When do tensor products of C*-algebras commute with colimits? 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. 
 A: 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. 
