Derivators - diagrams in homotopy category of chain complexes

$\require{AMScd}$ 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.

• Let me give an indirect argument to show that $K(A^J)$ is not the same as $K(A)^{J}$. Indeed, let $J=[1]=(0\to 1)$ be the category with two objects and one non-identity homomorphism between them. Then $K(A)^J$ is the category of morphisms in $K(A)$. As you probably know, $K(A)$ is a triangulated category, so for any morphism $\phi:X\to Y$ you can construct a mapping cone $C(\phi)$ such that $X\to Y\to C(\phi)\to$ is a triangle. Now, this construction is not functorial: if it were functorial, then $K(A)$ would be semisimple, and this is not the case in general. (it continues below...) – Simone Virili Nov 19 '16 at 16:26
• now, the mapping cone gives you a functor $K(A^{[1]})\to K(A)$, thus, whenever $K(A^{[1]})$ is equivalent to $K(A)^{[1]}$ you actually get functorial cones in your triangulated category $K(A)$. So, to find a counterexample you just need a category $A$ for which $K(A)$ is not semisimple... you have quite a lot of choices! – Simone Virili Nov 19 '16 at 16:28
• You can play similar games with $\mathbb N$ instead of $[1]$, in that case you will get that an equivalence of $K(A)^{\mathbb N}$ with $K(A^{\mathbb N})$ gives you that homotopy colimits are the same as (usual, categorical) colimits taken in $K(A)$, and this is of course not always the case. – Simone Virili Nov 19 '16 at 16:30
• If you want something more precise, you can prove that, given two objects $X=(X_0\to X_1)$ and $Y=(Y_0\to Y_1)$ in $K(A)^{[1]}$, then there is an exact sequence of Abelian groups as follows: $K(A)(\Sigma X_1,Y_1)\oplus K(A)(\Sigma X_0,Y_0)\to (\Sigma X_0,Y_1)\to K(A^{[1]})(X,Y)\to K(A)^{[1]}(X,Y)\to 0$. Thus, whenever the first map is not epi, you obtain that the natural functor $K(A^{[1]})\to K(A)^{[1]}$ is not faithful (it is always full and essentially surjective). – Simone Virili Nov 19 '16 at 16:53