Example 22.26.6 (Differential graded category of complexes). Let $\mathcal{B}$ be an additive category. We will construct a differential graded category $\text{Comp}^{dg}(\mathcal{B})$ over $R = \mathbf{Z}$ whose associated category of complexes is $\text{Comp}(\mathcal{B})$ and whose associated homotopy category is $K(\mathcal{B})$. As objects of $\text{Comp}^{dg}(\mathcal{B})$ we take complexes of $\mathcal{B}$. Given complexes $A^\bullet $ and $B^\bullet $ of $\mathcal{B}$, we sometimes also denote $A^\bullet $ and $B^\bullet $ the corresponding graded objects of $\mathcal{B}$ (i.e., forget about the differential). Using this abuse of notation, we set

as a graded $\mathbf{Z}$-module with notation and definitions as in Example 22.25.5. In other words, the $n$th graded piece is the abelian group of homogeneous morphism of degree $n$ of graded objects

Observe reversal of indices and observe we have a direct product and not a direct sum. For an element $f \in \mathop{\mathrm{Hom}}\nolimits ^ n(A^\bullet , B^\bullet )$ of degree $n$ we set

The sign is exactly as in More on Algebra, Section 15.71. To make sense of this we think of $\text{d}_ B$ and $\text{d}_ A$ as maps of graded objects of $\mathcal{B}$ homogeneous of degree $1$ and we use composition in the category $\text{Gr}^{gr}(\mathcal{B})$ on the right hand side. In terms of components, if $f = (f_{p, q})$ with $f_{p, q} : A^{-q} \to B^ p$ we have

Note that the first term of this expression is in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(A^{-q}, B^{p + 1})$ and the second term is in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(A^{-q - 1}, B^ p)$. The reader checks that

$\text{d}$ has square zero,

an element $f$ in $\mathop{\mathrm{Hom}}\nolimits ^ n(A^\bullet , B^\bullet )$ has $\text{d}(f) = 0$ if and only if the morphism $f : A^\bullet \to B^\bullet [n]$ of graded objects of $\mathcal{B}$ is actually a map of complexes,

in particular, the category of complexes of $\text{Comp}^{dg}(\mathcal{B})$ is equal to $\text{Comp}(\mathcal{B})$,

the morphism of complexes defined by $f$ as in (2) is homotopy equivalent to zero if and only if $f = \text{d}(g)$ for some $g \in \mathop{\mathrm{Hom}}\nolimits ^{n - 1}(A^\bullet , B^\bullet )$.

in particular, we obtain a canonical isomorphism

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{B})}(A^\bullet , B^\bullet ) \longrightarrow H^0(\mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{B})}(A^\bullet , B^\bullet )) \]and the homotopy category of $\text{Comp}^{dg}(\mathcal{B})$ is equal to $K(\mathcal{B})$.

Given complexes $A^\bullet $, $B^\bullet $, $C^\bullet $ we define composition

by composition $(g, f) \mapsto g \circ f$ in the graded category $\text{Gr}^{gr}(\mathcal{B})$, see Example 22.25.5. This defines a map of differential graded modules

as required in Definition 22.26.1 because

as desired.

## Comments (5)

Comment #286 by arp on

Comment #299 by Johan on

Comment #300 by arp on

Comment #301 by Johan on

Comment #304 by Johan on