Exercise 109.30.1. Let $\mathcal{A}$ be an abelian category. Let $I$ be a filtered object of $\mathcal{A}$. Assume that the filtration on $I$ is finite and that each $\text{gr}^ p(I)$ is an injective object of $\mathcal{A}$. Show that there exists an isomorphism $I \cong \bigoplus \text{gr}^ p(I)$ with filtration $F^ p(I)$ corresponding to $\bigoplus _{p' \geq p} \text{gr}^ p(I)$.

## 109.30 Filtered derived category

In order to do the exercises in this section, please read the material in Homology, Section 12.19. We will say $A$ is a filtered object of $\mathcal{A}$, to mean that $A$ comes endowed with a filtration $F$ which we omit from the notation.

Exercise 109.30.2. Let $\mathcal{A}$ be an abelian category. Let $I$ be a filtered object of $\mathcal{A}$. Assume that the filtration on $I$ is finite. Show the following are equivalent:

For any solid diagram

\[ \xymatrix{ A \ar[r]_\alpha \ar[d] & B \ar@{-->}[ld] \\ I & } \]of filtered objects with (i) the filtrations on $A$ and $B$ are finite, and (ii) $\text{gr}(\alpha )$ injective the dotted arrow exists making the diagram commute.

Each $\text{gr}^ p I$ is injective.

Note that given a morphism $\alpha : A \to B$ of filtered objects with finite filtrations to say that $\text{gr}(\alpha )$ injective is the same thing as saying that $\alpha $ is a *strict monomorphism* in the category $\text{Fil}(\mathcal{A})$. Namely, being a monomorphism means $\mathop{\mathrm{Ker}}(\alpha ) = 0$ and strict means that this also implies $\mathop{\mathrm{Ker}}(\text{gr}(\alpha )) = 0$. See Homology, Lemma 12.19.13. (We only use the term “injective” for a morphism in an abelian category, although it makes sense in any additive category having kernels.) The exercises above justifies the following definition.

Definition 109.30.3. Let $\mathcal{A}$ be an abelian category. Let $I$ be a filtered object of $\mathcal{A}$. Assume the filtration on $I$ is finite. We say $I$ is *filtered injective* if each $\text{gr}^ p(I)$ is an injective object of $\mathcal{A}$.

We make the following definition to avoid having to keep saying “with a finite filtration” everywhere.

Definition 109.30.4. Let $\mathcal{A}$ be an abelian category. We denote *$\text{Fil}^ f(\mathcal{A})$* the full subcategory of $\text{Fil}(\mathcal{A})$ whose objects consist of those $A \in \mathop{\mathrm{Ob}}\nolimits (\text{Fil}(\mathcal{A}))$ whose filtration is finite.

Exercise 109.30.5. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Let $A$ be an object of $\text{Fil}^ f(\mathcal{A})$. Show that there exists a strict monomorphism $\alpha : A \to I$ of $A$ into a filtered injective object $I$ of $\text{Fil}^ f(\mathcal{A})$.

Definition 109.30.6. Let $\mathcal{A}$ be an abelian category. Let $\alpha : K^\bullet \to L^\bullet $ be a morphism of complexes of $\text{Fil}(\mathcal{A})$. We say that $\alpha $ is a *filtered quasi-isomorphism* if for each $p \in \mathbf{Z}$ the morphism $\text{gr}^ p(K^\bullet ) \to \text{gr}^ p(L^\bullet )$ is a quasi-isomorphism.

Definition 109.30.7. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $ be a complex of $\text{Fil}^ f(\mathcal{A})$. We say that $K^\bullet $ is *filtered acyclic* if for each $p \in \mathbf{Z}$ the complex $\text{gr}^ p(K^\bullet )$ is acyclic.

Exercise 109.30.8. Let $\mathcal{A}$ be an abelian category. Let $\alpha : K^\bullet \to L^\bullet $ be a morphism of bounded below complexes of $\text{Fil}^ f(\mathcal{A})$. (Note the superscript $f$.) Show that the following are equivalent:

$\alpha $ is a filtered quasi-isomorphism,

for each $p \in \mathbf{Z}$ the map $\alpha : F^ pK^\bullet \to F^ pL^\bullet $ is a quasi-isomorphism,

for each $p \in \mathbf{Z}$ the map $\alpha : K^\bullet /F^ pK^\bullet \to L^\bullet /F^ pL^\bullet $ is a quasi-isomorphism, and

the cone of $\alpha $ (see Derived Categories, Definition 13.9.1) is a filtered acyclic complex.

Moreover, show that if $\alpha $ is a filtered quasi-isomorphism then $\alpha $ is also a usual quasi-isomorphism.

Exercise 109.30.9. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Let $A$ be an object of $\text{Fil}^ f(\mathcal{A})$. Show there exists a complex $I^\bullet $ of $\text{Fil}^ f(\mathcal{A})$, and a morphism $A[0] \to I^\bullet $ such that

each $I^ p$ is filtered injective,

$I^ p = 0$ for $p < 0$, and

$A[0] \to I^\bullet $ is a filtered quasi-isomorphism.

Exercise 109.30.10. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Let $K^\bullet $ be a bounded below complex of objects of $\text{Fil}^ f(\mathcal{A})$. Show there exists a filtered quasi-isomorphism $\alpha : K^\bullet \to I^\bullet $ with $I^\bullet $ a complex of $\text{Fil}^ f(\mathcal{A})$ having filtered injective terms $I^ n$, and bounded below. In fact, we may choose $\alpha $ such that each $\alpha ^ n$ is a strict monomorphism.

Exercise 109.30.11. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

of complexes of $\text{Fil}^ f(\mathcal{A})$. Assume $K^\bullet $, $L^\bullet $ and $I^\bullet $ are bounded below and assume each $I^ n$ is a filtered injective object. Also assume that $\alpha $ is a filtered quasi-isomorphism.

There exists a map of complexes $\beta $ making the diagram commute up to homotopy.

If $\alpha $ is a strict monomorphism in every degree then we can find a $\beta $ which makes the diagram commute.

Exercise 109.30.12. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $, $K^\bullet $ be complexes of $\text{Fil}^ f(\mathcal{A})$. Assume

$K^\bullet $ bounded below and filtered acyclic, and

$I^\bullet $ bounded below and consisting of filtered injective objects.

Then any morphism $K^\bullet \to I^\bullet $ is homotopic to zero.

Exercise 109.30.13. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

of complexes of $\text{Fil}^ f(\mathcal{A})$. Assume $K^\bullet $, $L^\bullet $ and $I^\bullet $ bounded below and each $I^ n$ a filtered injective object. Also assume $\alpha $ a filtered quasi-isomorphism. Any two morphisms $\beta _1, \beta _2$ making the diagram commute up to homotopy are homotopic.

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