Rotation axiom for the triangulation on a derivator I'm having some trouble following an argument in Moritz Groth's paper on Derivators, pointed derivators and stable derivators. More precisely, I'm currently stuck on the rotation axiom of the triangulated structure for stable derivators and I'm hopeful that someone familiar with the paper can help out.
Let $\mathbb{D}$ be a stable derivator and let $X \xrightarrow{f} Y \xrightarrow{g} Z \xrightarrow{h} \Sigma X$ be a distinguished triangle in $\mathbb{D}(e)$. We wish to prove that the rotation $Y \xrightarrow{g} Z \xrightarrow{h} \Sigma X \xrightarrow{-\Sigma f} \Sigma Y$ is distinguished.
To do so, Moritz considers $J \subseteq [2] \times [2]$ depicted as

Moreover, $i:[1] \to J$ classifies $(0,0) \to (1,0)$ and $j:J \hookrightarrow ([2] \times [2]) - \{(0,2) \} = K$. One checks that the underlying diagram of $j_! i_*(f)$ looks as follows:

At this point, I cannot see why the underlying map $\Sigma X \to \Sigma Y$ has to be $\Sigma f$ and how one formally rotates this rectangle on the right.
Groth is writing something about a unique natural transformation $(d^0 \times d^1) \Rightarrow (d^1 \times d^2)$ of functors $\ulcorner \to K$ and that $(d^0 \times d^1) : \ulcorner \to K$ differs from the usual inclusion by an automorphism. But I can't see how these help.
(Sorry, this is towards the end of the paper, so I'm really not sure how to make my question more transparent for more readers without it being infinitely long. But I hope that someone who has read Groth's paper sees this question.)
This is a repost from MSE: MSE/4555945.
 A: The $-\Sigma f$ bit follows from the following claims. Let me know if the last two seem to need further fleshing out. Keeping track of the differences in things that are all called $0$ is the trickiest part of the calculus of stable derivators.

*

*When you restrict $j_!i_*(f)$ along $d^1\times d^2,$ you get $(0,0)_*X.$

*When you restrict $j_!i_*(f)$ along $d^0\times d^1,$ you get $\sigma^*(0,0)_*Y,$ where $\sigma$ is the nontrivial automorphism of $\ulcorner.$

*By definition of the cogroup structure on suspended objects, a map $(0,0)_*X\to \sigma^*(0,0)_*Y$ evaluating to $f$ under $(0,0)^*$ induces $-\Sigma f$ when extended to the suspensions.

*Since the natural transformation $\alpha:d^1\times d^2\Rightarrow d^0\times d^1$ (pretty sure Groth has the variance backwards here) has $(0,0)\to (1,0)$ as its component at $(0,0),$ the component of $\alpha^*$ at $j_!i_*(f)$ evaluates to $f$ under $(0,0)^*.$
As for the formal reflection (not rotation!) of the rectangle on the right, observe that the rectangle defining the distinguished triangle generated by $g$ arises by restricting $j_!i_*(f)$ along the functor $t:[2]\times [1]\to K$ sending $(0,0)\mapsto (1,0),(0,1)\mapsto (2,0),(1,0)\mapsto (1,1),$ and so on. In particular, the composition of $t$ with $d^1:\ulcorner\to [2]\times [1],$ which is the standard identification of $(2,2)^*j_!i_*(f)$ with $\Sigma Y,$ sends $(1,0)$ to $(1,2)$ and $(0,1)$ to $(2,0).$ That is, the standard identification is equal to $\sigma\circ (d^0\times d^1).$
Note that if we rotated the $[2]\times [1]$-diagram by $90^\circ$ to be vertical, the induced map would be $\Sigma f$: in particular, this proves that there's no way to get the triangle-generating rectangles for $f$ and $g$ both into a connected planar diagram without flipping one of their orientations.
