The Stacks project

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

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 12.19.13. 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. $\square$

Proof. The pushout $f' : I \to I \amalg _ A B$ of $f$ by $u$ is a strict monomorphism, see Homology, Lemma 12.19.10. Hence the result follows formally from Lemma 13.26.3. $\square$

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 $u_ n : A/F^{n + 1}A \to I_ n$ with $I_ n$ an injective object. Set $I = \bigoplus _{a \leq n \leq b} I_ n$ with filtration $F^ pI = \bigoplus _{n \geq p} I_ n$ and set $u : A \to I$ equal to the direct sum of the maps $u_ n$. $\square$

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 12.19.13. $\square$

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 12.19.15. 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. $\square$

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 13.13.4. 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 $b \oplus c \circ \beta : M \to I^0 \oplus J^0$. $\square$

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 12.18.3. 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 12.25.4 to conclude that $\text{gr}(K^\bullet ) \to \text{gr}(\text{Tot}(I^{\bullet , \bullet }))$ is a quasi-isomorphism as desired. $\square$

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 12.19.15. 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. $\square$

Proof. Proof of (1). Note that

is a distinguished triangle in $K(\text{Fil}^ f(\mathcal{A}))$ (Lemma 13.9.14) and $C(\alpha )^\bullet$ is a filtered acyclic complex (Lemma 13.13.4). 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., $b = 0$. $\square$

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. $\square$

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 12.24. By Homology, Lemma 12.24.2 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 12.24.11 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 12.24.4. The last statement follows from this. $\square$