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Tag 0794

Chapter 13: Derived Categories > Section 13.28: Unbounded complexes

Proposition 13.28.2. Let $F : \mathcal{A} \to \mathcal{B}$ be a right exact functor of abelian categories. Let $\mathcal{P} \subset \mathop{\rm Ob}\nolimits(\mathcal{A})$ be a subset. Assume

  1. every object of $\mathcal{A}$ is a quotient of an element of $\mathcal{P}$,
  2. for any bounded above acyclic complex $P^\bullet$ of $\mathcal{A}$ with $P^n \in \mathcal{P}$ for all $n$ the complex $F(P^\bullet)$ is exact,
  3. $\mathcal{A}$ and $\mathcal{B}$ have colimits of systems over $\mathbf{N}$,
  4. colimits over $\mathbf{N}$ are exact in both $\mathcal{A}$ and $\mathcal{B}$, and
  5. $F$ commutes with colimits over $\mathbf{N}$.

Then $LF$ is defined on all of $D(\mathcal{A})$.

Proof. By (1) and Lemma 13.16.5 for any bounded above complex $K^\bullet$ there exists a quasi-isomorphism $P^\bullet \to K^\bullet$ with $P^\bullet$ bounded above and $P^n \in \mathcal{P}$ for all $n$. Suppose that $s : P^\bullet \to (P')^\bullet$ is a quasi-isomorphism of bounded above complexes consisting of objects of $\mathcal{P}$. Then $F(P^\bullet) \to F((P')^\bullet)$ is a quasi-isomorphism because $F(C(s)^\bullet)$ is acyclic by assumption (2). This already shows that $LF$ is defined on $D^{-}(\mathcal{A})$ and that a bounded above complex consisting of objects of $\mathcal{P}$ computes $LF$, see Lemma 13.15.15.

Next, let $K^\bullet$ be an arbitrary complex of $\mathcal{A}$. Choose a diagram $$ \xymatrix{ P_1^\bullet \ar[d] \ar[r] & P_2^\bullet \ar[d] \ar[r] & \ldots \\ \tau_{\leq 1}K^\bullet \ar[r] & \tau_{\leq 2}K^\bullet \ar[r] & \ldots } $$ as in Lemma 13.28.1. Note that the map $\mathop{\rm colim}\nolimits P_n^\bullet \to K^\bullet$ is a quasi-isomorphism because colimits over $\mathbf{N}$ in $\mathcal{A}$ are exact and $H^i(P_n^\bullet) = H^i(K^\bullet)$ for $n > i$. We claim that $$ F(\mathop{\rm colim}\nolimits P_n^\bullet) = \mathop{\rm colim}\nolimits F(P_n^\bullet) $$ (termwise colimits) is $LF(K^\bullet)$, i.e., that $\mathop{\rm colim}\nolimits P_n^\bullet$ computes $LF$. To see this, by Lemma 13.15.15, it suffices to prove the following claim. Suppose that $$ \mathop{\rm colim}\nolimits Q_n^\bullet = Q^\bullet \xrightarrow{~\alpha~} P^\bullet = \mathop{\rm colim}\nolimits P_n^\bullet $$ is a quasi-isomorphism of complexes, such that each $P_n^\bullet$, $Q_n^\bullet$ is a bounded above complex whose terms are in $\mathcal{P}$ and the maps $P_n^\bullet \to \tau_{\leq n}P^\bullet$ and $Q_n^\bullet \to \tau_{\leq n}Q^\bullet$ are quasi-isomorphisms. Claim: $F(\alpha)$ is a quasi-isomorphism.

The problem is that we do not assume that $\alpha$ is given as a colimit of maps between the complexes $P_n^\bullet$ and $Q_n^\bullet$. However, for each $n$ we know that the solid arrows in the diagram $$ \xymatrix{ & R^\bullet \ar@{..>}[d] \\ P_n^\bullet \ar[d] & L^\bullet \ar@{..>}[l] \ar@{..>}[r] & Q_n^\bullet \ar[d] \\ \tau_{\leq n}P^\bullet \ar[rr]^{\tau_{\leq n}\alpha} & & \tau_{\leq n}Q^\bullet } $$ are quasi-isomorphisms. Because quasi-isomorphisms form a multiplicative system in $K(\mathcal{A})$ (see Lemma 13.11.2) we can find a quasi-isomorphism $L^\bullet \to P_n^\bullet$ and map of complexes $L^\bullet \to Q_n^\bullet$ such that the diagram above commutes up to homotopy. Then $\tau_{\leq n}L^\bullet \to L^\bullet$ is a quasi-isomorphism. Hence (by the first part of the proof) we can find a bounded above complex $R^\bullet$ whose terms are in $\mathcal{P}$ and a quasi-isomorphism $R^\bullet \to L^\bullet$ (as indicated in the diagram). Using the result of the first paragraph of the proof we see that $F(R^\bullet) \to F(P_n^\bullet)$ and $F(R^\bullet) \to F(Q_n^\bullet)$ are quasi-isomorphisms. Thus we obtain a isomorphisms $H^i(F(P_n^\bullet)) \to H^i(F(Q_n^\bullet))$ fitting into the commutative diagram $$ \xymatrix{ H^i(F(P_n^\bullet)) \ar[r] \ar[d] & H^i(F(Q_n^\bullet)) \ar[d] \\ H^i(F(P^\bullet)) \ar[r] & H^i(F(Q^\bullet)) } $$ The exact same argument shows that these maps are also compatible as $n$ varies. Since by (4) and (5) we have $$ H^i(F(P^\bullet)) = H^i(F(\mathop{\rm colim}\nolimits P_n^\bullet)) = H^i(\mathop{\rm colim}\nolimits F(P_n^\bullet)) = \mathop{\rm colim}\nolimits H^i(F(P_n^\bullet)) $$ and similarly for $Q^\bullet$ we conclude that $H^i(\alpha) : H^i(F(P^\bullet) \to H^i(F(Q^\bullet)$ is an isomorphism and the claim follows. $\square$

    The code snippet corresponding to this tag is a part of the file derived.tex and is located in lines 8508–8526 (see updates for more information).

    \begin{proposition}
    \label{proposition-left-derived-exists}
    Let $F : \mathcal{A} \to \mathcal{B}$ be a right exact functor
    of abelian categories. Let $\mathcal{P} \subset \Ob(\mathcal{A})$ be a
    subset. Assume
    \begin{enumerate}
    \item every object of $\mathcal{A}$ is a quotient of an element of
    $\mathcal{P}$,
    \item for any bounded above acyclic complex $P^\bullet$ of
    $\mathcal{A}$ with $P^n \in \mathcal{P}$ for all $n$ the
    complex $F(P^\bullet)$ is exact,
    \item $\mathcal{A}$ and $\mathcal{B}$ have colimits
    of systems over $\mathbf{N}$,
    \item colimits over $\mathbf{N}$ are exact in both
    $\mathcal{A}$ and $\mathcal{B}$, and
    \item $F$ commutes with colimits over $\mathbf{N}$.
    \end{enumerate}
    Then $LF$ is defined on all of $D(\mathcal{A})$.
    \end{proposition}
    
    \begin{proof}
    By (1) and Lemma \ref{lemma-subcategory-left-resolution} for any bounded
    above complex $K^\bullet$ there exists a quasi-isomorphism
    $P^\bullet \to K^\bullet$ with $P^\bullet$ bounded above and
    $P^n \in \mathcal{P}$ for all $n$. Suppose that
    $s : P^\bullet \to (P')^\bullet$ is a quasi-isomorphism of bounded
    above complexes consisting of objects of $\mathcal{P}$. Then
    $F(P^\bullet) \to F((P')^\bullet)$ is a quasi-isomorphism because
    $F(C(s)^\bullet)$ is acyclic by assumption (2). This already shows that
    $LF$ is defined on $D^{-}(\mathcal{A})$ and that a bounded above
    complex consisting of objects of $\mathcal{P}$ computes $LF$, see
    Lemma \ref{lemma-find-existence-computes}.
    
    \medskip\noindent
    Next, let $K^\bullet$ be an arbitrary complex of $\mathcal{A}$.
    Choose a diagram
    $$
    \xymatrix{
    P_1^\bullet \ar[d] \ar[r] & P_2^\bullet \ar[d] \ar[r] & \ldots \\
    \tau_{\leq 1}K^\bullet \ar[r] & \tau_{\leq 2}K^\bullet \ar[r] & \ldots
    }
    $$
    as in Lemma \ref{lemma-special-direct-system}. Note that
    the map $\colim P_n^\bullet \to K^\bullet$ is a quasi-isomorphism
    because colimits over $\mathbf{N}$ in $\mathcal{A}$ are exact
    and $H^i(P_n^\bullet) = H^i(K^\bullet)$ for $n > i$. We claim that
    $$
    F(\colim P_n^\bullet) = \colim F(P_n^\bullet)
    $$
    (termwise colimits) is $LF(K^\bullet)$, i.e., that $\colim P_n^\bullet$
    computes $LF$. To see this, by Lemma \ref{lemma-find-existence-computes},
    it suffices to prove the following claim. Suppose that
    $$
    \colim Q_n^\bullet = Q^\bullet
    \xrightarrow{\ \alpha\ }
    P^\bullet = \colim P_n^\bullet
    $$
    is a quasi-isomorphism of complexes, such that each
    $P_n^\bullet$, $Q_n^\bullet$ is a bounded above complex whose terms are
    in $\mathcal{P}$ and the maps $P_n^\bullet \to \tau_{\leq n}P^\bullet$ and
    $Q_n^\bullet \to \tau_{\leq n}Q^\bullet$ are quasi-isomorphisms.
    Claim: $F(\alpha)$ is a quasi-isomorphism.
    
    \medskip\noindent
    The problem is that we do not assume that $\alpha$ is given as a colimit
    of maps between the complexes $P_n^\bullet$ and $Q_n^\bullet$. However,
    for each $n$ we know that the solid arrows in the diagram
    $$
    \xymatrix{
    & R^\bullet \ar@{..>}[d] \\
    P_n^\bullet \ar[d] &
    L^\bullet \ar@{..>}[l] \ar@{..>}[r] &
    Q_n^\bullet \ar[d] \\
    \tau_{\leq n}P^\bullet \ar[rr]^{\tau_{\leq n}\alpha} & &
    \tau_{\leq n}Q^\bullet
    }
    $$
    are quasi-isomorphisms. Because quasi-isomorphisms form a multiplicative
    system in $K(\mathcal{A})$ (see Lemma \ref{lemma-acyclic})
    we can find a quasi-isomorphism
    $L^\bullet \to P_n^\bullet$ and map of complexes $L^\bullet \to Q_n^\bullet$
    such that the diagram above commutes up to homotopy. Then
    $\tau_{\leq n}L^\bullet \to L^\bullet$ is a quasi-isomorphism.
    Hence (by the first part of the proof) we can find a bounded above
    complex $R^\bullet$ whose terms are in $\mathcal{P}$ and a quasi-isomorphism
    $R^\bullet \to L^\bullet$ (as indicated in the diagram). Using the result
    of the first paragraph of the proof we see that
    $F(R^\bullet) \to F(P_n^\bullet)$ and $F(R^\bullet) \to F(Q_n^\bullet)$
    are quasi-isomorphisms. Thus we obtain a isomorphisms
    $H^i(F(P_n^\bullet)) \to H^i(F(Q_n^\bullet))$ fitting into the commutative
    diagram
    $$
    \xymatrix{
    H^i(F(P_n^\bullet)) \ar[r] \ar[d] &
    H^i(F(Q_n^\bullet)) \ar[d] \\
    H^i(F(P^\bullet)) \ar[r] &
    H^i(F(Q^\bullet))
    }
    $$
    The exact same argument shows that these maps are also compatible
    as $n$ varies. Since by (4) and (5) we have
    $$
    H^i(F(P^\bullet)) =
    H^i(F(\colim P_n^\bullet)) =
    H^i(\colim F(P_n^\bullet)) = \colim H^i(F(P_n^\bullet))
    $$
    and similarly for $Q^\bullet$ we conclude that
    $H^i(\alpha) : H^i(F(P^\bullet) \to H^i(F(Q^\bullet)$ is an isomorphism
    and the claim follows.
    \end{proof}

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