Homotopy category of chain complexes

Remarks

Chain homotopy is an equivalence relation compatible with sums and composition, so $K(\mathcal{A})$ is again additive and composition is well defined. Example: in $K$ of abelian groups, the complex $C = (0 \to \Z \xrightarrow{\id} \Z \to 0)$, in degrees $1$ and $0$, is isomorphic to $0$, since $s_0 = \id \colon C_0 \to C_1$ gives $\id_C = d s + s d$. Non-example: the complex $\cdots \to \Z/4\Z \xrightarrow{\cdot 2} \Z/4\Z \xrightarrow{\cdot 2} \Z/4\Z \to \cdots$ is exact, but it is not isomorphic to $0$ in $K$: a contraction would split each short exact sequence $0 \to 2\Z/4\Z \to \Z/4\Z \to 2\Z/4\Z \to 0$, and $\Z/4\Z \not\cong \Z/2\Z \oplus \Z/2\Z$. If $\mathcal{A}$ is abelian, homology is a functor on $K(\mathcal{A})$. Bounded variants $K^b(\mathcal{A})$, $K^+(\mathcal{A})$ and $K^-(\mathcal{A})$ use only complexes that are zero in all but finitely many degrees, in low degrees, or in high degrees.