Definition 13.26.1. Let $\mathcal{A}$ be an abelian category. We say an object $I$ of $\text{Fil}^ f(\mathcal{A})$ is *filtered injective* if each $\text{gr}^ p(I)$ is an injective object of $\mathcal{A}$.

## 13.26 Filtered derived category and injective resolutions

Let $\mathcal{A}$ be an abelian category. In this section we will show that if $\mathcal{A}$ has enough injectives, then so does the category $\text{Fil}^ f(\mathcal{A})$ in some sense. One can use this observation to compute in the filtered derived category of $\mathcal{A}$.

The category $\text{Fil}^ f(\mathcal{A})$ is an example of an exact category, see Injectives, Remark 19.9.6. A special role is played by the strict morphisms, see Homology, Definition 12.16.3, i.e., the morphisms $f$ such that $\mathop{\mathrm{Coim}}(f) = \mathop{\mathrm{Im}}(f)$. We will say that a complex $A \to B \to C$ in $\text{Fil}^ f(\mathcal{A})$ is *exact* if the sequence $\text{gr}(A) \to \text{gr}(B) \to \text{gr}(C)$ is exact in $\mathcal{A}$. This implies that $A \to B$ and $B \to C$ are strict morphisms, see Homology, Lemma 12.16.15.

Lemma 13.26.2. Let $\mathcal{A}$ be an abelian category. An object $I$ of $\text{Fil}^ f(\mathcal{A})$ is filtered injective if and only if there exist $a \leq b$, injective objects $I_ n$, $a \leq n \leq b$ of $\mathcal{A}$ and an isomorphism $I \cong \bigoplus _{a \leq n \leq b} I_ n$ such that $F^ pI = \bigoplus _{n \geq p} I_ n$.

**Proof.**
Follows from the fact that any injection $J \to M$ of $\mathcal{A}$ is split if $J$ is an injective object. Details omitted.
$\square$

Lemma 13.26.3. Let $\mathcal{A}$ be an abelian category. Any strict monomorphism $u : I \to A$ of $\text{Fil}^ f(\mathcal{A})$ where $I$ is a filtered injective object is a split injection.

**Proof.**
Let $p$ be the largest integer such that $F^ pI \not= 0$. In particular $\text{gr}^ p(I) = F^ pI$. Let $I'$ be the object of $\text{Fil}^ f(\mathcal{A})$ whose underlying object of $\mathcal{A}$ is $F^ pI$ and with filtration given by $F^ nI' = 0$ for $n > p$ and $F^ nI' = I' = F^ pI$ for $n \leq p$. Note that $I' \to I$ is a strict monomorphism too. The fact that $u$ is a strict monomorphism implies that $F^ pI \to A/F^{p + 1}(A)$ is injective, see Homology, Lemma 12.16.13. Choose a splitting $s : A/F^{p + 1}A \to F^ pI$ in $\mathcal{A}$. The induced morphism $s' : A \to I'$ is a strict morphism of filtered objects splitting the composition $I' \to I \to A$. Hence we can write $A = I' \oplus \mathop{\mathrm{Ker}}(s')$ and $I = I' \oplus \mathop{\mathrm{Ker}}(s'|_ I)$. Note that $\mathop{\mathrm{Ker}}(s'|_ I) \to \mathop{\mathrm{Ker}}(s')$ is a strict monomorphism and that $\mathop{\mathrm{Ker}}(s'|_ I)$ is a filtered injective object. By induction on the length of the filtration on $I$ the map $\mathop{\mathrm{Ker}}(s'|_ I) \to \mathop{\mathrm{Ker}}(s')$ is a split injection. Thus we win.
$\square$

Lemma 13.26.4. Let $\mathcal{A}$ be an abelian category. Let $u : A \to B$ be a strict monomorphism of $\text{Fil}^ f(\mathcal{A})$ and $f : A \to I$ a morphism from $A$ into a filtered injective object in $\text{Fil}^ f(\mathcal{A})$. Then there exists a morphism $g : B \to I$ such that $f = g \circ u$.

**Proof.**
The pushout $f' : I \to I \amalg _ A B$ of $f$ by $u$ is a strict monomorphism, see Homology, Lemma 12.16.10. Hence the result follows formally from Lemma 13.26.3.
$\square$

Lemma 13.26.5. Let $\mathcal{A}$ be an abelian category with enough injectives. For any object $A$ of $\text{Fil}^ f(\mathcal{A})$ there exists a strict monomorphism $A \to I$ where $I$ is a filtered injective object.

**Proof.**
Pick $a \leq b$ such that $\text{gr}^ p(A) = 0$ unless $p \in \{ a, a + 1, \ldots , b\} $. For each $n \in \{ a, a + 1, \ldots , b\} $ choose an injection $u_ n : A/F^ nA \to I_ n$ with $I_ n$ and injective object. Set $I = \bigoplus _{a \leq n \leq b} I_ p$ with filtration $F^ pI = \bigoplus _{n \geq p} I_ n$ and set $u : A \to I$ equal to the direct sum of the maps $u_ n$.
$\square$

Lemma 13.26.6. Let $\mathcal{A}$ be an abelian category with enough injectives. For any object $A$ of $\text{Fil}^ f(\mathcal{A})$ there exists a filtered quasi-isomorphism $A[0] \to I^\bullet $ where $I^\bullet $ is a complex of filtered injective objects with $I^ n = 0$ for $n < 0$.

**Proof.**
First choose a strict monomorphism $u_0 : A \to I^0$ of $A$ into a filtered injective object, see Lemma 13.26.5. Next, choose a strict monomorphism $u_1 : \mathop{\mathrm{Coker}}(u_0) \to I^1$ into a filtered injective object of $\mathcal{A}$. Denote $d^0$ the induced map $I^0 \to I^1$. Next, choose a strict monomorphism $u_2 : \mathop{\mathrm{Coker}}(u_1) \to I^2$ into a filtered injective object of $\mathcal{A}$. Denote $d^1$ the induced map $I^1 \to I^2$. And so on. This works because each of the sequences

is short exact, i.e., induces a short exact sequence on applying $\text{gr}$. To see this use Homology, Lemma 12.16.13. $\square$

Lemma 13.26.7. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $f : A \to B$ be a morphism of $\text{Fil}^ f(\mathcal{A})$. Given filtered quasi-isomorphisms $A[0] \to I^\bullet $ and $B[0] \to J^\bullet $ where $I^\bullet , J^\bullet $ are complexes of filtered injective objects with $I^ n = J^ n = 0$ for $n < 0$, then there exists a commutative diagram

**Proof.**
As $A[0] \to I^\bullet $ and $C[0] \to J^\bullet $ are filtered quasi-isomorphisms we conclude that $a : A \to I^0$, $b : B \to J^0$ and all the morphisms $d_ I^ n$, $d_ J^ n$ are strict, see Homology, Lemma 13.14.4. We will inductively construct the maps $f^ n$ in the following commutative diagram

Because $A \to I^0$ is a strict monomorphism and because $J^0$ is filtered injective, we can find a morphism $f^0 : I^0 \to J^0$ such that $f^0 \circ a = b \circ f$, see Lemma 13.26.4. The composition $d_ J^0 \circ b \circ f$ is zero, hence $d_ J^0 \circ f^0 \circ a = 0$, hence $d_ J^0 \circ f^0$ factors through a unique morphism

As $\mathop{\mathrm{Im}}(d_ I^0) \to I^1$ is a strict monomorphism we can extend the displayed arrow to a morphism $f^1 : I^1 \to J^1$ by Lemma 13.26.4 again. And so on. $\square$

Lemma 13.26.8. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $0 \to A \to B \to C \to 0$ be a short exact sequence in $\text{Fil}^ f(\mathcal{A})$. Given filtered quasi-isomorphisms $A[0] \to I^\bullet $ and $C[0] \to J^\bullet $ where $I^\bullet , J^\bullet $ are complexes of filtered injective objects with $I^ n = J^ n = 0$ for $n < 0$, then there exists a commutative diagram

where the lower row is a termwise split sequence of complexes.

**Proof.**
As $A[0] \to I^\bullet $ and $C[0] \to J^\bullet $ are filtered quasi-isomorphisms we conclude that $a : A \to I^0$, $c : C \to J^0$ and all the morphisms $d_ I^ n$, $d_ J^ n$ are strict, see Homology, Lemma 13.14.4. We are going to step by step construct the south-east and the south arrows in the following commutative diagram

As $A \to B$ is a strict monomorphism, we can find a morphism $b : B \to I^0$ such that $b \circ \alpha = a$, see Lemma 13.26.4. As $A$ is the kernel of the strict morphism $I^0 \to I^1$ and $\beta = \mathop{\mathrm{Coker}}(\alpha )$ we obtain a unique morphism $\overline{b} : C \to I^1$ fitting into the diagram. As $c$ is a strict monomorphism and $I^1$ is filtered injective we can find $\delta ^0 : J^0 \to I^1$, see Lemma 13.26.4. Because $B \to C$ is a strict epimorphism and because $B \to I^0 \to I^1 \to I^2$ is zero, we see that $C \to I^1 \to I^2$ is zero. Hence $d_ I^1 \circ \delta ^0$ is zero on $C \cong \mathop{\mathrm{Im}}(c)$. Hence $d_ I^1 \circ \delta ^0$ factors through a unique morphism

As $I^2$ is filtered injective and $\mathop{\mathrm{Im}}(d_ J^0) \to J^1$ is a strict monomorphism we can extend the displayed morphism to a morphism $\delta ^1 : J^1 \to I^2$, see Lemma 13.26.4. And so on. We set $M^\bullet = I^\bullet \oplus J^\bullet $ with differential

Finally, the map $B[0] \to M^\bullet $ is given by $b \oplus c \circ \beta : M \to I^0 \oplus J^0$. $\square$

Lemma 13.26.9. Let $\mathcal{A}$ be an abelian category with enough injectives. For every $K^\bullet \in K^{+}(\text{Fil}^ f(\mathcal{A}))$ there exists a filtered quasi-isomorphism $K^\bullet \to I^\bullet $ with $I^\bullet $ bounded below, each $I^ n$ a filtered injective object, and each $K^ n \to I^ n$ a strict monomorphism.

**Proof.**
After replacing $K^\bullet $ by a shift (which is harmless for the proof) we may assume that $K^ n = 0$ for $n < 0$. Consider the short exact sequences

of the exact category $\text{Fil}^ f(\mathcal{A})$ and the maps $u_ i : \mathop{\mathrm{Coim}}(d_ K^ i) \to \mathop{\mathrm{Ker}}(d_ K^{i + 1})$. For each $i \geq 0$ we may choose filtered quasi-isomorphisms

with $I_{ker, i}^ n, I_{coim, i}^ n$ filtered injective and zero for $n < 0$, see Lemma 13.26.6. By Lemma 13.26.7 we may lift $u_ i$ to a morphism of complexes $u_ i^\bullet : I_{coim, i}^\bullet \to I_{ker, i + 1}^\bullet $. Finally, for each $i \geq 0$ we may complete the diagrams

with the lower sequence a termwise split exact sequence, see Lemma 13.26.8. For $i \geq 0$ set $d_ i : I_ i^\bullet \to I_{i + 1}^\bullet $ equal to $d_ i = \alpha _{i + 1} \circ u_ i^\bullet \circ \beta _ i$. Note that $d_ i \circ d_{i - 1} = 0$ because $\beta _ i \circ \alpha _ i = 0$. Hence we have constructed a commutative diagram

Here the vertical arrows are filtered quasi-isomorphisms. The upper row is a complex of complexes and each complex consists of filtered injective objects with no nonzero objects in degree $< 0$. Thus we obtain a double complex by setting $I^{a, b} = I_ a^ b$ and using

the map $d_ a^ b$ and using for

the map $d_{I_ a}^ b$. Denote $\text{Tot}(I^{\bullet , \bullet })$ the total complex associated to this double complex, see Homology, Definition 12.22.3. Observe that the maps $K^ n[0] \to I_ n^\bullet $ come from maps $K^ n \to I^{n, 0}$ which give rise to a map of complexes

We claim this is a filtered quasi-isomorphism. As $\text{gr}(-)$ is an additive functor, we see that $\text{gr}(\text{Tot}(I^{\bullet , \bullet })) = \text{Tot}(\text{gr}(I^{\bullet , \bullet }))$. Thus we can use Homology, Lemma 12.22.7 to conclude that $\text{gr}(K^\bullet ) \to \text{gr}(\text{Tot}(I^{\bullet , \bullet }))$ is a quasi-isomorphism as desired. $\square$

Lemma 13.26.10. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet , I^\bullet \in K(\text{Fil}^ f(\mathcal{A}))$. Assume $K^\bullet $ is filtered acyclic and $I^\bullet $ bounded below and consisting of filtered injective objects. Any morphism $K^\bullet \to I^\bullet $ is homotopic to zero: $\mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(K^\bullet , I^\bullet ) = 0$.

**Proof.**
Let $\alpha : K^\bullet \to I^\bullet $ be a morphism of complexes. Assume that $\alpha ^ j = 0$ for $j < n$. We will show that there exists a morphism $h : K^{n + 1} \to I^ n$ such that $\alpha ^ n = h \circ d$. Thus $\alpha $ will be homotopic to the morphism of complexes $\beta $ defined by

This will clearly prove the lemma (by induction). To prove the existence of $h$ note that $\alpha ^ n \circ d_ K^{n - 1} = 0$ since $\alpha ^{n - 1} = 0$. Since $K^\bullet $ is filtered acyclic we see that $d_ K^{n - 1}$ and $d_ K^ n$ are strict and that

is an exact sequence of the exact category $\text{Fil}^ f(\mathcal{A})$, see Homology, Lemma 12.16.15. Hence we can think of $\alpha ^ n$ as a map into $I^ n$ defined on $\mathop{\mathrm{Im}}(d_ K^ n)$. Using that $\mathop{\mathrm{Im}}(d_ K^ n) \to K^{n + 1}$ is a strict monomorphism and that $I^ n$ is filtered injective we may lift this map to a map $h : K^{n + 1} \to I^ n$ as desired, see Lemma 13.26.4. $\square$

Lemma 13.26.11. Let $\mathcal{A}$ be an abelian category. Let $I^\bullet \in K(\text{Fil}^ f(\mathcal{A}))$ be a bounded below complex consisting of filtered injective objects.

Let $\alpha : K^\bullet \to L^\bullet $ in $K(\text{Fil}^ f(\mathcal{A}))$ be a filtered quasi-isomorphism. Then the map

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(L^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(K^\bullet , I^\bullet ) \]is bijective.

Let $L^\bullet \in K(\mathcal{A})$. Then

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{DF(\mathcal{A})}(L^\bullet , I^\bullet ). \]

**Proof.**
Proof of (1). Note that

is a distinguished triangle in $K(\text{Fil}^ f(\mathcal{A}))$ (Lemma 13.9.14) and $C(f)^\bullet $ is a filtered acyclic complex (Lemma 13.14.4). Then

is an exact sequence of abelian groups, see Lemma 13.4.2. At this point Lemma 13.26.10 guarantees that the outer two groups are zero and hence $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$.

Proof of (2). Let $a$ be an element of the right hand side. We may represent $a = \gamma \alpha ^{-1}$ where $\alpha : K^\bullet \to L^\bullet $ is a filtered quasi-isomorphism and $\gamma : K^\bullet \to I^\bullet $ is a map of complexes. By part (1) we can find a morphism $\beta : L^\bullet \to I^\bullet $ such that $\beta \circ \alpha $ is homotopic to $\gamma $. This proves that the map is surjective. Let $b$ be an element of the left hand side which maps to zero in the right hand side. Then $b$ is the homotopy class of a morphism $\beta : L^\bullet \to I^\bullet $ such that there exists a filtered quasi-isomorphism $\alpha : K^\bullet \to L^\bullet $ with $\beta \circ \alpha $ homotopic to zero. Then part (1) shows that $\beta $ is homotopic to zero also, i.e., $b = 0$. $\square$

Lemma 13.26.12. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $\mathcal{I}^ f \subset \text{Fil}^ f(\mathcal{A})$ denote the strictly full additive subcategory whose objects are the filtered injective objects. The canonical functor

is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories. Furthermore the diagrams

are commutative, where $\mathcal{I} \subset \mathcal{A}$ is the strictly full additive subcategory whose objects are the injective objects.

**Proof.**
The functor $K^{+}(\mathcal{I}^ f) \to DF^{+}(\mathcal{A})$ is essentially surjective by Lemma 13.26.9. It is fully faithful by Lemma 13.26.11. It is an exact functor by our definitions regarding distinguished triangles. The commutativity of the squares is immediate.
$\square$

Remark 13.26.13. We can invert the arrow of the lemma only if $\mathcal{A}$ is a category in our sense, namely if it has a set of objects. However, suppose given a big abelian category $\mathcal{A}$ with enough injectives, such as $\textit{Mod}(\mathcal{O}_ X)$ for example. Then for any given set of objects $\{ A_ i\} _{i\in I}$ there is an abelian subcategory $\mathcal{A}' \subset \mathcal{A}$ containing all of them and having enough injectives, see Sets, Lemma 3.12.1. Thus we may use the lemma above for $\mathcal{A}'$. This essentially means that if we use a set worth of diagrams, etc then we will never run into trouble using the lemma.

Let $\mathcal{A}, \mathcal{B}$ be abelian categories. Let $T : \mathcal{A} \to \mathcal{B}$ be a left exact functor. (We cannot use the letter $F$ for the functor since this would conflict too much with our use of the letter $F$ to indicate filtrations.) Note that $T$ induces an additive functor

by the rule $T(A, F) = (T(A), F)$ where $F^ pT(A) = T(F^ pA)$ which makes sense as $T$ is left exact. (Warning: It may not be the case that $\text{gr}(T(A)) = T(\text{gr}(A))$.) This induces functors of triangulated categories

The filtered right derived functor of $T$ is the right derived functor of Definition 13.15.2 for this exact functor composed with the exact functor $K^{+}(\text{Fil}^ f(\mathcal{B})) \to DF^{+}(\mathcal{B})$ and the multiplicative set $\text{FQis}^{+}(\mathcal{A})$. Assume $\mathcal{A}$ has enough injectives. At this point we can redo the discussion of Section 13.20 to define the *filtered right derived functors*

of our functor $T$.

However, instead we will proceed as in Section 13.25, and it will turn out that we can define $RT$ even if $T$ is just additive. Namely, we first choose a quasi-inverse $j' : DF^{+}(\mathcal{A}) \to K^{+}(\mathcal{I}^ f)$ of the equivalence of Lemma 13.26.12. By Lemma 13.4.17 we see that $j'$ is an exact functor of triangulated categories. Next, we note that for a filtered injective object $I$ we have a (noncanonical) decomposition

by Lemma 13.26.2. Hence if $T$ is any additive functor $T : \mathcal{A} \to \mathcal{B}$ then we get an additive functor

by setting $T_{ext}(I) = \bigoplus T(I_ p)$ with $F^ pT_{ext}(I) = \bigoplus _{q \geq p} T(I_ q)$. Note that we have the property $\text{gr}(T_{ext}(I)) = T(\text{gr}(I))$ by construction. Hence we obtain a functor

which commutes with $\text{gr}$. Then we define (13.26.13.2) by the composition

Since $RT : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is computed by injective resolutions as well, see Lemmas 13.20.1, the commutation of $T$ with $\text{gr}$, and the commutative diagrams of Lemma 13.26.12 imply that

and

as functors $DF^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.

The filtered derived functor $RT$ (13.26.13.2) induces functors

Note that since $\text{Fil}^ f(\mathcal{A})$, and $\text{Comp}^{+}(\text{Fil}^ f(\mathcal{A}))$ are no longer abelian it does not make sense to say that $RT$ restricts to a $\delta $-functor on them. (This can be repaired by thinking of these categories as exact categories and formulating the notion of a $\delta $-functor from an exact category into a triangulated category.) But it does make sense, and it is true by construction, that $RT$ is an exact functor on the triangulated category $KF^{+}(\mathcal{A})$.

Lemma 13.26.14. Let $\mathcal{A}, \mathcal{B}$ be abelian categories. Let $T : \mathcal{A} \to \mathcal{B}$ be a left exact functor. Assume $\mathcal{A}$ has enough injectives. Let $(K^\bullet , F)$ be an object of $\text{Comp}^{+}(\text{Fil}^ f(\mathcal{A}))$. There exists a spectral sequence $(E_ r, d_ r)_{r\geq 0}$ consisting of bigraded objects $E_ r$ of $\mathcal{B}$ and $d_ r$ of bidegree $(r, - r + 1)$ and with

Moreover, this spectral sequence is bounded, converges to $R^*T(K^\bullet )$, and induces a finite filtration on each $R^ nT(K^\bullet )$. The construction of this spectral sequence is functorial in the object $K^\bullet $ of $\text{Comp}^{+}(\text{Fil}^ f(\mathcal{A}))$ and the terms $(E_ r, d_ r)$ for $r \geq 1$ do not depend on any choices.

**Proof.**
Choose a filtered quasi-isomorphism $K^\bullet \to I^\bullet $ with $I^\bullet $ a bounded below complex of filtered injective objects, see Lemma 13.26.9. Consider the complex $RT(K^\bullet ) = T_{ext}(I^\bullet )$, see (13.26.13.6). Thus we can consider the spectral sequence $(E_ r, d_ r)_{r \geq 0}$ associated to this as a filtered complex in $\mathcal{B}$, see Homology, Section 12.21. By Homology, Lemma 12.21.2 we have $E_1^{p, q} = H^{p + q}(\text{gr}^ p(T(I^\bullet )))$. By Equation (13.26.13.3) we have $E_1^{p, q} = H^{p + q}(T(\text{gr}^ p(I^\bullet )))$, and by definition of a filtered injective resolution the map $\text{gr}^ p(K^\bullet ) \to \text{gr}^ p(I^\bullet )$ is an injective resolution. Hence $E_1^{p, q} = R^{p + q}T(\text{gr}^ p(K^\bullet ))$.

On the other hand, each $I^ n$ has a finite filtration and hence each $T(I^ n)$ has a finite filtration. Thus we may apply Homology, Lemma 12.21.11 to conclude that the spectral sequence is bounded, converges to $H^ n(T(I^\bullet )) = R^ nT(K^\bullet )$ moreover inducing finite filtrations on each of the terms.

Suppose that $K^\bullet \to L^\bullet $ is a morphism of $\text{Comp}^{+}(\text{Fil}^ f(\mathcal{A}))$. Choose a filtered quasi-isomorphism $L^\bullet \to J^\bullet $ with $J^\bullet $ a bounded below complex of filtered injective objects, see Lemma 13.26.9. By our results above, for example Lemma 13.26.11, there exists a diagram

which commutes up to homotopy. Hence we get a morphism of filtered complexes $T(I^\bullet ) \to T(J^\bullet )$ which gives rise to the morphism of spectral sequences, see Homology, Lemma 12.21.4. The last statement follows from this. $\square$

Remark 13.26.15. As promised in Remark 13.21.4 we discuss the connection of the lemma above with the constructions using Cartan-Eilenberg resolutions. Namely, let $T : \mathcal{A} \to \mathcal{B}$ be a left exact functor of abelian categories, assume $\mathcal{A}$ has enough injectives, and let $K^\bullet $ be a bounded below complex of $\mathcal{A}$. We give an alternative construction of the spectral sequences ${}'E$ and ${}''E$ of Lemma 13.21.3.

First spectral sequence. Consider the “stupid” filtration on $K^\bullet $ obtained by setting $F^ p(K^\bullet ) = \sigma _{\geq p}(K^\bullet )$, see Homology, Section 12.14. Note that this stupid in the sense that $d(F^ p(K^\bullet )) \subset F^{p + 1}(K^\bullet )$, compare Homology, Lemma 12.21.3. Note that $\text{gr}^ p(K^\bullet ) = K^ p[-p]$ with this filtration. According to Lemma 13.26.14 there is a spectral sequence with $E_1$ term

as in the spectral sequence ${}'E_ r$. Observe moreover that the differentials $E_1^{p, q} \to E_1^{p + 1, q}$ agree with the differentials in $'{}E_1$, see Homology, Lemma 12.21.3 part (2) and the description of ${}'d_1$ in the proof of Lemma 13.21.3.

Second spectral sequence. Consider the filtration on the complex $K^\bullet $ obtained by setting $F^ p(K^\bullet ) = \tau _{\leq -p}(K^\bullet )$, see Homology, Section 12.14. The minus sign is necessary to get a decreasing filtration. Note that $\text{gr}^ p(K^\bullet )$ is quasi-isomorphic to $H^{-p}(K^\bullet )[p]$ with this filtration. According to Lemma 13.26.14 there is a spectral sequence with $E_1$ term

with $i = 2p + q$ and $j = -p$. (This looks unnatural, but note that we could just have well developed the whole theory of filtered complexes using increasing filtrations, with the end result that this then looks natural, but the other one doesn't.) We leave it to the reader to see that the differentials match up.

Actually, given a Cartan-Eilenberg resolution $K^\bullet \to I^{\bullet , \bullet }$ the induced morphism $K^\bullet \to sI^\bullet $ into the associated simple complex will be a filtered injective resolution for either filtration using suitable filtrations on $sI^\bullet $. This can be used to match up the spectral sequences exactly.

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