$\require{AMScd}$ Let $\mathcal{A}$ be an additive category and $K(\mathcal{A})$ be the homotopy category of $\mathcal{A}$, i.e. the category of chain complexes $Ch(\mathcal{A})$ over $\mathcal{A}$ localized at homotopy equivalences. (This is the same as the category with objects $Ch(\mathcal{A})$ and morphisms $Hom_{K(\mathcal{A})}(A,B) = Hom_{Ch(\mathcal{A})}(A,B)/\sim$, where $f\sim g$ if $f$ and $g$ are chain homotopic).

My main question is, if $D(J) = K(\mathcal{A}^J)$ defines a derivator. While I was trying to show this, I came to a point of showing that for $J$ any diagram category, the two categories $K(\mathcal{A}^J), K(\mathcal{A})^J$ are isomorphic.

I know that $Ch(\mathcal{A}^J) = Ch(\mathcal{A})^J$. I assume that this is not true. It is easy to construct a functor $K(\mathcal{A}^J) \to K(\mathcal{A})^J$, but it is not clear how to construct an inverse.

My problem is that for two objects $F^n,G^n \colon J \to \mathcal{A}$ in $Ch(\mathcal{A}^J)$ (i.e. sequences of functors) and two homotopic morphisms $\eta,\nu\colon F\Rightarrow G$ the homotopy is a natural transformation $h^n\colon F^n \Rightarrow G^{n-1}$ such that for each $m\colon j\to j'$ in $J$, we have $ \eta^n(m) - \nu^n(m) = dh^n(m) - h^{n+1}(m)d$.

On the other hand in $K(\mathcal{A})^J$ two morphisms $\eta',\nu'$ of the functors $F,G$ from above are homotopic, if levelwise there exists a homotopy, i.e. for $m\colon j\to j'$ in $J$, the diagrams

\begin{CD} F^n(j) @>F^n(m)>> F^n(j')\\ @V \eta'_j V\nu'_j V @V \eta'_{j'} V\nu'_{j'} V\\ G^n(j) @>>G^n(m)> G^n(j') \end{CD}

commutes up to homotopy for each $n$. But it is unclear, why these levelwise homotopies should come from an homotopy which is also a natural transformation.