TITLE: Is 'the' homotopy colimit of a sequence of exact triangles an exact triangle?
QUESTION [9 upvotes]: 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.

REPLY [5 votes]: 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.