Triangulated structure on the opposite category

category-theoryhomological-algebratriangulated-categories

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

where

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?

Best Answer

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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