilenberg resolutions (Derived Categories, Section 13.21).

Lemma 19.13.7. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $K^\bullet $ be a filtered complex of $\mathcal{A}$, see Homology, Definition 12.21.1. Then there exists a morphism $j : K^\bullet \to J^\bullet $ of filtered complexes of $\mathcal{A}$ such that

  1. $J^ n$, $F^ pJ^ n$, $J^ n/F^ pJ^ n$ and $F^ pJ^ n/F^{p'}J^ n$ are injective objects of $\mathcal{A}$,

  2. $J^\bullet $, $F^ pJ^\bullet $, $J^\bullet /F^ pJ^\bullet $, and $F^ pJ^\bullet /F^{p'}J^\bullet $ are K-injective complexes,

  3. $j$ induces quasi-isomorphisms $K^\bullet \to J^\bullet $, $F^ pK^\bullet \to F^ pJ^\bullet $, $K^\bullet /F^ pK^\bullet \to J^\bullet /F^ pJ^\bullet $, and $F^ pK^\bullet /F^{p'}K^\bullet \to F^ pJ^\bullet /F^{p'}J^\bullet $.

Proof. By Theorem 19.12.6 we obtain quasi-isomorphisms $i : K^\bullet \to I^\bullet $ and $i^ p : F^ pK^\bullet \to I^{p, \bullet }$ as well as commutative diagrams

\[ \vcenter { \xymatrix{ K^\bullet \ar[d]_ i & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^\bullet & I^{p, \bullet } \ar[l]_{\alpha ^ p} } } \quad \text{and}\quad \vcenter { \xymatrix{ F^{p'}K^\bullet \ar[d]_{i^{p'}} & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^{p', \bullet } & I^{p, \bullet } \ar[l]_{\alpha ^{p p'}} } } \quad \text{for }p' \leq p \]

such that $\alpha ^ p \circ \alpha ^{p' p} = \alpha ^{p'}$ and $\alpha ^{p'p''} \circ \alpha ^{pp'} = \alpha ^{pp''}$. The problem is that the maps $\alpha ^ P : I^{p, \bullet } \to I^\bullet $ need not be injective. For each $p$ we choose an injection $t^ p : I^{p, \bullet } \to J^{p, \bullet }$ into an acyclic K-injective complex $J^{p, \bullet }$ whose terms are injective objects of $\mathcal{A}$ (first map to the cone on the identity and then use the theorem). Choose a map of complexes $s^ p : I^\bullet \to J^{p, \bullet }$ such that the following diagram commutes

\[ \xymatrix{ K^\bullet \ar[d]_ i & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^\bullet \ar[rd]_{s^ p} & I^{p, \bullet } \ar[d]^{t^ p} \\ & J^{p, \bullet } } \]

This is possible: the composition $F^ pK^\bullet \to J^{p, \bullet }$ is homotopic to zero because $J^{p, \bullet }$ is acyclic and K-injective (Derived Categories, Lemma 13.29.2). Since the objects $J^{p, n - 1}$ are injective and since $F^ pK^ n \to K^ n \to I^ n$ are injective morphisms, we can lift the maps $F^ pK^ n \to J^{p, n - 1}$ giving the homotopy to a map $h^ n : I^ n \to J^{p, n - 1}$. Then we set $s^ p$ equal to $h \circ \text{d} + \text{d} \circ h$. (Warning: It will not be the case that $t^ p = s^ p \circ \alpha ^ p$, so we have to be careful not to use this below.)

Consider

\[ J^\bullet = I^\bullet \times \prod \nolimits _ p J^{p, \bullet } \]

Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma 19.13.4) and since $J^{p, \bullet }$ is isomorphic to $0$ in $D(\mathcal{A})$ we see that $J^\bullet \to I^\bullet $ is an isomorphism in $D(\mathcal{A})$. Consider the map

\[ j = i \times (s^ p \circ i)_{p \in \mathbf{Z}} : K^\bullet \longrightarrow I^\bullet \times \prod \nolimits _ p J^{p, \bullet } = J^\bullet \]

By our remarks above this is a quasi-isomorphism. It is also injective. For $p \in \mathbf{Z}$ we let $F^ pJ^\bullet \subset J^\bullet $ be

\[ \mathop{\mathrm{Im}}\left( \alpha ^ p \times (t^{p'} \circ \alpha ^{pp'})_{p' \leq p} : I^{p, \bullet } \to I^\bullet \times \prod \nolimits _{p' \leq p} J^{p', \bullet } \right) \times \prod \nolimits _{p' > p} J^{p', \bullet } \]

This complex is isomorphic to the complex $I^{p, \bullet } \times \prod _{p' > p} J^{p, \bullet }$ as $\alpha ^{pp} = \text{id}$ and $t^ p$ is injective. Hence $F^ pJ^\bullet $ is quasi-isomorphic to $I^{p, \bullet }$ (argue as above). We have $j(F^ pK^\bullet ) \subset F^ pJ^\bullet $ because of the commutativity of the diagram above. The corresponding map of complexes $F^ pK^\bullet \to F^ pJ^\bullet $ is a quasi-isomorphism by what we just said. Finally, to see that $F^{p + 1}J^\bullet \subset F^ pJ^\bullet $ use that $\alpha ^{p + 1p} \circ \alpha ^{pp'} = \alpha ^{p + 1p'}$ and the commutativity of the first displayed diagram in the first paragraph of the proof.

We claim that $j : K^\bullet \to J^\bullet $ is a solution to the problem posed by the lemma. Namely, $F^ pJ^ n$ is an injective object of $\mathcal{A}$ because it is isomorphic to $I^{p, n} \times \prod _{p' > p} J^{p', n}$ and products of injectives are injective. Then the injective map $F^ pJ^ n \to J^ n$ splits and hence the quotient $J^ n/F^ pJ^ n$ is injective as well as a direct summand of the injective object $J^ n$. Similarly for $F^ pJ^ n/F^{p'}J^ n$. This in particular means that $0 \to F^ pJ^\bullet \to J^\bullet \to J^\bullet /F^ pJ^\bullet \to 0$ is a termwise split short exact sequence of complexes, hence defines a distinguished triangle in $K(\mathcal{A})$ by fiat. Since $J^\bullet $ and $F^ pJ^\bullet $ are K-injective complexes we see that the same is true for $J^\bullet /F^ pJ^\bullet $ by Derived Categories, Lemma 13.29.3. A similar argument shows that $F^ pJ^\bullet /F^{p'}J^\bullet $ is K-injective. By construction $j : K^\bullet \to J^\bullet $ and the induced maps $F^ pK^\bullet \to F^ pJ^\bullet $ are quasi-isomorphisms. Using the long exact cohomology sequences of the complexes in play we find that the same holds for $K^\bullet /F^ pK^\bullet \to J^\bullet /F^ pJ^\bullet $ and $F^ pK^\bullet /F^{p'}K^\bullet \to F^ pJ^\bullet /F^{p'}J^\bullet $. $\square$

Lemma 19.13.8. Let $\mathcal{A}$ be a Grothendieck abelian category. Suppose given an object $E \in D(\mathcal{A})$ and an inverse system $\{ E^ i\} _{i \in \mathbf{Z}}$ of objects of $D(\mathcal{A})$ over $\mathbf{Z}$ together with a compatible system of maps $E^ i \to E$. Picture:

\[ \ldots \to E^{i + 1} \to E^ i \to E^{i - 1} \to \ldots \to E \]

Then there exists a filtered complex $K^\bullet $ of $\mathcal{A}$ (Homology, Definition 12.21.1) such that $K^\bullet $ represents $E$ and $F^ iK^\bullet $ represents $E^ i$ compatibly with the given maps.

Proof. By Theorem 19.12.6 we can choose a K-injective complex $I^\bullet $ representing $E$ all of whose terms $I^ n$ are injective objects of $\mathcal{A}$. Choose a complex $G^{0, \bullet }$ representing $E^0$. Choose a map of complexes $\varphi ^0 : G^{0, \bullet } \to I^\bullet $ representing $E^0 \to E$. For $i > 0$ we inductively represent $E^ i \to E^{i - 1}$ by a map of complexes $\delta : G^{i, \bullet } \to G^{i - 1, \bullet }$ and we set $\varphi ^ i = \delta \circ \varphi ^{i - 1}$. For $i < 0$ we inductively represent $E^{i + 1} \to E^ i$ by a termwise injective map of complexes $\delta : G^{i + 1, \bullet } \to G^{i, \bullet }$ (for example you can use Derived Categories, Lemma 13.9.6). Claim: we can find a map of complexes $\varphi ^ i : G^{i, \bullet } \to I^\bullet $ representing the map $E^ i \to E$ and fitting into the commutative diagram

\[ \xymatrix{ G^{i + 1, \bullet } \ar[r]_\delta \ar[d]_{\varphi ^{i + 1}} & G^{i, \bullet } \ar[ld]^{\varphi ^ i} \\ I^\bullet } \]

Namely, we first choose any map of complexes $\varphi : G^{i, \bullet } \to I^\bullet $ representing the map $E^ i \to E$. Then we see that $\varphi \circ \delta $ and $\varphi ^{i + 1}$ are homotopic by some homotopy $h^ p : G^{i + 1, p} \to I^{p - 1}$. Since the terms of $I^\bullet $ are injective and since $\delta $ is termwise injective, we can lift $h^ p$ to $(h')^ p : G^{i, p} \to I^{p - 1}$. Then we set $\varphi ^ i = \varphi + h' \circ d + d \circ h'$ and we get what we claimed.

Next, we choose for every $i$ a termwise injective map of complexes $a^ i : G^{i, \bullet } \to J^{i, \bullet }$ with $J^{i, \bullet }$ acyclic, K-injective, with $J^{i, p}$ injective objects of $\mathcal{A}$. To do this first map $G^{i, \bullet }$ to the cone on the identity and then apply the theorem cited above. Arguing as above we can find maps of complexes $\delta ' : J^{i, \bullet } \to J^{i - 1, \bullet }$ such that the diagrams

\[ \xymatrix{ G^{i, \bullet } \ar[r]_\delta \ar[d]_{a^ i} & G^{i - 1, \bullet } \ar[d]^{a^{i - 1}} \\ J^{i, \bullet } \ar[r]^{\delta '} & J^{i - 1, \bullet } } \]

commute. (You could also use the functoriality of cones plus the functoriality in the theorem to get this.) Then we consider the maps

\[ \xymatrix{ G^{i + 1, \bullet } \times \prod \nolimits _{p > i + 1} J^{p, \bullet } \ar[r] \ar[rd] & G^{i, \bullet } \times \prod \nolimits _{p > i} J^{p, \bullet } \ar[r] \ar[d] & G^{i - 1, \bullet } \times \prod \nolimits _{p > i - 1} J^{p, \bullet } \ar[ld] \\ & I^\bullet \times \prod \nolimits _ p J^{p, \bullet } } \]

Here the arrows on $J^{p, \bullet }$ are the obvious ones (identity or zero). On the factor $G^{i, \bullet }$ we use $\delta : G^{i, \bullet } \to G^{i - 1, \bullet }$, the map $\varphi ^ i : G^{i, \bullet } \to I^\bullet $, the zero map $0 : G^{i, \bullet } \to J^{p, \bullet }$ for $p > i$, the map $a^ i : G^{i, \bullet } \to J^{p, \bullet }$ for $p = i$, and $(\delta ')^{i - p} \circ a^ i = a^ p \circ \delta ^{i - p} : G^{i, \bullet } \to J^{p, \bullet }$ for $p < i$. We omit the verification that all the arrows in the diagram are termwise injective. Thus we obtain a filtered complex. Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma 19.13.4) and because $J^{p, \bullet }$ is zero in $D(\mathcal{A})$ we conclude this diagram represents the given diagram in the derived category. This finishes the proof. $\square$

Lemma 19.13.9. In the situation of Lemma 19.13.8 assume we have a second inverse system $\{ (E')^ i\} _{i \in \mathbf{Z}}$ and a compatible system of maps $(E')^ i \to E$. Then there exists a bi-filtered complex $K^\bullet $ of $\mathcal{A}$ such that $K^\bullet $ represents $E$, $F^ iK^\bullet $ represents $E^ i$, and $(F')^ iK^\bullet $ represents $(E')^ i$ compatibly with the given maps.

Proof. Using the lemma we can first choose $K^\bullet $ and $F$. Then we can choose $(K')^\bullet $ and $F'$ which work for $\{ (E')^ i\} _{i \in \mathbf{Z}}$ and the maps $(E')^ i \to E$. Using Lemma 19.13.7 we can assume $K^\bullet $ is a K-injective complex. Then we can choose a map of complexes $(K')^\bullet \to K^\bullet $ corresponding to the given identifications $(K')^\bullet \cong E \cong K^\bullet $. We can additionally choose a termwise injective map $(K')^\bullet \to J^\bullet $ with $J^\bullet $ acyclic and K-injective. (To do this first map $(K')^\bullet $ to the cone on the identity and then apply Theorem 19.12.6.) Then $(K')^\bullet \to K^\bullet \times J^\bullet $ and $K^\bullet \to K^\bullet \times J^\bullet $ are both termwise injective and quasi-isomorphisms (as the product represents $E$ by Lemma 19.13.4). Then we can simply take the images of the filtrations on $K^\bullet $ and $(K')^\bullet $ under these maps to conclude. $\square$


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