Triangulated structure on the opposite category 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, they dont 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?
 A: 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.)
