Exercise 111.30.1. 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 $\bigoplus _{p' \geq p} \text{gr}^ p(I)$.

## 111.30 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 111.30.2. 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 111.30.3. 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 111.30.4. 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 111.30.5. 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 111.30.6. 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 111.30.7. 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 111.30.8. 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 111.30.9. 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 111.30.10. 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 111.30.11. 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 111.30.12. 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 111.30.13. 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.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)