Let $(\mathsf{K},T)$ be a triangulated category and $D:\mathsf{K}\to\mathsf{K}^\text{op}$ be the usual contravariant functor which sends each object to itself and inverts all the arrows. As usual, we define a translation functor on $\mathsf{K}^\text{op}$ as $T^{\text{op}}:=D\circ T^{-1}\circ D^{-1}.$

Now, the book Derived Categories by A. Yekutieli defines a distinguished triangle on $\mathsf{K}^\text{op}$ to be a triangle


is a distinguished triangle in $\mathsf{K}$. In Categories and Sheaves by M. Kashiwara and P. Schapira, Remark 10.1.10.(ii), they don't put a minus sign on the $D(-T^{-1}(\rho))$, it is just $D(T^{-1}(\rho))$.

I think both definitions give natural triangulated structures on the opposite category and I don't see why they would coincide.

We know that $\hom(-,M):\mathsf{K}\to\mathsf{Ab}^\text{op}$ is a cohomological functor. My question is: I would hope that it is also a cohomological functor if seen as $\mathsf{K}^{\text{op}}\to\mathsf{Ab}$. Is this true for both definitions of the triangulated structure?

  • $\begingroup$ I don't understand the definition of $D$ already. $\endgroup$ May 29, 2021 at 10:25
  • $\begingroup$ @MartinBrandenburg $D$ sends an object $L\in \mathsf{K}$ to itself and a morphism $\varphi:L\to M$ to $\varphi^{\text{op}}:M\to L$. $\endgroup$
    – Gabriel
    May 29, 2021 at 10:27
  • $\begingroup$ Does the sign really change the notion of cohomological functor ? I mean, in a long exact sequence, you can replace any morphism by its opposite and the sequence will remain exact. $\endgroup$
    – Roland
    May 31, 2021 at 21:29
  • $\begingroup$ @Roland Well... maybe, I think. The problem is not on the sign by itself, is that it changes the collection of distinguished triangles. $\endgroup$
    – Gabriel
    Jun 2, 2021 at 9:24
  • $\begingroup$ Yes it changes the collection of distinguished triangles, we need to be careful about that and my guess is that there is an error in one of them (I would have put a sign, because $T^{-1}(N)\xrightarrow{-T^{-1}(\rho)} L\xrightarrow{\varphi} M\xrightarrow{\psi} N$ is distinguished and so I would want that the same triangle seen in the opposite category is distinguished, but I am not 100% sure about that). My previous comment adresses your question : if you are only interested in long exact sequences with $Hom(-,M)$, well I don't think the sign matter. $\endgroup$
    – Roland
    Jun 2, 2021 at 9:44

1 Answer 1


I will use Chapter 10, Triangulated Categories, from Categories and Sheaves, Kashiwara + Shapira as reference.

Exercise 10.10, page 266, claims that the triangulated categories:

  • $(D, T)$, a reference triangulated category, and
  • $(D^{\text{ant}}, T)$, a slightly changed version of the reference triangulated category $(D,T)$, with same objects, same morphisms, same translation functor $T$, but with distinguished triangles (d.t. for short) given by $(f,g,h)$ is d.t. in $D$ iff $(f,g,-h)$ is d.t. in $D^{\text{ant}}$, are equivalent as triangulated categories.

The equivalence functor is the identity on objects, and on morphisms sends $f$ to $-f$. Then together with a d.t. $(f,g,h)$ in $D$ we have in $D^{\text{ant}}$ the d.t. $(-f,-g,-h)$, and then any other triangle $(-\epsilon_f f, -\epsilon_g g, -\epsilon_h h)$ with $\epsilon_f \epsilon_g \epsilon_h =1$, see also Remark 10.1.10 (page 245). Use for instance

$\require{AMScd}$ \begin{CD} X @>f>> Y @>g>> Z @>h>> TX \\ @V1VV @V-1VV @V1VV @V1VV \\ X @>>-f> Y @>>-g> Z @>>h> TX \\ \end{CD}

and the fact that a triangle isomorphic to a d.t. is also a d.t. - so exactness in $D$ has a corresponding exactness in $D^{\text{ant}}$.

So up to (minus-twist-)equivalence the two definitions of a triangulated structure on the the opposite category coincide.

The question related to the "Hom's", whether they are cohomological functors is, as i understand the question, a corrolary of Proposition 10.1.13 (applied for the opposite structure).

(Note that pairwise $f^*$ and $(-f)^*$ have same kernels and images, and $g_*$ and $(-g)_*$ have same kernels and images, since composition is bilinear in additive categories. So the minus-twist-equivalence does not change the cohomology in the end.)


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