nfinite chain of adjoint morphisms.\[item:sf5\]

Combining \[thm:stable-lim-III,thm:stab-op\], we see that \[item:sf1\] implies \[item:sf4\], which clearly implies \[item:sf3\], while \[item:sf3\] implies \[item:sf2\] since the cone is a composite of a right extension by zero with a pushout. And \[item:sf2\] implies \[item:sf1\] by \[thm:stable-lim-III\]\[item:sl6\], since right adjoints preserve all limits, so the first four statements are equivalent. The equivalence of \[item:sf1\] with \[item:sf2a\], \[item:sf3a\], and \[item:sf4a\] is dual. Evidently \[item:sf5\] implies \[item:sf2\]. And conversely, if is a stable derivator, then by there are natural isomorphisms $$\Sigma F\toiso C\qquad\text{and}\qquad F\toiso\Omega C.$$ Since $\Sigma$ and $\Omega$ are equivalences in stable derivators (), this shows that the outer morphisms in the adjoint $7$-tuple match up to an equivalence. This implies that the adjoint $7$-tuple can be extended to a doubly-infinite chain of adjoint morphisms and that this chain has period six (in the obvious sense).

We conclude by offering a first interpretation and visualization of this chain of morphisms.

Let be a stable derivator. Then a few additional adjoint morphisms in the doubly-infinite sequence extending are given by: $$\ldots\dashv\pi^\ast\Omega\dashv \Sigma 0^\ast\dashv 0_\ast\Omega\dashv C\dashv 1_!\dashv 1^\ast\dashv \pi^\ast\dashv 0^\ast\dashv 0_\ast\dashv F\dashv 1_!\Sigma\dashv \Omega 1^\ast\dashv \pi^\ast\Sigma\dashv\ldots$$ In fact, this is immediate from the proof of .

In order to not get lost in all these morphisms, let us recall that Barratt–Puppe sequences in a stable derivator can be thought of as refinements of the more classical distinguished triangles. More precisely, associated to $(f\colon x\to y)\in{\sD}^{[1]}$ there is the Barratt–Puppe sequence $BP(f)$ generated by $f$. This is a coherent diagram as in which vanishes on the boundary stripes and which makes all squares bicartesian.

$$\vcenter{
\xymatrix@-1pc{
\ar@{}[dr]|{\ddots}&\ar@{}[