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 $z\in F \otimes B?$

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!