## Homotopy Equivalence of Mapping Cones

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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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. – Fernando Muro Jan 21 at 20:55

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. – Marc Hoyois Jan 22 at 2:38