Lemma 13.11.1. Let $\mathcal{A}$ be an abelian category. The functor

is homological.

In this section we construct the derived category of an abelian category $\mathcal{A}$ by inverting the quasi-isomorphisms in $K(\mathcal{A})$. Before we do this recall that the functors $H^ i : \text{Comp}(\mathcal{A}) \to \mathcal{A}$ factor through $K(\mathcal{A})$, see Homology, Lemma 12.13.11. Moreover, in Homology, Definition 12.14.8 we have defined identifications $H^ i(K^\bullet [n]) = H^{i + n}(K^\bullet )$. At this point it makes sense to redefine

\[ H^ i(K^\bullet ) = H^0(K^\bullet [i]) \]

in order to avoid confusion and possible sign errors.

Lemma 13.11.1. Let $\mathcal{A}$ be an abelian category. The functor

\[ H^0 : K(\mathcal{A}) \longrightarrow \mathcal{A} \]

is homological.

**Proof.**
Because $H^0$ is a functor, and by our definition of distinguished triangles it suffices to prove that given a termwise split short exact sequence of complexes $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ the sequence $H^0(A^\bullet ) \to H^0(B^\bullet ) \to H^0(C^\bullet )$ is exact. This follows from Homology, Lemma 12.13.12.
$\square$

In particular, this lemma implies that a distinguished triangle $(X, Y, Z, f, g, h)$ in $K(\mathcal{A})$ gives rise to a long exact cohomology sequence

13.11.1.1

\begin{equation} \label{derived-equation-long-exact-cohomology-sequence-D} \xymatrix{ \ldots \ar[r] & H^ i(X) \ar[r]^{H^ i(f)} & H^ i(Y) \ar[r]^{H^ i(g)} & H^ i(Z) \ar[r]^{H^ i(h)} & H^{i + 1}(X) \ar[r] & \ldots } \end{equation}

see (13.3.5.1). Moreover, there is a compatibility with the long exact sequence of cohomology associated to a short exact sequence of complexes (insert future reference here). For example, if $(A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta )$ is the distinguished triangle associated to a termwise split exact sequence of complexes (see Definition 13.9.9), then the cohomology sequence above agrees with the one defined using the snake lemma, see Homology, Lemma 12.13.12 and for agreement of sequences, see Homology, Lemma 12.14.11.

Recall that a complex $K^\bullet $ is *acyclic* if $H^ i(K^\bullet ) = 0$ for all $i \in \mathbf{Z}$. Moreover, recall that a morphism of complexes $f : K^\bullet \to L^\bullet $ is a *quasi-isomorphism* if and only if $H^ i(f)$ is an isomorphism for all $i$. See Homology, Definition 12.13.10.

Lemma 13.11.2. Let $\mathcal{A}$ be an abelian category. The full subcategory $\text{Ac}(\mathcal{A})$ of $K(\mathcal{A})$ consisting of acyclic complexes is a strictly full saturated triangulated subcategory of $K(\mathcal{A})$. The corresponding saturated multiplicative system (see Lemma 13.6.10) of $K(\mathcal{A})$ is the set $\text{Qis}(\mathcal{A})$ of quasi-isomorphisms. In particular, the kernel of the localization functor $Q : K(\mathcal{A}) \to \text{Qis}(\mathcal{A})^{-1}K(\mathcal{A})$ is $\text{Ac}(\mathcal{A})$ and the functor $H^0$ factors through $Q$.

**Proof.**
We know that $H^0$ is a homological functor by Lemma 13.11.1. Thus this lemma is a special case of Lemma 13.6.11.
$\square$

Definition 13.11.3. Let $\mathcal{A}$ be an abelian category. Let $\text{Ac}(\mathcal{A})$ and $\text{Qis}(\mathcal{A})$ be as in Lemma 13.11.2. The *derived category of $\mathcal{A}$* is the triangulated category

\[ D(\mathcal{A}) = K(\mathcal{A})/\text{Ac}(\mathcal{A}) = \text{Qis}(\mathcal{A})^{-1} K(\mathcal{A}). \]

We denote $H^0 : D(\mathcal{A}) \to \mathcal{A}$ the unique functor whose composition with the quotient functor gives back the functor $H^0$ defined above. Using Lemma 13.6.4 we introduce the strictly full saturated triangulated subcategories $D^{+}(\mathcal{A}), D^{-}(\mathcal{A}), D^ b(\mathcal{A})$ whose sets of objects are

\[ \begin{matrix} \mathop{\mathrm{Ob}}\nolimits (D^{+}(\mathcal{A})) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (D(\mathcal{A})) \mid H^ n(X) = 0\text{ for all }n \ll 0\}
\\ \mathop{\mathrm{Ob}}\nolimits (D^{-}(\mathcal{A})) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (D(\mathcal{A})) \mid H^ n(X) = 0\text{ for all }n \gg 0\}
\\ \mathop{\mathrm{Ob}}\nolimits (D^ b(\mathcal{A})) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (D(\mathcal{A})) \mid H^ n(X) = 0\text{ for all }|n| \gg 0\}
\end{matrix} \]

The category $D^ b(\mathcal{A})$ is called the *bounded derived category* of $\mathcal{A}$.

If $K^\bullet $ and $L^\bullet $ are complexes of $\mathcal{A}$ then we sometimes say “$K^\bullet $ is *quasi-isomorphic* to $L^\bullet $” to indicate that $K^\bullet $ and $L^\bullet $ are isomorphic objects of $D(\mathcal{A})$.

Remark 13.11.4. In this chapter, we consistently work with “small” abelian categories (as is the convention in the Stacks project). For a “big” abelian category $\mathcal{A}$, it isn't clear that the derived category $D(\mathcal{A})$ exists, because it isn't clear that morphisms in the derived category are sets. In fact, in general they aren't, see Examples, Lemma 109.61.1. However, if $\mathcal{A}$ is a Grothendieck abelian category, and given $K^\bullet , L^\bullet $ in $K(\mathcal{A})$, then by Injectives, Theorem 19.12.6 there exists a quasi-isomorphism $L^\bullet \to I^\bullet $ to a K-injective complex $I^\bullet $ and Lemma 13.31.2 shows that

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

which is a set. Some examples of Grothendieck abelian categories are the category of modules over a ring, or more generally the category of sheaves of modules on a ringed site.

Each of the variants $D^{+}(\mathcal{A}), D^{-}(\mathcal{A}), D^ b(\mathcal{A})$ can be constructed as a localization of the corresponding homotopy category. This relies on the following simple lemma.

Lemma 13.11.5. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $ be a complex.

If $H^ n(K^\bullet ) = 0$ for all $n \ll 0$, then there exists a quasi-isomorphism $K^\bullet \to L^\bullet $ with $L^\bullet $ bounded below.

If $H^ n(K^\bullet ) = 0$ for all $n \gg 0$, then there exists a quasi-isomorphism $M^\bullet \to K^\bullet $ with $M^\bullet $ bounded above.

If $H^ n(K^\bullet ) = 0$ for all $|n| \gg 0$, then there exists a commutative diagram of morphisms of complexes

\[ \xymatrix{ K^\bullet \ar[r] & L^\bullet \\ M^\bullet \ar[u] \ar[r] & N^\bullet \ar[u] } \]where all the arrows are quasi-isomorphisms, $L^\bullet $ bounded below, $M^\bullet $ bounded above, and $N^\bullet $ a bounded complex.

**Proof.**
Pick $a \ll 0 \ll b$ and set $M^\bullet = \tau _{\leq b}K^\bullet $, $L^\bullet = \tau _{\geq a}K^\bullet $, and $N^\bullet = \tau _{\leq b}L^\bullet = \tau _{\geq a}M^\bullet $. See Homology, Section 12.15 for the truncation functors.
$\square$

To state the following lemma denote $\text{Ac}^{+}(\mathcal{A})$, $\text{Ac}^{-}(\mathcal{A})$, resp. $\text{Ac}^ b(\mathcal{A})$ the intersection of $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, resp. $K^ b(\mathcal{A})$ with $\text{Ac}(\mathcal{A})$. Denote $\text{Qis}^{+}(\mathcal{A})$, $\text{Qis}^{-}(\mathcal{A})$, resp. $\text{Qis}^ b(\mathcal{A})$ the intersection of $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, resp. $K^ b(\mathcal{A})$ with $\text{Qis}(\mathcal{A})$.

Lemma 13.11.6. Let $\mathcal{A}$ be an abelian category. The subcategories $\text{Ac}^{+}(\mathcal{A})$, $\text{Ac}^{-}(\mathcal{A})$, resp. $\text{Ac}^ b(\mathcal{A})$ are strictly full saturated triangulated subcategories of $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, resp. $K^ b(\mathcal{A})$. The corresponding saturated multiplicative systems (see Lemma 13.6.10) are the sets $\text{Qis}^{+}(\mathcal{A})$, $\text{Qis}^{-}(\mathcal{A})$, resp. $\text{Qis}^ b(\mathcal{A})$.

The kernel of the functor $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is $\text{Ac}^{+}(\mathcal{A})$ and this induces an equivalence of triangulated categories

\[ K^{+}(\mathcal{A})/\text{Ac}^{+}(\mathcal{A}) = \text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{A}) \]The kernel of the functor $K^{-}(\mathcal{A}) \to D^{-}(\mathcal{A})$ is $\text{Ac}^{-}(\mathcal{A})$ and this induces an equivalence of triangulated categories

\[ K^{-}(\mathcal{A})/\text{Ac}^{-}(\mathcal{A}) = \text{Qis}^{-}(\mathcal{A})^{-1}K^{-}(\mathcal{A}) \longrightarrow D^{-}(\mathcal{A}) \]The kernel of the functor $K^ b(\mathcal{A}) \to D^ b(\mathcal{A})$ is $\text{Ac}^ b(\mathcal{A})$ and this induces an equivalence of triangulated categories

\[ K^ b(\mathcal{A})/\text{Ac}^ b(\mathcal{A}) = \text{Qis}^ b(\mathcal{A})^{-1}K^ b(\mathcal{A}) \longrightarrow D^ b(\mathcal{A}) \]

**Proof.**
The initial statements follow from Lemma 13.6.11 by considering the restriction of the homological functor $H^0$. The statement on kernels in (1), (2), (3) is a consequence of the definitions in each case. Each of the functors is essentially surjective by Lemma 13.11.5. To finish the proof we have to show the functors are fully faithful. We first do this for the bounded below version.

Suppose that $K^\bullet , L^\bullet $ are bounded above complexes. A morphism between these in $D(\mathcal{A})$ is of the form $s^{-1}f$ for a pair $f : K^\bullet \to (L')^\bullet $, $s : L^\bullet \to (L')^\bullet $ where $s$ is a quasi-isomorphism. This implies that $(L')^\bullet $ has cohomology bounded below. Hence by Lemma 13.11.5 we can choose a quasi-isomorphism $s' : (L')^\bullet \to (L'')^\bullet $ with $(L'')^\bullet $ bounded below. Then the pair $(s' \circ f, s' \circ s)$ defines a morphism in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$. Hence the functor is “full”. Finally, suppose that the pair $f : K^\bullet \to (L')^\bullet $, $s : L^\bullet \to (L')^\bullet $ defines a morphism in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$ which is zero in $D(\mathcal{A})$. This means that there exists a quasi-isomorphism $s' : (L')^\bullet \to (L'')^\bullet $ such that $s' \circ f = 0$. Using Lemma 13.11.5 once more we obtain a quasi-isomorphism $s'' : (L'')^\bullet \to (L''')^\bullet $ with $(L''')^\bullet $ bounded below. Thus we see that $s'' \circ s' \circ f = 0$ which implies that $s^{-1}f$ is zero in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$. This finishes the proof that the functor in (1) is an equivalence.

The proof of (2) is dual to the proof of (1). To prove (3) we may use the result of (2). Hence it suffices to prove that the functor $\text{Qis}^ b(\mathcal{A})^{-1}K^ b(\mathcal{A}) \to \text{Qis}^{-}(\mathcal{A})^{-1}K^{-}(\mathcal{A})$ is fully faithful. The argument given in the previous paragraph applies directly to show this where we consistently work with complexes which are already bounded above. $\square$

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