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Consider the following situation in some triangulated category: We are given a collection of distinguished triangles $A_n \to B_n \to C_n \to A_n[1]$ indexed by the natural numbers, together with maps (in the obvious sense) from the n'th triangle to the (n+1)'st. If $A \to B \to C$ is the colimit of this system of triangles, is it also a distinguished triangle?

It would be really interesting to have a proof or a counterexample, or possible a proof depending on some additional hypotheses on the triangulated category in question.

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  • $\begingroup$ A colimit in your category is a limit in the opposite category, so perhaps we can find a counterexample in the opposite category of complexes of abelian groups. $\endgroup$
    – thel
    Commented Jul 26, 2010 at 3:17
  • $\begingroup$ I am not completely sure what you mean by colimit here. Are you assuming that your triangulated category has a model so that you can take a homotopy colimit? $\endgroup$
    – Greg Stevenson
    Commented Jul 26, 2010 at 11:26
  • $\begingroup$ Actually I was not completely sure myself what I meant. I am trying to understand this for a specific application and wasn't sure which case I need. When writing the question I was thinking of a situation where the colimit happens to exist, but I also had in mind a homotopy colimit. Now I learnt from Tom that they coincide, which was a big surprise. About your last question: As far as I understand one can define sequential hocolim in any triangulated category, without assuming existence of a model, as a cone of the shift map on the direct sum of all terms, or something like that. $\endgroup$
    – Andreas Holmstrom
    Commented Jul 26, 2010 at 21:01
  • $\begingroup$ Yes, provided one has countable coproducts one can always define linear countable homotopy colimits as you say. This does not give the correct definition for larger cardinalities of indexing set though even in the linear case. If one has a countable sequence of triangles I believe it is true that the homotopy colimits of their terms in this sense do form a triangle. One should be able to obtain this fairly directly from the 3x3 Lemma (as in May's The additivity of traces in triangulated categories). In particular it doesn't use anything more than the octahedral axiom. $\endgroup$
    – Greg Stevenson
    Commented Jul 26, 2010 at 21:30
  • $\begingroup$ Aha, I guess I never had to think of uncountable homotopy colimits so far ;-) and thanks, I will check the reference to May. $\endgroup$
    – Andreas Holmstrom
    Commented Jul 27, 2010 at 0:33

3 Answers 3

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I strongly suspect that the answer is 'no'. You see: when you use the triangulated version of the definition of (countable) homotopy limits, you question becomes a partial case of the following one: can a morphism of distinguished triangles be completed to a $3\times3$-commutative diagram whose rows and columns are distinguished triangles? The answer to the latter question is 'no' (though any commutative square could be completed to such a diagram); looking at it one can probably construct a counterexample for your original question.

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this is true in the topological setting. cofibrations can be thought of as colimits, they are actually colimits of a diagram (i have been told, but i cant recall the example) and colimits commute with colimits. In the setting i am thinking of cofibrations are the distinguished triangles. So i doubt you will find a counterexample in general since the result is definitely true for at least one triangulated category. Unfortunately, i do not see how this could be extended to other triangulated categories.

It does not seem like a result that would be true in general given my above reasons for believing the result. I will ask about the diagram it is the colimit/homotopy colimit of.

Is there a particular triangulated category you are interested in?

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  • $\begingroup$ Yes, I'm interested in the motivic stable homotopy category over some base scheme. Most statements that hold in the topological setting should hold there as well. $\endgroup$
    – Andreas Holmstrom
    Commented Jul 26, 2010 at 7:08
  • $\begingroup$ Maybe an argument in this spirit could work, but one question would be the following: Does homotopy colimits commute with homotopy colimits in general? $\endgroup$
    – Andreas Holmstrom
    Commented Jul 26, 2010 at 9:06
  • $\begingroup$ i would imagine that homotopy colimits better commute with homotopy colimits. if they don't then we are in trouble. have you looked at any dugger-isaksen stuff? i think it might be the case that in a triangulate model category the statement you ask about is in fact true, so that would pass through to your setting. However, all this is speculation that should be cleared up by someone more knowledgeable, if one of the obvious parties doesnt check in before i get a response from someone i will post it in my answer. $\endgroup$
    – Sean Tilson
    Commented Jul 26, 2010 at 17:23
  • $\begingroup$ Tom is one of the obvious parties... $\endgroup$
    – Sean Tilson
    Commented Jul 26, 2010 at 17:23
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    $\begingroup$ I think that in your answer you only consider morphisms of inductive systems in the homotopy category that could be lifted to the level of spaces. For a morphism of inductive systems that does not possess such a lift, the answer is probably "no" in general. $\endgroup$
    – Mikhail Bondarko
    Commented Aug 9, 2010 at 9:56
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(Deligne has an assertion that a certain triangle in the ind-category of a triangulated category can be written as a limit of distinguished triangles there; see SGA 4, XVII, Lemme 1.2.2.1, http://www.msri.org/publications/books/sga/sga/4-3/4-3t_275.html . Does not seem to answer the question, though.)

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