Good morphisms of distinguished triangles: can Neeman's method be applied to the motivic stable homotopy category?

It is well known that non-uniqueness of a cone for a morphism in a triangulated category $C$ makes constructing exact functors (of triangulated categories) a difficult task. In section 3 of his "Some new axioms for triangulated categories" Neeman proposes an alternative axiomatics of triangulated categories (that includes a certain notion of 'good morphism' of distinguished triangles). I wonder whether Neeman's methods can be applied to the motivic stable homotopy category $SH$ of Morel and Voevodsky? Can any better methods for 'rigidifying' (for example, higher categories?) be applied to $SH$?

Upd. It seems that I need an category $G$ whose objects and morphism are morally distinguished triangles in $C$ and their morphisms + some extra data. Any $C$-morphism can be extended to an object of $G$, any commutative square could be extended to a $G$-morphism, whereas for any $G$-morphism of distinguished triangles $X\to Y\stackrel{v}{\to} Z\stackrel{w}{\to} X[1]$ and $X'\to Y'\stackrel{v}{\to} Z'\stackrel{w'}{\to} X'[1]$ that is $0$ on $X$ and $Y$ the corresponinding morphism from $Z$ to $Z'$ equals $v'\circ \theta \circ w$ for some $\theta\in C(X[1],Y')$ (this is the correction of the Neeman's axiom GTR2).

Denis-Charles Cisinski has written some very interesting comments on the relations between various types of 'nice' triangulated categories; yet I wonder which texts treat this topic.

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Doesn't the motivic stable homotopy category come from some nice stable model category (or stable $infty$-category)? If so it should satisfy any reasonable proposed axioms for a triangulated category, since cones upstairs in the model category (or $\infty$-category) are as unique as it makes sense to make them. –  Dylan Wilson Nov 9 '12 at 21:32
Dear Dylan Wilson, you are probably right. Yet I would prefer to have a proof of this statement. Does you argument yield a way to define (functorially) 'good morphisms of triangles' (so that a commutative square could be extended to a morphism of triangles in a way that is 'unique up to a homotopy')? –  Mikhail Bondarko Nov 9 '12 at 21:54
See the remark at the end of section 3 of the referenced paper of Neeman. The argument he gives showing that the stable homotopy category of spectra is "new triangulated" should work equally well in any stable model category (possibly with a few functorial cofibrant or fibrant replacements thrown in). –  Dylan Wilson Nov 9 '12 at 23:00
In the following list, each element defines an object of the next kind in the list: Stable model category, Stable $\infty$-category, Grothendieck's triangulated derivator (aka stable homotopy theory in the sense of Heller), filtered triangulated category in the sense of Beilinson (aka $f$-categories), Neeman's triangulated category, Verdier's classical triangulated category. And $SH$ is obtained from a stable model category by construction, isn't it? So you can pick up your favourite structure to work with. –  Denis-Charles Cisinski Nov 10 '12 at 0:49
Another comment: the principle of Neeman's axioms is that we want enough structure to be able to define the cone of a morphism of triangles. But, whenever your triangulated category comes from something in the list above between a stable model category and a filtered triangulated category (included), there is much more than Neeman is asking: there is a genuine triangulated category of distinguished triangles (and this is why Neeman's axioms are very obviously true in this case). –  Denis-Charles Cisinski Nov 10 '12 at 0:56