If $(\text{id}_A\otimes \text{ev}_x)(z)= 0$ for all $x \in X$. Do we have $z=0$? 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 $z \in M(A \otimes C(X))$.
 A: 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 and references therein.  A simple proof can be found in Ryan's book.  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.

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, $z=0$.
