Let $A$ and $B$ be $C^\*$-algebras. Given $f \in B^\*$, we can form the right slice map
$$\iota \otimes f: A \otimes B \to A: a \otimes b \mapsto af(b)$$
which extends uniquely to a bounded linear map
$$\iota \otimes f: M(A \otimes B) \to M(A)$$
that is strictly continuous on the unit ball.


Assume $X \in M(A \otimes B)$ satisfies $(\iota \otimes f)(X)=0$ for all $f \in A^\*$. Can we conclude that $X=0?$


**Attempt**: When $B$ is unital, we can proceed as follows: if $a\in A$
$$0 = (\iota\otimes f)(X)a = (\iota \otimes f)(X(a \otimes 1\_B))$$ for all $f \in A^\*$, so since $X(a \otimes 1\_B) \in A \otimes B$ we conclude that $X(a \otimes 1\_B)=0$. Hence, $X(A \otimes B) = 0$ which implies $X=0$.


How to deal with the case that $B$ is non-unital?



$(\iota \otimes f)(X) = 0$ for all $f \in B^*$ implies $X=0$
From comments, it seems that the OP is using the "abstract" definition of multipliers (compare below). A good reference is indeed the appendix of [arXiv:funct-an/9707009](https://arxiv.org/abs/funct-an/9707009). Let's use some remarks from there (bottom of page 38) to show that $\iota\otimes f:A\otimes B\rightarrow A$ is indeed strict. By Cohen--Hewitt factorisation, we can find $g\in B^\*, c\in B$ with $f = cg$, and so
$$ (\iota\otimes f)(a\otimes b) = f(b) a = g(bc) a = (\iota\otimes g)(a\otimes bc) $$
Thus if $(u\_i)$ is a bounded net in $A\otimes B$ converging strictly to $0$, for $a\in A$ we have that
$$ (\iota\otimes f)(u\_i) a = (\iota\otimes g)(u\_i(a\otimes c)) \rightarrow 0 $$
because $u\_i(a\otimes c)\rightarrow 0$ in norm.


So we form the strict extension to $M(A\otimes B)$. If $(u\_i)$ is a bounded net in $A\otimes B$ converging strictly to $X\in M(A\otimes B)$ then by definition of the strict extension (or by strict continuity),
$$ (\iota\otimes f)(X) a = \lim\_i (\iota\otimes f)(u\_i) a \qquad (a\in A). $$
However, this is equal to
$$ \lim\_i (\iota\otimes g)(u\_i(a\otimes c)) = (\iota\otimes g)(X(a\otimes c)). $$
So if $(\iota\otimes f)(X)=0$ for all $f$, then $(\iota\otimes g)(X(a\otimes c)) = 0$ for all $g,c$ and $a$, and so $X(a\otimes c)=0$ for all $a,c$ so $X=0$.


(I think of this as the "factorisation trick". Aside from CP maps, most examples of strict linear maps seem to feature some notion of "factorisation".)




---


Taka's comment was to use a representation of $A\otimes B$ on a Hilbert space. This is the "centraliser" picture of multipliers: if $A\subseteq\mathcal B(H), B\subseteq\mathcal B(K)$ acting non-degenerately, then $A\otimes B\subseteq\mathcal B(H\otimes K)$ non-degenerately and
$$ M(A\otimes B) \cong \{ T\in\mathcal B(H\otimes K) : Tu, uT\in A\otimes B \ (u\in A\otimes B) \}. $$
This is independent of the representations chosen, so let's suppose that $f\in B^\*$ is the restriction of $\omega\_{\xi,\eta}$ to $B$. Then we have a natural notion of what $(\iota\otimes f)$ is acting on $M(A\otimes B)$: just the restriction of $\iota\otimes\omega\_{\xi,\eta}:\mathcal B(H\otimes K) \rightarrow \mathcal B(H)$. Of course, you'd need to check that this gave the same definition as before. The required result is now obvious.


