This tag has label etale-cohomology-section-derived-categories, it is called Derived categories in the Stacks project and it points to
The corresponding content:
41.72. Derived categories
To set up notation, let $\mathcal{A}$ be an abelian category. Let $\text{Comp}(\mathcal{A})$ be the abelian category of complexes in $\mathcal{A}$. Let $K(\mathcal{A})$ be the category of complexes up to homotopy, with objects equal to complexes in $\mathcal{A}$ and objects equal to homotopy classes of morphisms of complexes. This is not an abelian category. Loosely speaking, $D(A)$ is defined to be the category obtained by inverting all quasi-isomorphisms in $\text{Comp}(\mathcal{A})$ or, equivalently, in $K(\mathcal{A})$. Moreover, we can define $\text{Comp}^+(\mathcal{A}), K^+(\mathcal{A}), D^+(\mathcal{A})$ analogously using only bounded below complexes. Similarly, we can define $\text{Comp}^-(\mathcal{A}), K^-(\mathcal{A}), D^-(\mathcal{A})$ using bounded above complexes, and we can define $\text{Comp}^b(\mathcal{A}), K^b(\mathcal{A}), D^b(\mathcal{A})$ using bounded complexes.
Remark 41.72.1. Notes on derived categories.
- There are some set-theoretical problems when $\mathcal{A}$ is somewhat arbitrary, which we will happily disregard.
- The categories $K(A)$ and $D(A)$ may be endowed with the structure of triangulated category, but we will not need these structures in the following discussion.
- The categories $\text{Comp}(\mathcal{A})$ and $K(\mathcal{A})$ can also be defined when $\mathcal{A}$ is an additive category.
The homology functor $H^i: \text{Comp}(\mathcal{A}) \to \mathcal{A}$ taking a complex $K^\bullet \mapsto H^i(K^\bullet)$ extends to functors $H^i: K(\mathcal{A}) \to \mathcal{A}$ and $H^i: D(\mathcal{A}) \to \mathcal{A}$.
Lemma 41.72.2. An object $E$ of $D(\mathcal{A})$ is contained in $D^+(\mathcal{A})$ if and only if $H^i(E) =0 $ for all $i \ll 0$. Similar statements hold for $D^-$ and $D^+$.
Proof. Hint: use truncation functors. See Derived Categories, Lemma 12.10.5. $\square$
Lemma 41.72.3. Morphisms between objects in the derived category.
- Let $I^\bullet \in \text{Comp}^+(\mathcal{A})$ with $I^n$ injective for all $n \in \mathbf{Z}$. Then $$ \mathop{\rm Hom}\nolimits_{D(\mathcal{A})}(K^\bullet, I^\bullet) = \mathop{\rm Hom}\nolimits_{K(\mathcal{A})}(K^\bullet, I^\bullet). $$
- Let $P^\bullet \in \text{Comp}^-(\mathcal{A})$ with $P^n$ is projective for all $n \in \mathbf{Z}$. Then $$ \mathop{\rm Hom}\nolimits_{D(\mathcal{A})}(P^\bullet, K^\bullet) = \mathop{\rm Hom}\nolimits_{K(\mathcal{A})}(P^\bullet, K^\bullet). $$
- If $\mathcal{A}$ has enough injectives and $\mathcal{I} \subset \mathcal{A}$ is the additive subcategory of injectives, then $ D^+(\mathcal{A})\cong K^+(\mathcal{I}) $ (as triangulated categories).
- If $\mathcal{A}$ has enough projectives and $\mathcal{P} \subset \mathcal{A}$ is the additive subcategory of projectives, then $ D^-(\mathcal{A}) \cong K^-(\mathcal{P}). $
Proof. Omitted. $\square$
Definition 41.72.4. Let $F: \mathcal{A} \to \mathcal{B}$ be a left exact functor and assume that $\mathcal{A}$ has enough injectives. We define the total right derived functor of $F$ as the functor $RF: D^+(\mathcal{A}) \to D^+(\mathcal{B})$ fitting into the diagram $$ \xymatrix{ D^+(\mathcal{A}) \ar[r]^{RF} & D^+(\mathcal{B}) \\ K^+(\mathcal I) \ar[u] \ar[r]^F & K^+(\mathcal{B}). \ar[u] } $$ This is possible since the left vertical arrow is invertible by the previous lemma. Similarly, let $G: \mathcal{A} \to \mathcal{B}$ be a right exact functor and assume that $\mathcal{A}$ has enough projectives. We define the total right derived functor of $G$ as the functor $LG: D^-(\mathcal{A}) \to D^-(\mathcal{B})$ fitting into the diagram $$ \xymatrix{ D^-(\mathcal{A}) \ar[r]^{LG} & D^-(\mathcal{B}) \\ K^-(\mathcal{P}) \ar[u] \ar[r]^G & K^-(\mathcal{B}). \ar[u] } $$ This is possible since the left vertical arrow is invertible by the previous lemma.
Remark 41.72.5. In these cases, it is true that $R^iF(K^\bullet) = H^i(RF(K^\bullet))$, where the left hand side is defined to be $i$th homology of the complex $F(K^\bullet)$.
\section{Derived categories}
\label{section-derived-categories}
\noindent
To set up notation, let $\mathcal{A}$ be an abelian category. Let
$\text{Comp}(\mathcal{A})$ be the abelian category of complexes in
$\mathcal{A}$. Let $K(\mathcal{A})$ be the category of complexes up to
homotopy, with objects equal to complexes in $\mathcal{A}$ and objects equal to
homotopy classes of morphisms of complexes. This is not an abelian category.
Loosely speaking, $D(A)$ is defined to be the category obtained by inverting
all quasi-isomorphisms in $\text{Comp}(\mathcal{A})$ or, equivalently, in
$K(\mathcal{A})$. Moreover, we can define $\text{Comp}^+(\mathcal{A}),
K^+(\mathcal{A}), D^+(\mathcal{A})$ analogously using only bounded below
complexes. Similarly, we can define $\text{Comp}^-(\mathcal{A}),
K^-(\mathcal{A}), D^-(\mathcal{A})$ using bounded above complexes, and we can
define $\text{Comp}^b(\mathcal{A}), K^b(\mathcal{A}), D^b(\mathcal{A})$ using
bounded complexes.
\begin{remark}
\label{remarks-derived-categories}
Notes on derived categories.
\begin{enumerate}
\item
There are some set-theoretical problems when $\mathcal{A}$ is somewhat
arbitrary, which we will happily disregard.
\item
The categories $K(A)$ and $D(A)$ may be endowed with the structure of
triangulated category, but we will not need these structures in the following
discussion.
\item
The categories $\text{Comp}(\mathcal{A})$ and $K(\mathcal{A})$ can also be
defined when $\mathcal{A}$ is an additive category.
\end{enumerate}
\end{remark}
\noindent
The homology functor $H^i: \text{Comp}(\mathcal{A}) \to \mathcal{A}$ taking a
complex $K^\bullet \mapsto H^i(K^\bullet)$ extends to functors $H^i:
K(\mathcal{A}) \to \mathcal{A}$ and $H^i: D(\mathcal{A}) \to \mathcal{A}$.
\begin{lemma}
\label{lemma-when-in-bounded}
An object $E$ of $D(\mathcal{A})$ is contained in $D^+(\mathcal{A})$ if and
only if $H^i(E) =0 $ for all $i \ll 0$. Similar statements hold for $D^-$ and
$D^+$.
\end{lemma}
\begin{proof}
Hint: use truncation functors. See
Derived Categories, Lemma \ref{derived-lemma-bounded-derived}.
\end{proof}
\begin{lemma}
\label{lemma-derived-categories}
Morphisms between objects in the derived category.
\begin{enumerate}
\item
Let $I^\bullet \in \text{Comp}^+(\mathcal{A})$ with $I^n$ injective for all
$n \in \mathbf{Z}$. Then
$$
\Hom_{D(\mathcal{A})}(K^\bullet, I^\bullet)
=
\Hom_{K(\mathcal{A})}(K^\bullet, I^\bullet).
$$
\item
Let $P^\bullet \in \text{Comp}^-(\mathcal{A})$ with $P^n$ is projective for all
$n \in \mathbf{Z}$. Then
$$
\Hom_{D(\mathcal{A})}(P^\bullet, K^\bullet)
=
\Hom_{K(\mathcal{A})}(P^\bullet, K^\bullet).
$$
\item
If $\mathcal{A}$ has enough injectives and $\mathcal{I} \subset \mathcal{A}$
is the additive subcategory of injectives, then
$
D^+(\mathcal{A})\cong K^+(\mathcal{I})
$
(as triangulated categories).
\item
If $\mathcal{A}$ has enough projectives and $\mathcal{P} \subset \mathcal{A}$
is the additive subcategory of projectives, then
$
D^-(\mathcal{A}) \cong K^-(\mathcal{P}).
$
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-derived-functor}
Let $F: \mathcal{A} \to \mathcal{B}$ be a left exact functor and assume that
$\mathcal{A}$ has enough injectives. We define the {\it total right derived
functor of $F$} as the functor $RF: D^+(\mathcal{A}) \to D^+(\mathcal{B})$
fitting into the diagram
$$
\xymatrix{
D^+(\mathcal{A}) \ar[r]^{RF} & D^+(\mathcal{B}) \\
K^+(\mathcal I) \ar[u] \ar[r]^F & K^+(\mathcal{B}). \ar[u]
}
$$
This is possible since the left vertical arrow is invertible by the previous
lemma. Similarly, let $G: \mathcal{A} \to \mathcal{B}$ be a right exact
functor and assume that $\mathcal{A}$ has enough projectives. We define the
{\it total right derived functor of $G$} as the functor $LG: D^-(\mathcal{A})
\to D^-(\mathcal{B})$ fitting into the diagram
$$
\xymatrix{
D^-(\mathcal{A}) \ar[r]^{LG} & D^-(\mathcal{B}) \\
K^-(\mathcal{P}) \ar[u] \ar[r]^G & K^-(\mathcal{B}). \ar[u]
}
$$
This is possible since the left vertical arrow is invertible by the previous
lemma.
\end{definition}
\begin{remark}
\label{remark-cohomology-of-derived-functor}
In these cases, it is true that $R^iF(K^\bullet) = H^i(RF(K^\bullet))$, where
the left hand side is defined to be $i$th homology of the complex
$F(K^\bullet)$.
\end{remark}
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\cite[\href{http://stacks.math.columbia.edu/tag/03T3}{Tag 03T3}]{stacks-project}
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