Is there an accessible proof for the following fact?



> 
> If $A=C\_0(X)$ with $X$ locally compact Hausdorff and $B$ is a
> $C^\ast$-algebra then $M(A\otimes B)$ is the set of bounded strictly
> continuous functions $X \to M(B)$.
> 
> 
> 


Denote the set of bounded strictly continuous functions by $C\_b^s (X, M(B))$.


Thanks to the hint in the comments, we can say the following:


Given $x \in X$, there is a mapping
$$\pi\_x: C\_0(X) \otimes B \to B: f \otimes b \mapsto f(x)b$$
which extends to a map
$$\pi\_x: M(C\_0(X) \otimes B) \to M(B)$$
and this allows us to define
$$M(C\_0(X) \otimes B) \to C\_b^s(X,M(B)): L \mapsto (x \mapsto \pi\_x(L))$$


Why is this an isomorphism of $C^\*$-algebras, i.e. why is it injective and surjective?



Identifying the multiplier $C^*$-algebra $M(C_0(X) \otimes B)$
Indeed, Blackadar appears to have made a mistake here. But the reference he gives is good, and seems to give both a correct statement, and a proof:


Akemann, Charles A.; Pedersen, Gert K.; Tomiyama, Jun
Multipliers of C∗-algebras.
J. Functional Analysis 13 (1973), 277–301.


[MR470685](https://mathscinet.ams.org/mathscinet-getitem?mr=470685) [DOI: 10.1016/0022-1236(73)90036-0](https://doi.org/10.1016/0022-1236(73)90036-0)


See Corollary 3.4.


**Edit:** How I would approach this. Firstly, understand carefully the proof that $C\_0(X) \otimes B \cong C\_0(X,B)$. Much the same ideas are used for the multiplier algebra case. In particular, for $f\in C\_0(X), b\in B$ we identify $f\otimes b$ with the continuous map $X\rightarrow B; x\mapsto f(x) b$.


I would look at $\Phi: C^b\_{str}(X, M(B)) \rightarrow M(C\_0(X,B))$ defined by pointwise multiplication:


* First show this is well-defined. This is easy, as for $F\in C^b\_{str}(X, M(B))$ we have that $F(f\otimes b) \in C\_0(X,B)$. Then copy the proof that $C\_0(X) \otimes B \cong C\_0(X,B)$ to show that $F$ does multiply $C\_0(X,B)$ into itself.
* $\Phi$ is clearly injective.
* To show $\Phi$ is surjective, argue as in the OP: given $L\in M(C\_0(X,B))$ we define $F(x) = \pi\_x(L)$. Then $X\rightarrow B; x \mapsto \pi\_x(L) f(x) b
= \pi\_x(L(f\otimes b))$ is continuous, for each $f\in C\_0(X)$ and $b\in B$. This is enough to show that $F$ is continuous, for the strict topology
* Checking that $\Phi$ is a $\*$-homomorphism is routine.


