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Tag 06XX

Chapter 13: Derived Categories > Section 13.28: Unbounded complexes

Lemma 13.28.1. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits(\mathcal{A})$ be a subset. Assume that every object of $\mathcal{A}$ is a quotient of an element of $\mathcal{P}$. Let $K^\bullet$ be a complex. There exists a commutative diagram $$ \xymatrix{ P_1^\bullet \ar[d] \ar[r] & P_2^\bullet \ar[d] \ar[r] & \ldots \\ \tau_{\leq 1}K^\bullet \ar[r] & \tau_{\leq 2}K^\bullet \ar[r] & \ldots } $$ in the category of complexes such that

  1. the vertical arrows are quasi-isomorphisms,
  2. $P_n^\bullet$ is a bounded above complex with terms in $\mathcal{P}$,
  3. the arrows $P_n^\bullet \to P_{n + 1}^\bullet$ are termwise split injections and each cokernel $P^i_{n + 1}/P^i_n$ is an element of $\mathcal{P}$.

Proof. By Lemma 13.16.5 any bounded above complex has a resolution by a bounded above complex whose terms are in $\mathcal{P}$. Thus we obtain the first complex $P_1^\bullet$. By induction it suffices, given $P_1^\bullet, \ldots, P_n^\bullet$ to construct $P_{n + 1}^\bullet$ and the maps $P_n^\bullet \to P_{n + 1}^\bullet$ and $P_{n + 1}^\bullet \to \tau_{\leq n + 1}K^\bullet$. Consider the cone $C_1^\bullet$ of the composition $P_n^\bullet \to \tau_{\leq n}K^\bullet \to \tau_{\leq n + 1}K^\bullet$. This fits into the distinguished triangle $$ P_n^\bullet \to \tau_{\leq n + 1}K^\bullet \to C_1^\bullet \to P_n^\bullet[1] $$ Note that $C_1^\bullet$ is bounded above, hence we can choose a quasi-isomorphism $Q^\bullet \to C_1^\bullet$ where $Q^\bullet$ is a bounded above complex whose terms are elements of $\mathcal{P}$. Take the cone $C_2^\bullet$ of the map of complexes $Q^\bullet \to P_n^\bullet[1]$ to get the distinguished triangle $$ Q^\bullet \to P_n^\bullet[1] \to C_2^\bullet \to Q^\bullet[1] $$ By the axioms of triangulated categories we obtain a map of distinguished triangles $$ \xymatrix{ P_n^\bullet \ar[r] \ar[d] & C_2^\bullet[-1] \ar[r] \ar[d] & Q^\bullet \ar[r] \ar[d] & P_n^\bullet[1] \ar[d] \\ P_n^\bullet \ar[r] & \tau_{\leq n + 1}K^\bullet \ar[r] & C_1^\bullet \ar[r] & P_n^\bullet[1] } $$ in the triangulated category $K(\mathcal{A})$. Set $P_{n + 1}^\bullet = C_2^\bullet[-1]$. Note that (3) holds by construction. Choose an actual morphism of complexes $f : P_{n + 1}^\bullet \to \tau_{\leq n + 1}K^\bullet$. The left square of the diagram above commutes up to homotopy, but as $P_n^\bullet \to P_{n + 1}^\bullet$ is a termwise split injection we can lift the homotopy and modify our choice of $f$ to make it commute. Finally, $f$ is a quasi-isomorphism, because both $P_n^\bullet \to P_n^\bullet$ and $Q^\bullet \to C_1^\bullet$ are. $\square$

    The code snippet corresponding to this tag is a part of the file derived.tex and is located in lines 8429–8451 (see updates for more information).

    \begin{lemma}
    \label{lemma-special-direct-system}
    Let $\mathcal{A}$ be an abelian category. Let
    $\mathcal{P} \subset \Ob(\mathcal{A})$ be a subset.
    Assume that every object of $\mathcal{A}$ is a quotient of an
    element of $\mathcal{P}$. Let $K^\bullet$ be a complex.
    There exists a commutative diagram
    $$
    \xymatrix{
    P_1^\bullet \ar[d] \ar[r] & P_2^\bullet \ar[d] \ar[r] & \ldots \\
    \tau_{\leq 1}K^\bullet \ar[r] & \tau_{\leq 2}K^\bullet \ar[r] & \ldots
    }
    $$
    in the category of complexes such that
    \begin{enumerate}
    \item the vertical arrows are quasi-isomorphisms,
    \item $P_n^\bullet$ is a bounded above complex with terms in
    $\mathcal{P}$,
    \item the arrows $P_n^\bullet \to P_{n + 1}^\bullet$
    are termwise split injections and each cokernel
    $P^i_{n + 1}/P^i_n$ is an element of $\mathcal{P}$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    By
    Lemma \ref{lemma-subcategory-left-resolution}
    any bounded above complex has a resolution by a bounded above complex
    whose terms are in $\mathcal{P}$. Thus we obtain the first complex
    $P_1^\bullet$. By induction it suffices, given
    $P_1^\bullet, \ldots, P_n^\bullet$ to construct
    $P_{n + 1}^\bullet$ and the maps
    $P_n^\bullet \to P_{n + 1}^\bullet$ and
    $P_{n + 1}^\bullet \to \tau_{\leq n + 1}K^\bullet$.
    Consider the cone $C_1^\bullet$ of the composition
    $P_n^\bullet \to \tau_{\leq n}K^\bullet \to \tau_{\leq n + 1}K^\bullet$.
    This fits into the distinguished triangle
    $$
    P_n^\bullet \to \tau_{\leq n + 1}K^\bullet \to C_1^\bullet \to P_n^\bullet[1]
    $$
    Note that $C_1^\bullet$ is bounded above, hence we can choose a
    quasi-isomorphism $Q^\bullet \to C_1^\bullet$ where $Q^\bullet$ is a
    bounded above complex whose terms are elements of $\mathcal{P}$.
    Take the cone $C_2^\bullet$ of the map of complexes
    $Q^\bullet \to P_n^\bullet[1]$ to get the
    distinguished triangle
    $$
    Q^\bullet \to P_n^\bullet[1] \to C_2^\bullet \to Q^\bullet[1]
    $$
    By the axioms of triangulated categories we obtain a map
    of distinguished triangles
    $$
    \xymatrix{
    P_n^\bullet \ar[r] \ar[d] &
    C_2^\bullet[-1] \ar[r] \ar[d] &
    Q^\bullet \ar[r] \ar[d] &
    P_n^\bullet[1] \ar[d] \\
    P_n^\bullet \ar[r] &
    \tau_{\leq n + 1}K^\bullet \ar[r] &
    C_1^\bullet \ar[r] &
    P_n^\bullet[1]
    }
    $$
    in the triangulated category $K(\mathcal{A})$.
    Set $P_{n + 1}^\bullet = C_2^\bullet[-1]$.
    Note that (3) holds by construction.
    Choose an actual morphism of complexes
    $f : P_{n + 1}^\bullet \to \tau_{\leq n + 1}K^\bullet$.
    The left square of the diagram above commutes up to homotopy, but as
    $P_n^\bullet \to P_{n + 1}^\bullet$ is a termwise split injection
    we can lift the homotopy and modify our choice of $f$ to make it commute.
    Finally, $f$ is a quasi-isomorphism, because both $P_n^\bullet \to P_n^\bullet$
    and $Q^\bullet \to C_1^\bullet$ are.
    \end{proof}

    Comments (5)

    Comment #2466 by anonymous on March 26, 2017 a 3:10 pm UTC

    Is there a reason item (2) does not state that all $P_n$ are bounded above with terms in $\mathcal{P}$?

    Same in Tag 070F.

    Comment #2475 by Tanya Kaushal Srivastava on April 5, 2017 a 11:44 am UTC

    A small typo in the proof, we want to construct the map from $P^{\bullet}_{n+1} \rightarrow \tau_{\leq n+1}K^{\bullet}$ as we already have the map $P^{\bullet}_{n} \rightarrow \tau_{\leq n+1}K^{\bullet}$ just by composition.

    Comment #2502 by Johan (site) on April 13, 2017 a 11:52 pm UTC

    @#2475: Sorry, I do not understand. @#2466: Thanks, fixed here.

    Comment #2512 by Tanya Kaushal Srivastava on April 15, 2017 a 9:42 am UTC

    Rephrasing: Why do we want to construct a map from $P_n^{\bullet} \rightarrow \tau_{\leq n +1}K^{\bullet}$ but not just $P_{n+1}^{\bullet} \rightarrow \tau_{\leq n +1}K^{\bullet}$? (Line 4 of proof)

    Comment #2555 by Johan (site) on May 25, 2017 a 5:55 pm UTC

    @#2512: OK, now I see it. Thanks. Fixed here.

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