Let $\mathcal{A}$ be an abelian category, it is known that any complex $A^{\bullet}$ admits a distinguished triangle
$$B^{\bullet}\rightarrow A^{\bullet}\rightarrow C^{\bullet}\rightarrow B^{\bullet}[1]$$
in the unbounded derived category $D(\mathcal{A})$ for some $B^{\bullet}\in D^-(\mathcal{A})$ and $C^{\bullet}\in D^+(\mathcal{A})$.


In other words, any complex can be decomposed into a complex bounded above and a complex bounded below.


Can we do the same thing in the unbounded homotopy category $K(\mathcal{A})$?



Decompose an unbounded (cochain) complex in the homotopy category
Yes. Let
$$\tau^{\leq0}A^\bullet:= \cdots\to A^{-2}\to A^{-1}\to\ker(d^0)\to0\to\cdots$$
be the usual truncation of $A^\bullet$. Then the mapping cone of the inclusion map $\tau^{\leq0}A^\bullet\to A^\bullet$ is homotopy equivalent to
$$\rho^{>0}A^\bullet:=\cdots\to0\to\ker(d^0)\to A^0\to A^1\to\cdots,$$
so there is a distinguished triangle
$$\tau^{\leq0}A^\bullet\to A^\bullet\to\rho^{>0}A^\bullet\to(\tau^{\leq0}A^\bullet)[1]$$
in $K(\mathcal{A})$.


Or another way to see this triangle: If the contractible complex
$$K^\bullet:=\cdots\to0\to\ker(d^0)\xrightarrow{\sim}\ker(d^0)\to0\to\cdots$$
is concentrated in degrees $-1$ and $0$, then there is a degreewise split short exact sequence
$$0\to\tau^{\leq0}A^\bullet\to A^\bullet\oplus K^\bullet\to\rho^{>0}A^\bullet\to0.$$


