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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 $K(\mathcal{A})$?

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    $\begingroup$ No because there cannot be non trivial chain maps from C to B (let alone shifted) so if there was such a triangle A would split as a direct sum of B and C and this doesn’t happen in general. $\endgroup$
    – Fernando Muro
    Commented Mar 8, 2023 at 0:47
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    $\begingroup$ @FernandoMuro I think you may have misread the question. The OP only asks for $B^\bullet$ and $C^\bullet$ to be bounded above/below, but not bounded in any particular degree. $\endgroup$
    – Jeremy Rickard
    Commented Mar 8, 2023 at 8:46
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    $\begingroup$ @JeremyRickard thanks for the heads-up. I somehow read it as asking about a canonical t-structure. $\endgroup$
    – Fernando Muro
    Commented Mar 8, 2023 at 8:57

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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 $$0\to\tau^{\leq0}A^\bullet\to A^\bullet\oplus K^\bullet\to\rho^{>0}A^\bullet\to0.$$

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  • $\begingroup$ Do you have a reference for the homotopy equivalence $\operatorname{Cone}(\tau^{\leq 0}A^\bullet \to A^\bullet) \simeq \rho^{>0}A^\bullet$? I tried doing something like this, but couldn't find the correct homotopy... $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 8, 2023 at 12:26
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    $\begingroup$ @R.vanDobbendeBruyn I don’t know a reference, sorry. The way I saw this was not by writing down explicit homotopies, but by the fact that the obvious map from $\rho^{>0}A^\bullet$ to the mapping cone is split monic in each degree, and its cokernel is contractible (being the cone of the identity map on $\cdots\to A^{-2}\to A^{-1}\to 0\to\cdots$. I’m sure you can figure out explicit homotopies from that. $\endgroup$
    – Jeremy Rickard
    Commented Mar 8, 2023 at 13:29

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