## 64.7 Filtered derived category

It turns out we have to do it all again and build the filtered derived category also.

Definition 64.7.1. Let $\mathcal{A}$ be an abelian category.

Let $\text{Fil}(\mathcal{A})$ be the category of filtered objects $(A, F)$ of $\mathcal{A}$, where $F$ is a filtration of the form

\[ A \supset \ldots \supset F^ n A \supset F^{n+1}A \supset \ldots \supset 0. \]

This is an additive category.

We denote $\text{Fil}^ f(\mathcal{A})$ the full subcategory of $\text{Fil}(\mathcal{A})$ whose objects $(A, F)$ have finite filtration. This is also an additive category.

An object $I \in \text{Fil}^ f(\mathcal{A})$ is called *filtered injective* (respectively *projective*) provided that $\text{gr}^ p(I) = \text{gr}_ F^ p(I) = F^ pI/F^{p+1}I$ is injective (resp. projective) in $\mathcal{A}$ for all $p$.

The category of complexes $\text{Comp}(\text{Fil}^ f(\mathcal{A})) \supset \text{Comp}^+(\text{Fil}^ f(\mathcal{A}))$ and its homotopy category $K(\text{Fil}^ f(\mathcal{A})) \supset K^+(\text{Fil}^ f(\mathcal A))$ are defined as before.

A morphism $\alpha : K^\bullet \to L^\bullet $ of complexes in $\text{Comp}(\text{Fil}^ f(\mathcal{A}))$ is called a *filtered quasi-isomorphism* provided that

\[ \text{gr}^ p(\alpha ): \text{gr}^ p(K^\bullet ) \to \text{gr}^ p(L^\bullet ) \]

is a quasi-isomorphism for all $p \in \mathbf{Z}$.

We define $DF(\mathcal{A})$ (resp. $DF^+(\mathcal{A})$) by inverting the filtered quasi-isomorphisms in $K(\text{Fil}^ f(\mathcal{A}))$ (resp. $K^+(\text{Fil}^ f(\mathcal{A}))$).

Lemma 64.7.2. If $\mathcal{A}$ has enough injectives, then $DF^+(\mathcal{A}) \cong K^+(\mathcal{I})$, where $\mathcal{I}$ is the full additive subcategory of $\text{Fil}^ f(\mathcal{A})$ consisting of filtered injective objects. Similarly, if $\mathcal{A}$ has enough projectives, then $DF^-(\mathcal{A}) \cong K^-(\mathcal{P})$, where $\mathcal P$ is the full additive subcategory of $\text{Fil}^ f(\mathcal{A})$ consisting of filtered projective objects.

**Proof.**
Omitted.
$\square$

## Comments (2)

Comment #5000 by Lenny Taelman on

Comment #5001 by Johan on