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I've been working through some of the early parts of Neeman's book on triangulated categories, and he mentions that he does not know of a pre-triangulated category that is not triangulated. Is this still an open question? Actually, let me be a bit more specific and break this into two parts:

  1. The usual axioms for a triangulated category are known to be redundant. Do we know of a list of independent axioms? (Peter May wrote something about this, and it's the best I've seen but it still doesn't give independence results for the new list of axioms.)
  2. Assuming it is known that, say, the octahedral axiom is independent from the others, what is an example of a pre-triangulated category that is not a triangulated category?

It appears that if there is such an example, it would have to be very artificial... all of the pre-triangulated categories appearing in nature are automatically triangulated. (I haven't read all of May's article but it looks like he explains how one usually goes about proving that the octahedral axiom holds for a given category, so perhaps this would give some clues for situations where this would fail.)

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    $\begingroup$ I've never thought hard about producing a counterexample but I have discussed this with a couple of people relatively recently and they weren't aware of any examples of a category which was pre-triangulated but not triangulated. $\endgroup$
    – Greg Stevenson
    Commented Jul 14, 2010 at 23:48
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    $\begingroup$ Part 2 of this question came up in a local reading group - even though it's not so recent, I hope people won't mind my attempting to bump it up again. Is there still no known example, abstract or otherwise? $\endgroup$
    – Jan Grabowski
    Commented May 19, 2014 at 15:28

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Don't know if you're still interested in this question, but for anyone who is curious about it, the answer seems to be that there are no pretriangulated categories that are not triangulated.

Macioca Pre-triangulated categories are triangulated. Unfortunately, the article is removed by the author with the comment: "There is a serious error at the end of the proof of 3.4 which cannot be fixed. There is a counterexample to the statement of 3.4 due to Canonaco and Kunzer. Lemma 3.5 also needs stronger hypotheses."

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    $\begingroup$ Wow, I find it amazing that such a basic result took 50 years for anyone to figure out! $\endgroup$
    – Eric Wofsey
    Commented Jun 3, 2015 at 1:47
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    $\begingroup$ I think it's a great, major achievement! $\endgroup$
    – Fernando Muro
    Commented Jun 3, 2015 at 4:04
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    $\begingroup$ Whoa! I definitely didn't expect that result to be true- very cool! $\endgroup$
    – Dylan Wilson
    Commented Jun 3, 2015 at 8:00
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    $\begingroup$ The author has apparently removed the preprint from the arxiv (see arxiv.org/abs/1506.00887v2), leaving the following comment: "There is a serious error at the end of the proof of 3.4 which cannot be fixed. Thee [sic] is a counterexample to the statement of 3.4 due to Canonaco and Kunzer. Lemma 3.5 also needs stronger hypotheses. Many thanks to those whonsent [sic] comments, counterexamples and encouragement" $\endgroup$
    – Ricardo Andrade
    Commented Jun 17, 2015 at 14:37
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    $\begingroup$ Perhaps all the excitement was premature... $\endgroup$
    – Ricardo Andrade
    Commented Jun 17, 2015 at 14:41
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I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake.

Lemma 3.5, Counterexample - take group $G=$ sequences of integer numbers $=\mathbb{Z}^\mathbb{N}$.

$\ell$ and $m$ are right and left shift respectively.

Then $m\circ\ell=id$, so $Image(e)=G$.

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  • $\begingroup$ Why is this a counterexample? Take the isomorphism asked of in the lemma to be $l$. $\endgroup$
    – Dylan Wilson
    Commented Jun 8, 2015 at 15:25
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    $\begingroup$ well, it is not isomorphism ) right shift don't have sequence 1 0 0 0 ... in it's image $\endgroup$
    – Tomas Paul
    Commented Jun 8, 2015 at 15:31
  • $\begingroup$ although I'm not sure if this is very important - this lemma is used to construct automorphism of some Ext, which is zero in many cases $\endgroup$
    – Tomas Paul
    Commented Jun 8, 2015 at 16:27
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    $\begingroup$ (I have informed the author of the paper) $\endgroup$
    – André Henriques
    Commented Jun 8, 2015 at 16:39
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Yes, I'm about to post some fixes. My interest in it was really just the k-linear case. In fact, that example is not a counterexample to the main result because the map omega_* cannot be an injection (as the spectral sequence implies that the kernel and cokernel must either both not vanish or both vanish). However,in replacing Lemma 3.5 I need an assumption on the enrichment. This means that the result is still open for the general case. It seems to work for k-linear, divisible, finitely generated and, as it were, co-finitely generated (ie the duals of finitely generated groups). The assumptions are not likely to be necessary though and I would expect that someone better equipped to deal with the infinite group case might be able to fix it.

There are a couple of of other blunders at the end. In particular, there is some mistake in the uniqueness result. I've removed it in the meantime.

I'll post the fixes on arxiv in a day or so. Antony Maciocia

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I think there do exists some pretriangulated categories which are not triangulated. There are at least two methods to produce such examples, although I don't have any concrete example.

The first is algebraic. Let $\mathcal{A}$ be an abelian category, $\mathcal{B}$ a functorially finite subcategory. Then the quotient category $\mathcal{A}/\mathcal{B}$ is a pretriangulated category, but it is usually not triangulated.

The second is topological. Take a Quillen model category which is also an additive category, but which is not a stable model category. Then the homotopy category of this model category is a pretriangualted category which is not triangulated.

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    $\begingroup$ I'm highly skeptical of both of these 'examples'. $\endgroup$
    – Dylan Wilson
    Commented Mar 7, 2017 at 1:08
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    $\begingroup$ The quotient of an abelian category by a functorially finite subcategory is not in general pretriangulated. For example, the zero subcategory is always functorially finite. $\endgroup$
    – Jeremy Rickard
    Commented Mar 7, 2017 at 9:33
  • $\begingroup$ I am very sorry that I misunderstood the question. I had another definition in head, that is, pretriangulated categories are additive categories which are both left triangulated categories and right triangulated categories etc. Very sorry for this wrong answer! $\endgroup$
    – motivique
    Commented Mar 11, 2017 at 13:27

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