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## Tag 06XW

### 13.28. Unbounded complexes

A reference for the material in this section is [Spaltenstein]. The following lemma is useful to find ''good'' left resolutions of unbounded complexes.

Lemma 13.28.1. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{P} \subset \mathop{\rm Ob}\nolimits(\mathcal{A})$ 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. $\square$

In some cases we can use the lemma above to show that a left derived functor is everywhere defined.

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. $\square$

Lemma 13.28.3. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{I} \subset \mathop{\rm Ob}\nolimits(\mathcal{A})$ be a subset. Assume that every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$. Let $K^\bullet$ be a complex. There exists a commutative diagram $$\xymatrix{ \ldots \ar[r] & \tau_{\geq -2}K^\bullet \ar[r] \ar[d] & \tau_{\geq -1}K^\bullet \ar[d] \\ \ldots \ar[r] & I_2^\bullet \ar[r] & I_1^\bullet }$$ in the category of complexes such that

1. the vertical arrows are quasi-isomorphisms,
2. $I_n^\bullet$ is a bounded below complex with terms in $\mathcal{I}$,
3. the arrows $I_{n + 1}^\bullet \to I_n^\bullet$ are termwise split surjections and $\mathop{\rm Ker}(I^i_{n + 1} \to I^i_n)$ is an element of $\mathcal{I}$.

Proof. This lemma is dual to Lemma 13.28.1. $\square$

Lemma 13.28.4. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{A}$ be functors of abelian categories such that $F$ is a right adjoint to $G$. Let $K^\bullet$ be a complex of $\mathcal{A}$ and let $M^\bullet$ be a complex of $\mathcal{B}$. If $RF$ is defined at $K^\bullet$ and $LG$ is defined at $M^\bullet$, then there is a canonical isomorphism $$\mathop{\rm Hom}\nolimits_{D(\mathcal{B})}(M^\bullet, RF(K^\bullet)) = \mathop{\rm Hom}\nolimits_{D(\mathcal{A})}(LG(M^\bullet), K^\bullet)$$ This isomorphism is functorial in both variables on the triangulated subcategories of $D(\mathcal{A})$ and $D(\mathcal{B})$ where $RF$ and $LG$ are defined.

Proof. Since $RF$ is defined at $K^\bullet$, we see that the rule which assigns to a quasi-isomorphism $s : K^\bullet \to I^\bullet$ the object $F(I^\bullet)$ is essentially constant as an ind-object of $D(\mathcal{B})$ with value $RF(K^\bullet)$. Similarly, the rule which assigns to a quasi-isomorphism $t : P^\bullet \to M^\bullet$ the object $G(P^\bullet)$ is essentially constant as a pro-object of $D(\mathcal{A})$ with value $LG(M^\bullet)$. Thus we have \begin{align*} \mathop{\rm Hom}\nolimits_{D(\mathcal{B})}(M^\bullet, RF(K^\bullet)) & = \mathop{\rm colim}\nolimits_{s : K^\bullet \to I^\bullet} \mathop{\rm Hom}\nolimits_{D(\mathcal{B})}(M^\bullet, F(I^\bullet)) \\ & = \mathop{\rm colim}\nolimits_{s : K^\bullet \to I^\bullet} \mathop{\rm colim}\nolimits_{t : P^\bullet \to M^\bullet} \mathop{\rm Hom}\nolimits_{K(\mathcal{B})}(P^\bullet, F(I^\bullet)) \\ & = \mathop{\rm colim}\nolimits_{t : P^\bullet \to M^\bullet} \mathop{\rm colim}\nolimits_{s : K^\bullet \to I^\bullet} \mathop{\rm Hom}\nolimits_{K(\mathcal{B})}(P^\bullet, F(I^\bullet)) \\ & = \mathop{\rm colim}\nolimits_{t : P^\bullet \to M^\bullet} \mathop{\rm colim}\nolimits_{s : K^\bullet \to I^\bullet} \mathop{\rm Hom}\nolimits_{K(\mathcal{A})}(G(P^\bullet), I^\bullet) \\ & = \mathop{\rm colim}\nolimits_{s : K^\bullet \to I^\bullet} \mathop{\rm Hom}\nolimits_{D(\mathcal{A})}(G(P^\bullet), K^\bullet) \\ & = \mathop{\rm Hom}\nolimits_{D(\mathcal{A})}(LG(M^\bullet), K^\bullet) \end{align*} The first equality holds by Categories, Lemma 4.22.6. The second equality holds by the definition of morphisms in $D(\mathcal{B})$. The third equality holds by Categories, Lemma 4.14.9. The fourth equality holds because $F$ and $G$ are adjoint. The fifth equality holds by definition of morphism in $D(\mathcal{A})$. The sixth equality holds by Categories, Lemma 4.22.7. We omit the proof of functoriality. $\square$

The following lemma is an example of why it is easier to work with unbounded derived categories. Namely, without having the unbounded derived functors, the lemma could not even be stated.

Lemma 13.28.5. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{A}$ be functors of abelian categories such that $F$ is a right adjoint to $G$. If the derived functors $RF : D(\mathcal{A}) \to D(\mathcal{B})$ and $LG : D(\mathcal{B}) \to D(\mathcal{A})$ exist, then $RF$ is a right adjoint to $LG$.

Proof. Immediate from Lemma 13.28.4. $\square$

The code snippet corresponding to this tag is a part of the file derived.tex and is located in lines 8410–8717 (see updates for more information).

\section{Unbounded complexes}
\label{section-unbounded}

\noindent
A reference for the material in this section is \cite{Spaltenstein}.
The following lemma is useful to find good'' left resolutions of
unbounded complexes.

\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}

\noindent
In some cases we can use the lemma above to show that a left derived
functor is everywhere defined.

\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}

\begin{lemma}
\label{lemma-special-inverse-system}
Let $\mathcal{A}$ be an abelian category. Let
$\mathcal{I} \subset \Ob(\mathcal{A})$ be a subset.
Assume that every object of $\mathcal{A}$ is a subobject of an
element of $\mathcal{I}$. Let $K^\bullet$ be a complex.
There exists a commutative diagram
$$\xymatrix{ \ldots \ar[r] & \tau_{\geq -2}K^\bullet \ar[r] \ar[d] & \tau_{\geq -1}K^\bullet \ar[d] \\ \ldots \ar[r] & I_2^\bullet \ar[r] & I_1^\bullet }$$
in the category of complexes such that
\begin{enumerate}
\item the vertical arrows are quasi-isomorphisms,
\item $I_n^\bullet$ is a bounded below complex with terms in $\mathcal{I}$,
\item the arrows $I_{n + 1}^\bullet \to I_n^\bullet$ are termwise split
surjections and $\Ker(I^i_{n + 1} \to I^i_n)$ is an element of $\mathcal{I}$.
\end{enumerate}
\end{lemma}

\begin{proof}
This lemma is dual to
Lemma \ref{lemma-special-direct-system}.
\end{proof}

\begin{lemma}
Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{A}$
be functors of abelian categories such that $F$ is a right adjoint to $G$.
Let $K^\bullet$ be a complex of $\mathcal{A}$ and let $M^\bullet$ be
a complex of $\mathcal{B}$. If $RF$ is defined at $K^\bullet$
and $LG$ is defined at $M^\bullet$, then there is a canonical isomorphism
$$\Hom_{D(\mathcal{B})}(M^\bullet, RF(K^\bullet)) = \Hom_{D(\mathcal{A})}(LG(M^\bullet), K^\bullet)$$
This isomorphism is functorial in both variables on the triangulated
subcategories of $D(\mathcal{A})$ and $D(\mathcal{B})$
where $RF$ and $LG$ are defined.
\end{lemma}

\begin{proof}
Since $RF$ is defined at $K^\bullet$, we
see that the rule which assigns to a quasi-isomorphism
$s : K^\bullet \to I^\bullet$ the object $F(I^\bullet)$ is essentially
constant as an ind-object of $D(\mathcal{B})$ with value $RF(K^\bullet)$.
Similarly, the rule which assigns to a quasi-isomorphism
$t : P^\bullet \to M^\bullet$ the object $G(P^\bullet)$ is
essentially constant as a pro-object of $D(\mathcal{A})$
with value $LG(M^\bullet)$. Thus we have
\begin{align*}
\Hom_{D(\mathcal{B})}(M^\bullet, RF(K^\bullet))
& =
\colim_{s : K^\bullet \to I^\bullet}
\Hom_{D(\mathcal{B})}(M^\bullet, F(I^\bullet)) \\
& =
\colim_{s : K^\bullet \to I^\bullet}
\colim_{t : P^\bullet \to M^\bullet}
\Hom_{K(\mathcal{B})}(P^\bullet, F(I^\bullet)) \\
& =
\colim_{t : P^\bullet \to M^\bullet}
\colim_{s : K^\bullet \to I^\bullet}
\Hom_{K(\mathcal{B})}(P^\bullet, F(I^\bullet)) \\
& =
\colim_{t : P^\bullet \to M^\bullet}
\colim_{s : K^\bullet \to I^\bullet}
\Hom_{K(\mathcal{A})}(G(P^\bullet), I^\bullet) \\
& =
\colim_{s : K^\bullet \to I^\bullet}
\Hom_{D(\mathcal{A})}(G(P^\bullet), K^\bullet) \\
& = \Hom_{D(\mathcal{A})}(LG(M^\bullet), K^\bullet)
\end{align*}
The first equality holds by
Categories, Lemma \ref{categories-lemma-characterize-essentially-constant-ind}.
The second equality holds by the definition of morphisms in
$D(\mathcal{B})$. The third equality holds by
Categories, Lemma \ref{categories-lemma-colimits-commute}.
The fourth equality holds because $F$ and $G$ are adjoint.
The fifth equality holds by definition of morphism
in $D(\mathcal{A})$. The sixth equality holds by
Categories, Lemma \ref{categories-lemma-characterize-essentially-constant-pro}.
We omit the proof of functoriality.
\end{proof}

\noindent
The following lemma is an example of why it is easier to work
with unbounded derived categories. Namely, without having the
unbounded derived functors, the lemma could not even be stated.

\begin{lemma}
Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{A}$
be functors of abelian categories such that $F$ is a right adjoint to $G$.
If the derived functors $RF : D(\mathcal{A}) \to D(\mathcal{B})$ and
$LG : D(\mathcal{B}) \to D(\mathcal{A})$ exist, then
$RF$ is a right adjoint to $LG$.
\end{lemma}

\begin{proof}
\end{proof}

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