## 61.6 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 morphisms 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.

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 61.6.2. 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 13.11.5.
$\square$

Lemma 61.6.3. 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

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ). \]

Let $P^\bullet \in \text{Comp}^-(\mathcal{A})$ with $P^ n$ is projective for all $n \in \mathbf{Z}$. Then

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(P^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , K^\bullet ). \]

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 61.6.4. 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 left 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.

## Comments (4)

Comment #14 by Emmanuel Kowalski on

Comment #21 by Johan on

Comment #2167 by Alex on

Comment #2196 by Johan on