# stacks/stacks-project

A way to think about filtered complexes

 @@ -2238,6 +2238,142 @@ \section{Additional remarks on Grothendieck abelian categories} $F^pK^\bullet/F^{p'}K^\bullet \to F^pJ^\bullet/F^{p'}J^\bullet$. \end{proof} \begin{lemma} \label{lemma-represent-by-filtered-complex} Let $\mathcal{A}$ be a Grothendieck abelian category. Suppose given an object $E \in D(\mathcal{A})$ and an inverse system $\{E^i\}_{i \in \mathbf{Z}}$ of objects of $D(\mathcal{A})$ over $\mathbf{Z}$ together with a compatible system of maps $E^i \to E$. Picture: $$\ldots \to E^{i + 1} \to E^i \to E^{i - 1} \to \ldots \to E$$ Then there exists a filtered complex $K^\bullet$ of $\mathcal{A}$ (Homology, Definition \ref{homology-definition-filtered-complex}) such that $K^\bullet$ represents $E$ and $F^iK^\bullet$ represents $E^i$ compatibly with the given maps. \end{lemma} \begin{proof} By Theorem \ref{theorem-K-injective-embedding-grothendieck} we can choose a K-injective complex $I^\bullet$ representing $E$ all of whose terms $I^n$ are injective objects of $\mathcal{A}$. Choose a complex $G^{0, \bullet}$ representing $E^0$. Choose a map of complexes $\varphi^0 : G^{0, \bullet} \to I^\bullet$ representing $E^0 \to E$. For $i > 0$ we inductively represent $E^i \to E^{i - 1}$ by a map of complexes $\delta : G^{i, \bullet} \to G^{i - 1, \bullet}$ and we set $\varphi^i = \delta \circ \varphi^{i - 1}$. For $i < 0$ we inductively represent $E^{i + 1} \to E^i$ by a termwise injective map of complexes $\delta : G^{i + 1, \bullet} \to G^{i, \bullet}$ (for example you can use Derived Categories, Lemma \ref{derived-lemma-make-injective}). Claim: we can find a map of complexes $\varphi^i : G^{i, \bullet} \to I^\bullet$ representing the map $E^i \to E$ and fitting into the commutative diagram $$\xymatrix{ G^{i + 1, \bullet} \ar[r]_\delta \ar[d]_{\varphi^{i + 1}} & G^{i, \bullet} \ar[ld]^{\varphi^i} \\ I^\bullet }$$ Namely, we first choose any map of complexes $\varphi : G^{i, \bullet} \to I^\bullet$ representing the map $E^i \to E$. Then we see that $\varphi \circ \delta$ and $\varphi^{i + 1}$ are homotopic by some homotopy $h^p : G^{i + 1, p} \to I^{p - 1}$. Since the terms of $I^\bullet$ are injective and since $\delta$ is termwise injective, we can lift $h^p$ to $(h')^p : G^{i, p} \to I^{p - 1}$. Then we set $\varphi^i = \varphi + h' \circ d + d \circ h'$ and we get what we claimed. \medskip\noindent Next, we choose for every $i$ a termwise injective map of complexes $a^i : G^{i, \bullet} \to J^{i, \bullet}$ with $J^{i, \bullet}$ acyclic, K-injective, with $J^{i, p}$ injective objects of $\mathcal{A}$. To do this first map $G^{i, \bullet}$ to the cone on the identity and then apply the theorem cited above. Arguing as above we can find maps of complexes $\delta' : J^{i, \bullet} \to J^{i - 1, \bullet}$ such that the diagrams $$\xymatrix{ G^{i, \bullet} \ar[r]_\delta \ar[d]_{a^i} & G^{i - 1, \bullet} \ar[d]^{a^{i - 1}} \\ J^{i, \bullet} \ar[r]^{\delta'} & J^{i - 1, \bullet} }$$ commute. (You could also use the functoriality of cones plus the functoriality in the theorem to get this.) Then we consider the maps $$\xymatrix{ G^{i + 1, \bullet} \times \prod\nolimits_{p > i + 1} J^{p, \bullet} \ar[r] \ar[rd] & G^{i, \bullet} \times \prod\nolimits_{p > i} J^{p, \bullet} \ar[r] \ar[d] & G^{i - 1, \bullet} \times \prod\nolimits_{p > i - 1} J^{p, \bullet} \ar[ld] \\ & I^\bullet \times \prod\nolimits_p J^{p, \bullet} }$$ Here the arrows on $J^{p, \bullet}$ are the obvious ones (identity or zero). On the factor $G^{i, \bullet}$ we use $\delta : G^{i, \bullet} \to G^{i - 1, \bullet}$, the map $\varphi^i : G^{i, \bullet} \to I^\bullet$, the zero map $0 : G^{i, \bullet} \to J^{p, \bullet}$ for $p > i$, the map $a^i : G^{i, \bullet} \to J^{p, \bullet}$ for $p = i$, and $(\delta')^{i - p} \circ a^i = a^p \circ \delta^{i - p} : G^{i, \bullet} \to J^{p, \bullet}$ for $p < i$. We omit the verification that all the arrows in the diagram are termwise injective. Thus we obtain a filtered complex. Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma \ref{lemma-derived-products}) and because $J^{p, \bullet}$ is zero in $D(\mathcal{A})$ we conclude this diagram represents the given diagram in the derived category. This finishes the proof. \end{proof} \begin{lemma} \label{lemma-represent-by-filtered-complex-bis} In the situation of Lemma \ref{lemma-represent-by-filtered-complex} assume we have a second inverse system $\{(E')^i\}_{i \in \mathbf{Z}}$ and a compatible system of maps $(E')^i \to E$. Then there exists a bi-filtered complex $K^\bullet$ of $\mathcal{A}$ such that $K^\bullet$ represents $E$, $F^iK^\bullet$ represents $E^i$, and $(F')^iK^\bullet$ represents $(E')^i$ compatibly with the given maps. \end{lemma} \begin{proof} Using the lemma we can first choose $K^\bullet$ and $F$. Then we can choose $(K')^\bullet$ and $F'$ which work for $\{(E')^i\}_{i \in \mathbf{Z}}$ and the maps $(E')^i \to E$. Using Lemma \ref{lemma-K-injective-embedding-filtration} we can assume $K^\bullet$ is a K-injective complex. Then we can choose a map of complexes $(K')^\bullet \to K^\bullet$ corresponding to the given identifications $(K')^\bullet \cong E \cong K^\bullet$. We can additionally choose a termwise injective map $(K')^\bullet \to J^\bullet$ with $J^\bullet$ acyclic and K-injective. (To do this first map $(K')^\bullet$ to the cone on the identity and then apply Theorem \ref{theorem-K-injective-embedding-grothendieck}.) Then $(K')^\bullet \to K^\bullet \times J^\bullet$ and $K^\bullet \to K^\bullet \times J^\bullet$ are both termwise injective and quasi-isomorphisms (as the product represents $E$ by Lemma \ref{lemma-derived-products}). Then we can simply take the images of the filtrations on $K^\bullet$ and $(K')^\bullet$ under these maps to conclude. \end{proof}