Tensor product commutes with direct limits Let $(A_n, \phi_n)$ be an inductive system of $C^*$ algebras and let $B$ be an arbitary $C^*$ algebra.
Is it true that $(\varinjlim A_n)\otimes B \cong \varinjlim (A_n \otimes B)$? 
I'm using the notation $\otimes$ for the spatial norm.
If not, please give me a counterexample, 
but if it is true let me prove this by myself...

 Thank you!
 A: Let $(A_n,\varphi_{n+1,n})_{n\in \mathbb{N}}$ be an inductive system of $C^*$-algebras with faithful connecting maps and let $B$ be an arbitrary $C^*$ algebra. Then $(\varinjlim A_n)\otimes B \cong \varinjlim (A_n\otimes B)$.
Remark: If the connecting maps are not faithful, the claim is not true.
Proof: Denote by $(A,\{\varphi^n\})$ the inductive limit of $A_1\xrightarrow{\varphi_{2,1}} A_2 \xrightarrow{\varphi_{3,2}}A_3 \xrightarrow{\varphi_{4,3}}\ldots$. For each $n\in \mathbb{N}$ we have a unique $*$-homomorphism $\varphi_{n+1,n}\otimes id_B:A_n\otimes B \to A_{n+1}\otimes B$ such that $(\varphi_{n+1,n}\otimes id_B)(a\otimes b)=\varphi_{n+1,n}(a)\otimes b$ for all $a\in A_n, b\in B$.
As $\varphi_{n+1,n}$ and $id_B$ are injective for all $n$, also $\varphi_{n+1,n}\otimes id_B$ is faithful (By exercises 3.3.5 and 3.4.1 - Brown&Ozawa).
We have an inductive system
$$A_1\otimes B \xrightarrow{\varphi_{2,1}\otimes id} A_2\otimes B \xrightarrow{\varphi_{3,2}\otimes id}A_3\otimes B \xrightarrow{\varphi_{4,3}\otimes id}\ldots$$ Denote by $(C, \{\psi^n\})$ the inductive limit.
We need to show that $C\cong A\otimes B$.
For all $n\in \mathbb{N}$ we have a commutative diagram:

(We check it commutes by verifying it on elementary tensors and then using linearity and continuity).
By the universal property of inductive limits there exists a unique $*$-homomorphism $\lambda:C\to A\otimes B$ making the diagram:

commutative. It is left to show that $\lambda$ is a $*$-isomorphism.
$\lambda$ is injective: We have $C=\overline{\cup_n{\psi^n(A_n\otimes B)}}$ and $A=\overline{\cup_n{\varphi^n(A_n)}}$. It suffices to check that $\lambda$ is injective on each $\psi^n(A_n\otimes B)$.
But, $\overline{\psi^n(A_n\odot B)}=\psi^n(A_n\otimes B)$. So, it suffices to show $\lambda$ is isometric on $\psi^n(A_n\odot B)$, and this is immediate by the commutativity of the diagram and the fact that $\varphi^n\otimes id$ is isometric. 
$\lambda$ is surjective: As the range of $\lambda$ is closed, it suffices for any $\epsilon>0$ and $y\in A\otimes B$ to find $x\in C$ s.t. $||\lambda(x)-y||<\epsilon$. So, let $z\in A\odot B$ be s.t. $||z-y||_{min}<\epsilon/2$.
Write $z=\sum_{i=1}^{m} s_i\otimes b_i$, where $s_i\in A$ and $b_i\in B$.
We can find some $n$ and $a_1,...,a_m\in A_n$ s.t. 
$||\varphi^n(a_i)-s_i||<\frac{\epsilon}{2 max_i{||b_i||} m}$ for all $1\leq i\leq m$.
Therefore, $||z-\sum_{i=1}^{m} \varphi^n(a_i)\otimes b_i||=||\sum_{i=1}^{m}(s_i-\varphi^n(a_i))\otimes b_i||\leq \sum_{i=1}^{m} ||s_i-\varphi^n(a_i)|| ||b_i||<\epsilon/2$.
Combining the two, we have: $||y-\sum_{i=1}^{m} \varphi^n(a_i)\otimes b_i||<\epsilon$ and $\lambda(\psi^n(\sum_{i=1}^{m} a_i\otimes b_i))=\varphi^n\otimes id (\sum_{i=1}^{m} a_i\otimes b_i)=\sum_{i=1}^{m} \varphi^n(a_i)\otimes b_i$ ,
as required.
I hope my solution is correct, any comments would be appreciated.
