The Stacks project

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

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

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

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

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. $\square$

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

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 13.31.5. $\square$

Proof. Observe that $RF$ exists as $\mathcal{A}$ has enough K-injectives (Theorem 19.12.6 and Derived Categories, Lemma 13.31.6). 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 13.34.1 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 13.34.2 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. $\square$

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

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

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 13.31.2). 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

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

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

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 13.31.3. 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 $. $\square$

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

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

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

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. $\square$

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. $\square$