I'm confused about the relation between signs and duality, or possibly about some essential definitions. Let me present an example of my confusion. $ \require{AMScd} \DeclareMathOperator{\op}{{op}} \DeclareMathOperator{\Hom}{{Hom}} \newcommand{\Z}{\mathbb Z} \newcommand{\eps}{\epsilon} $
Let $R$ be a ring and let $D(R)$ be the derived category of bounded complexes of $R$-modules. How is $D(R)^{\op}$ triangulated? All the sources I've found use this definition:
If $K$ is a triangulated category, then $K^{\op}$ is canonically a triangulated category by making $A^{\op}[1]:= (A[-1])^{\op}$ and saying that a triangle is distinguished iff its opposite is distinguished.
It seems to me that with this definition $R\Hom(\bullet,R)$ cannot be an exact functor. Consider the following example, with $R = \mathbb Z$. Consider the following distinguished triangle, where the columns represent complexes, and the bottom row sits in degree $0$: \begin{CD} (1) @. 0 @>{}>> 0 @>{}>> \Z @>{\epsilon_2}>> \Z\\ @. @VVV @VVV @V{\epsilon_13}VV @VVV \\ @. \Z @>{3}>> \Z @>{1}>> \Z @>>> 0. \end{CD} The signs $\epsilon_1,\epsilon_2\in \{\pm 1\}$ depend on your convention. For example, if you go by the Stacks Project or http://www-users.math.umn.edu/~tlawson/papers/signs.pdf , $\epsilon_1=\eps_2=1$. Suppose that the triangle resulting from applying $R\Hom(\bullet,\Z)$ was distinguished, i.e. the following triangle, where now the top row is in degree $0$: $$ \begin{CD} 0 @>{}>> \Z @>{1}>> \Z @>{3}>> \Z\\ @VVV @V{\epsilon_13}VV @VVV @VVV \\ \Z @>{\eps_2}>> \Z @>>> 0 @>>> 0. \end{CD} $$ By applying axiom TR2 twice to the triangle (1), we can see that this triangle is also distinguished. It also has the top row sitting in degree $0$. $$ \begin{CD} 0 @>{}>> \Z @>{-\eps_2}>> \Z @>{3}>> \Z\\ @VVV @V{-\epsilon_13}VV @VVV @VVV \\ \Z @>{-1}>> \Z @>>> 0 @>>> 0. \end{CD} $$ Now we apply TR3. Let $C_n$ be the complex $\Z\overset{n}\to\Z$. If the above two triangles are distinguished, then there must be a map filling this diagram: $$ \begin{CD} (2) @. \Z[-1] @>{\eps_2}>> C_{\eps_13} @>{1}>> \Z @>{3}>> \Z\\ @. @V{1}VV @V{\alpha}VV @V{\exists?}VV @V{1}VV \\ @. \Z[-1] @>{-1}>> C_{-\eps_13} @>{-\eps_2}>> \Z @>{3}>> \Z. \end{CD} $$ Here $\alpha$ is the map $$ \begin{CD} \Z @>{\eps_13}>> \Z\\ @V{\eps_2}VV @V{-\eps_2}VV \\ \Z @>{-\eps_13}>> \Z. \end{CD} $$ There is no arrow filling diagram (2): the right hand square forces it to be $1$, and the left hand square forces it to be $-1$.
Where am I going wrong? If the definitions need to change, what is the (a?) most reasonable convention to take?
This question is related: $D$ is triangulated. Then opposite functor $op:D\to D^{op}$ 's image is triangulated? . This might also be related: https://stacks.math.columbia.edu/tag/08I0 .