Is 'the' homotopy colimit of a sequence of exact triangles an exact triangle?

Question

Let $$ \begin{array}{rccccl} A_0&\to& B_0&\to& C_0&\to\\ \downarrow & &\downarrow&&\downarrow\\ A_1&\to& B_1&\to& C_1&\to\\ \downarrow & &\downarrow&&\downarrow\\ \vdots & &\vdots&&\vdots\\ \end{array} $$ be a commutative diagram in a triangulated category such that all the rows are exact triangles. One defines the homotopy colimit of a sequence $A_0\to A_1\to\ldots$ as a third object in an exact triangle $$ \coprod A_n\xrightarrow{1-shift}\coprod A_n\to \operatorname{hocolim}_nA_* $$ which is unique up to (non-unique) isomorphism. For a morphism of diagrams $A_*\to B_*$ (as above), one gets a (non-unique) morphism $\operatorname{hocolim}_nA_*\to \operatorname{hocolim}_nB_*$.

Given the diagram above, is there a sequence of morphisms $$ \operatorname{hocolim}_nA_*\to \operatorname{hocolim}_nB_*\to \operatorname{hocolim}_nC_*\to $$ which is an exact triangle? I am especially interested in the case that $B_*$ is a constant diagram.

I was trying to use the 9-lemma for triangulated categories, but this explicitely constructs the morphisms $\coprod C_n\to \coprod C_n\to \tilde C$ and I cannot prove, why the first map has to be $1-shift$ and therefore why $\tilde C$ has to be the homotopy colimit of $C_0\to C_1\to\ldots$. This question is strongly related, I think.

Answer 1

I'll assume that you at least want your triangulated category to have the property that a (countable) coproduct of exact triangles is an exact triangle?

Even then, I think this is probably not true in a general triangulated category, but it is in many (most?) familiar examples. For example, in the derived category (of an abelian category where coproducts of exact sequences are exact), it's possible to "rigidify" the original directed system of exact triangles to get a directed system of short exact sequences of complexes, so that your question becomes a question about genuine colimits of complexes, rather than homotopy colimits in the derived category.

Answer 2

The answer is yes if you assume that your triangulated category is the basis of a triangulated derivator, see Appendix 2 in

MR3031826 Keller, Bernhard(F-PARIS7-IMJ); Nicolás, Pedro(E-MURC2ED) Weight structures and simple dg modules for positive dg algebras. (English summary) Int. Math. Res. Not. IMRN 2013, no. 5, 1028–1078.

Among all possible enhancements of triangulated categories, derivators are the closest to the original concept. I mean, it's known how to construct a triangulated derivator out of any other kind of enhancement, but not the other way around. The reason that derivators suffice is that the poset $\mathbb N$ is a free category.

Answer 3

If your triangulated category arises a homotopy category of a stable $\infty$-category (as most well-known examples do), and the commutative chain of squares $A_i \to B_i$ arises from a diagram in the $\infty$-category (as many natural diagrams do), the answer is yes. An exact triangle is given by a hocolim of the diagram $0 \leftarrow A_i \to B_i$, and hocolims commute past each other, so

$$ hocolim ( 0 \leftarrow hocolim_{\mathbb{Z}_{\geq 0}} A_i \to hocolim_{\mathbb{Z}_{\geq 0}} B_i) \simeq hocolim_{\mathbb{Z}_{\geq 0}} hocolim ( 0 \leftarrow A_i \to B_i) \simeq hocolim_{\mathbb{Z}_{\geq 0}} C_i $$

i.e., we have an exact triangle

$$ hocolim_{\mathbb{Z}_{\geq 0}} A_i \to hocolim_{\mathbb{Z}_{\geq 0}} B_i \to hocolim_{\mathbb{Z}_{\geq 0}} C_i $$