MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 '13 at 20:55
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! – user30838 Jan 21 '13 at 23:28
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 '13 at 2:38
Model category arguments, and I presume also for triangulated categories, tend to be non explicit on homotopies. So I would expect my answer to 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. – Ronnie Brown May 14 '13 at 6:53
@Ronnie, I thing that model category arguments are usually very explicit about homotopies. – Fernando Muro May 28 '13 at 9:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.