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Suppose we have an additive category $\mathcal{A}$ and we consider the homotopy category of chain complexes in $\mathcal{A}$, denoted by $\mathcal{K}(\mathcal{A})$. If we have $X_1, X_1', X_2, X_2' \in \mathcal{K}(\mathcal{A})$ such that $ \Psi_i:X_i \to X_i'$ is a homotopy equivalence for $i = 1,2$, and $f:X_1 \to X_2$ a morphism, then what is the induced homotopy equivalence on the mapping cones $\Gamma(f:X_1 \to X_2)$ and $\Gamma(\Psi_2f\Psi_1': X_1' \to X_2')$?

(Where $\Psi_1'$ is the map such that $\Psi_1'\Psi_1 - id_{X_1} = \partial(\psi_1)$ for some homotopy $\psi_1$, and with the similar condition for $\Psi_1\Psi_1'$).

I know that such a map should exist by the axioms of a triangulated category and the 5-lemma, but I can't seem to pin down the actual map in terms of the information already present. Ultimately the issue is that part of the map between cones from $X_1$ to $X_2'$ and $X_1'$ to $X_2$ seem to need closed corrections which are complicating things.

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    $\begingroup$ Constructing explicitly the map may be a mess, but it is completely elementary. An excersise, in order to fully understand why the homotopy category is triangulated. $\endgroup$
    – Fernando Muro
    Commented Jan 21, 2013 at 20:55
  • $\begingroup$ Up to choice of homotopy, it's pretty clear how to construct the maps from looking at the induced homotopy exact squares. I'm specifically having issues with the fact that for us to have a homotopy equivalence of cones that the choice of homotopy is important (or I'm missing something completely). I'll go back and think about the proof more and thanks! $\endgroup$
    – user30838
    Commented Jan 21, 2013 at 23:28
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    $\begingroup$ You indeed only get a canonically induced map of cones once you choose the homotopy $\psi_1$. That's why taking cones is not a functorial construction in a triangulated category. To make the cone construction functorial you must consider the map $\psi_1$ as part of the data defining a commutative square, which is what higher category theory does. $\endgroup$
    – Marc Hoyois
    Commented Jan 22, 2013 at 2:38
  • $\begingroup$ Model category arguments, and I presume also for triangulated categories, tend to be non explicit on homotopies. So I would expect my answer to mathoverflow.net/questions/130116 could be relevant, since it gives explicit desription of homotopies in an analogous situation, relevant to, indeed used as a basis for a basis for, gluing homotopy equivalances. $\endgroup$
    – Ronnie Brown
    Commented May 14, 2013 at 6:53
  • $\begingroup$ @Ronnie, I thing that model category arguments are usually very explicit about homotopies. $\endgroup$
    – Fernando Muro
    Commented May 28, 2013 at 9:10

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