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.
Remark 64.6.1. 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) are endowed with the structure of a triangulated category.
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 64.6.2. 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^ b.
Proof. Hint: use truncation functors. See Derived Categories, Lemma 13.11.5. \square
Lemma 64.6.3. 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
Let P^\bullet \in \text{Comp}^-(\mathcal{A}) with P^ n is projective for all n \in \mathbf{Z}. Then
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 64.6.4. 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
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
This is possible since the left vertical arrow is invertible by the previous lemma.
Remark 64.6.5. 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 ith homology of the complex F(K^\bullet ).
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