Let $I$ be a filtered poset, which you should think of as being huge. Let $A\_i$ be an $I$-diagram of $C^{\star}$-algebras and let $A$ be the colimit of this diagram; if necessary, we can also assume that all structure maps $A\_i \rightarrow A\_j$ are inclusions. Given a nuclear $C^{\star}$-algebra $N$, is it then true that tensoring with $N$ commutes with the colimit, i.e. that $A \otimes N$ is the colimit of $A\_i \otimes N$? In particular, I am interested in the case $N = C(X)$ for some compact space $X$. 


I expect this to be true (and seem to have a proof for C(X) at least), but have not been able to find anything in the literature, so a reference would be welcome. 



When do tensor products of C*-algebras commute with colimits?
If $N$ is exact and the tensor products are minimal then $A\otimes N$ is the colimit of the $A\_i\otimes N$'s. Say the connecting maps are $\phi\_{i,j}$. Then to check that $A\otimes N$
is the colimit of $\{A\_i\otimes N,\phi\_{i,j}\otimes\mathrm{id}\_N\}$ two properties must be verified:


(1) the union of the ranges of the maps $\phi\_{i,\infty}\otimes \mathrm{id}\_N$ is dense in $A\otimes N$, 


(2) for each $i$, $\ker (\phi\_{i,\infty}\otimes \mathrm{id}\_N)=\overline{\bigcup\_{j>i} \ker({\phi\_{i,j}\otimes \mathrm{id}\_N})}$. 


The first property is straightforward, since the span of the elementary tensors $\phi\_{i,\infty}(a)\otimes n$ is dense. The second property follows from the fact that $\ker(\phi\_{i,\infty}\otimes \mathrm{id}\_N)=\ker(\phi\_{i,\infty})\otimes N$ and the correspoding property (2) for the colimit of the $A\_i$'s. 


That $\ker(\phi\otimes \mathrm{id}\_N)=\ker(\phi)\otimes N$ follows, if $\phi$ is surjective, by exactness of $N$, if $\phi$ is injective, by the minimality of $\otimes$
(the minimal tensor product behaves well with respect to inclusions), and the general case is a composite of those two. 


