Let $A$ be a $C^\*$-algebra (not necessarily unital). Let $X$ be a compact Hausdorff space. We can consider the minimal $C^\*$-tensor product $A \otimes C(X)$. On this space, we can consider the slice map
$$\text{id}\_A\otimes \text{ev}\_x: A \otimes C(X) \to A: a \otimes f \mapsto f(x)a$$


Suppose that $(\text{id}\_A\otimes \text{ev}\_x)(z)= 0$ for all $x \in X$. Do we have $z=0$?


We can also further extend the slice map further on the multiplier $C^\*$-algebra


$$\text{id}\_A\otimes \text{ev}\_x: M(A \otimes C(X)) \to M(A)$$


Is the same (stronger) statement still true?


If $z$ is an algebraic tensor this is easy: write $z= \sum\_i a\_i \otimes f\_i$ where the $a\_i$'s are chosen to be linearly independent. Then from the assumption we have $\sum\_i f\_i(x) a\_i = 0$ for all $x \in X$ and hence by linear independence $f\_i = 0$ for all $i$ whence $z=0$. But I don't see how to treat the cases that $z \in A \otimes C(X)$ or $z \in M(A \otimes C(X))$.



If $(\text{id}_A\otimes \text{ev}_x)(z)= 0$ for all $x \in X$. Do we have $z=0$?
The answer is "yes".


For $A\otimes C(X)$ we have the standard identification with $C(X,A)$ the space of (bounded) continuous maps $X\rightarrow A$ with the sup norm. Here $a\otimes f$ is identified with the function $x\mapsto f(x)a$. Then $(\operatorname{id}\_A\otimes \operatorname{ev}\_x)$ is identified with the map $C(X,A)\rightarrow A$ given by evaluation at $x$. The result follows.


For a proof, see [Theorem II.9.4.4 in Blackadar's book](https://packpages.unr.edu/media/1224/cycr.pdf) and references therein. A simple proof can be found in [Ryan's book](https://link.springer.com/book/10.1007/978-1-4471-3903-4). I am not aware of an elementary proof online; but comments are very welcome!


For the multiplier algebra case, there is the (less standard) identification of $M(A\otimes C(X)) = M(C(X,A))$ with the space of bounded, strictly-continuous maps $X\rightarrow M(A)$. I am afraid I am now aware of a good reference; but the proof is not so hard if you understand the proof that $A\otimes C(X) \cong C(X,A)$. Anyway, once you have this identification, the result also follows.




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Here's a more abstract argument. Let $A,B$ be $C^\*$-algebras and consider the tensor product $A\otimes B$. If we assume $A$ or $B$ is *nuclear* then it doesn't matter which tensor product I choose; otherwise I will assume the *minimal* tensor product. Let $X\subseteq B^\*$ be a subset which has weak$^\*$-dense linear span. We will show that for $z\in A\otimes B$, if $(\operatorname{id}\_A\otimes\phi)(z) = 0$ for all $\phi\in X$, then $z=0$.


By construction of the minimal tensor product, we may suppose there are Hilbert spaces $H,K$ so that $A\subseteq\mathcal B(H)$ and $B\subseteq\mathcal B(H)$, and then $A\otimes B$ can be identified with the norm-closed linear span of operators $a\otimes b$ in $\mathcal B(H\otimes K)$, for $a\in A,b\in B$.


For such a $z\in A\otimes B$ we have that
$$ 0 = \omega\_{\xi,\eta}\big( (\operatorname{id}\_A\otimes\phi)(z) \big)
= \phi\big( (\omega\_{\xi,\eta}\otimes\operatorname{id}\_B)(z) \big). $$
(Notice there is a bit of circularity here: to even *define* what $\operatorname{id}\_A\otimes\phi$ is, I need this sort of argument.)
This holds for all $\phi\in X$ so by weak$^\*$ linear density, it follows that $(\omega\_{\xi,\eta}\otimes\operatorname{id}\_B)(z)=0$ for all $\xi,\eta$. A similar argument then establishes that
$$ z(\xi\otimes\alpha)=0 \qquad (\xi\in H, \alpha\in K). $$
That is, $z=0$.


