## The residue class functor from a Frobenius category to its stable category induces a functor on cube-shaped diagrams - is it dense?

Let $\mathcal{E}$ be a Frobenius category, i.e. an exact category with sufficiently many bijective objects. (Such as e.g. the category of complexes over an additive category.)

Let $\underline{\mathcal{E}}$ be its stable category, defined as the factor category of $\mathcal{E}$ modulo its full subcategory of bijective objects. We have the residue class functor $R:\mathcal{E}\to\underline{\mathcal{E}}$. (For complexes, this is the canonical functor from the category of complexes to its homotopy category.)

Consider the category $D := \Delta_1\times\Delta_1\times\Delta_1$, i.e. a cube. Denote by $\mathcal{E}(D)$ the category of functors from $D$ to $\mathcal{E}$, i.e. the category of $\mathcal{E}$-valued diagrams of shape $D$. Etc.

Consider the functor $R(D):\mathcal{E}(D)\to\underline{\mathcal{E}}(D)$, obtained by pointwise application of $R$. Is this functor $R(D)$ dense, i.e. surjective on isoclasses? In other words, given a stably commutative cube-shaped diagram, does there exist a strictly commutative cube-shaped diagram isomorphic to it in $\underline{\mathcal{E}}(D)$?

My guess would be: no. But I was unable to construct a counterexample. I was searching in $\mathcal{E} = {\bf Z}/p^a\text{-mod}$.

The paper [1] deals with the question in greater generality, but it doesn't seem to give an explicit answer to my question. (This question can be generalised: one does not have to stick to the Frobenius case, and one can ask for obstructions depending on $D$ against the density of $R(D)$.)

[1] Dwyer, W. G.; Kan, D. M.; Smith, J. H., Homotopy commutative diagrams and their realizations, J. Pure Appl. Alg. 57, p. 5-24, 1989.

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