Let $A,B$ be $C^\*$-algebras and $E$ be a right $A$-Hilbert $C^\*$-module. We can form the Hilbert $A\otimes B$ (minimal tensor product) module $E \otimes B$. If $\omega \in B^\*$, there is a unique bounded map
$$\iota \otimes \omega: E \otimes B \to E$$
extending $\iota \odot \omega$.


Now, let $F$ be a closed submodule of $E$, and assume that $z\in E \otimes B$ satisfies
$$(\iota \otimes \omega)(z)\in F$$
for all $\omega \in B^\*$. Is it true that $z\in F \otimes B?$


I'm not even sure this is true if $E= A$ is a $C^\*$-algebra!



Property that follows from conditions involving slice maps on Hilbert module
Theorem 8 from the paper "A Pathology in the Ideal Space of $L(H)\otimes L(H)$" by Simon Wassermann states that there exists an element $x \in B(H)\otimes B(H)$ such that all slices belong to the algebra of compact operators, but $x \notin K(H) \otimes B(H)$, so in general it does not hold.


The property you are after is much more common in the setting of von Neumann algebras.


