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A way to think about filtered complexes

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aisejohan committed Jun 12, 2018
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@@ -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}

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