Tensor product of multiplier algebras is contained in multiplier of tensor product. I want to realize the following: Given two $C^*$-algebras $A$ and $B$ and with "$\otimes$" denoting the minimal (i.e., spatial) tensor product. We have that
$$M(A)\otimes M(B)\subseteq M(A\otimes B)$$
How do I show this? I suppose that one can realize $A\otimes B$ as an essential ideal in $M(A)\otimes M(B)$, but I am unsure how to proceed.
Any help is appreciated.
 A: Claim: There is a unique $*$-homomorphism $\iota: M(A)\otimes M(B) \to M(A\otimes B)$ that is the identity on $A\odot B$. Moreover, $\iota$ is injective and unital.
To prove the claim, let us recall the following:
Theorem: If $I$ is an ideal in a $C^*$-algebra $C$, there is a unique $*$-homomorphism $\phi: C \to M(I)$ such that the diagram

commutes. Moreover, if $I$ is essential in $C$, then $\phi$ is injective.
Proof: (sketch) Realise $M(I)$ as double centralisers. Given $c \in C$, the pair $(L_c, R_c)\in \operatorname{End}(I)\oplus\operatorname{End}(I)$ is an element of $M(I)$, and $\phi: C \to M(I): c \mapsto (L_c, R_c)$ is the desired map.
Finally, assume that $I$ is essential in $C$. Since $\phi\vert_I$ is injective, we have $I \cap \ker(\phi)=0$. Since $I$ is essential, this is only possible if $\ker(\phi)=0$. Thus $\phi$ is injective. $\quad \square$
Apply this result with $I=A\otimes B$, which is an essential ideal in $C= M(A)\otimes M(B)$.
A: Considering a faithful, non-degenerate representation of $A$ on a Hilbert space $H$, we may assume without loss of
generality that $A\subseteq \mathscr B(H)$, and similarly that $B\subseteq \mathscr B(K)$, where $K$ is another Hilbert space.
It is well known that $M(A)$ is faithfully represented in $\mathscr B(H)$ via the so called double centralizers.  Similarly
$M(B)\subseteq \mathscr B(K)$.
Since all tensor products considered are spacial, then both $A\otimes B$ and  $M(A)\otimes M(B)$ are faithfully represented in
$\mathscr B(H\otimes K)$, with
$A\otimes B$ being an ideal in $M(A)\otimes M(B)$.  Consequently every element of $M(A)\otimes M(B)$ acts (concretely) as a multiplier of
$A\otimes B$, hence leading up to a *-homomorphism
$$
  M(A)\otimes M(B) \to  M(A\otimes B).
  $$
The main outstanding point is then to show that this map is one-to-one, which is to say that   $A\otimes B$ is an essential
ideal in $M(A)\otimes M(B)$.  In order to show this, suppose that $T$ is an element of $M(A)\otimes M(B)$ such that $T\cdot (A\otimes B)=0$, and
we must show that $T=0$.
Actually,  for all we care, $T$ could be any bounded operator on $H\otimes K$, whatsoever.
By assumption we have for all $a$ in $A$, $b$ in $B$, $\xi \in  H$, and $\eta \in  K$, that
$$
  0  = T(a\otimes b)(\xi \otimes \eta ) =
  T\big (a(\xi )\otimes b(\eta )\big ).
  $$
Now, taking into account that $A$ and $B$ are non-degenerate algebras of operators, the linear span of the vectors of the form
$a(\xi )\otimes b(\eta )$ is dense in $H\otimes K$, so we deduce that $T=0$, as desired.
