Does the stable category of a nice exact category embed in (the underlying category of) a derivator?

In Derivators, Pointed Derivators, and Stable Derivators, Moritz Groth gives as an example of a non-invertible morphism with trivial cone an inclusion $f:X\to I$. Here $X$ is an object of injective dimension $1$ in an exact category $\mathcal{E}$ with enough injectives, $I$ is an injective, and the stable category $\underline{\mathcal{E}}$ has a "suspended structure" given in the same way as the triangulation of the stable category of a Frobenius category but without the invertibility of suspension.

This is fine as far as it goes, but it's not obvious to me that this is an example of a cone in a pointed derivator, what he's just defined, and Groth doesn't give any indication of how to see it as such.

Question: how can I realize this example as a cone in a derivator?

Having looked around, I don't see any evidence that $\underline{\mathcal{E}}$ is known to be the underlying category of a derivator, even with extra assumptions a la a paper of Stovicek to make $\mathcal{E}$ more like a Grothendieck category. Is it, in fact, such an underlying category, given some extra assumptions? Or can we, perhaps, give it an exact embedding in a homotopy category, and extend to a derivator in that way?

• Is your question answered by the paper webusers.imj-prg.fr/~bernhard.keller/publ/…? – AAK Feb 23 '15 at 5:24
• @Adeel No, but thanks-I should have mentioned that paper, which gives a derivator with underlying category the derived category of an exact category, not the stable category-there are no chain complexes involved in the latter. – Kevin Arlin Feb 23 '15 at 5:27
• I would check with Cisinski's paper whether the axioms of a derivable category holds. I don't have it at hand now. – Fernando Muro Jul 5 '15 at 21:12

This is not exactly answering your question but: a Frobenius category $\mathcal{E}$ (with suitable assumptions) can be given the structure of a model category whose homotopy category is the stable category $\underline{\mathcal{E}}$. See James Gillespie's http://arxiv.org/abs/1009.3574. Now take the derivator associated with this model category structure (again, I am not so sure about the required assumptions). Its underlying category will be the stable category $\underline{\mathcal{E}}$.