I'm trying to learn Triangulated Category by reading some online notes and I have a question about the "turning the Triangles" Axiom. Below is the Axiom from the notes
Now, given a distinguished triangle XYZ, this axiom allows us to "turn the triangle" clockwise and still get a distinguished triangle T(X)YZ.
I wonder if there is a way to "turn the triangle" XYZ counterclockwise to get another distinguished triangle?
In the notes, the translation functor $T$ is assumed to be an automorphism of the category $\mathscr{C}$ i.e there exists a functor $T^{-1}: \mathscr{C} \to \mathscr{C}$ such that $T \circ T^{-1}$ and $T^{-1} \circ T$ are naturally isomorphic to the identity functors.
I tried to use this fact so $T \circ T^{-1}(Z) \cong Z$ along with relating $h$ to $T \circ T^{-1}(h)$ to show that the triangle $XY(T \circ T^{-1})(Z)$ is isomorphic to the original triangle XYZ and then use the backward direction of the axiom above. But I can't seem to show that they are isomorphic.
Thank you.