Throughout let $B$ be a stable C*-algebra, i.e. $B\cong B\otimes K$, where $K$ is the C*-algebra of compact operators on an infinite dimensional separable Hilbert space. It is well-known that any countably generated Hilbert $B$-module $X$ is singly generated, i.e. there exists a positive element $b\in B$ such that $X\cong \overline{bB}$.

Assume the following as the definition of c.p.c. order zero maps between C*-algebras.

**Definition** A completely positive linear map $\phi:A\to B$ between C*-algebras has order zero if there exist a positive element $h\in\mathcal M(C)\cap C'$ and a $*$-homomorphism $\pi:A\to\mathcal M(C)\cap\{h\}'$ such that $\Vert h\Vert = 1$ and
$$\phi(a) = h\pi(a)=\pi(a)h$$
for any $a\in A$, where $C = C^*(\phi(A))\subset B$, i.e. the C*-algebra generated by the image of $\phi$ and $\mathcal M(C)$ is the multiplier algebra of $C$.

Let $A$ be a separable C*-algebra and let $\phi:A\to B$ be a c.p.c. order zero map. One can construct a Hilbert $B$-module $X_\phi$ out of $\phi$ by setting $$X_\phi:=\overline{\phi(A)B}.$$ Conversely, given a countably generated Hilbert $B$-module $X$ and a separable C*-algebra $A$, when is that there exists a c.p.c. order zero map $\phi:A\to B$ such that $X\cong\overline{\phi(A)B}$?

As a special case, suppose that $X$ is the inductive limit of a sequence of isometric inclusions of modules $\overline{\phi_n(A)B}\hookrightarrow\overline{\phi_{n+1}(A)B}$, where $\{\phi_n\}_{n\in\mathbb N}$ is a sequence of c.p.c. order zero maps. Is there a c.p.c. order zero map $\phi$ such that $X\cong\overline{\phi(A)B}$? I believe this last question has a positive answer when the connecting maps "commute" with the c.p.c. order zero maps, so I would rather be interested in the case where there are no *a priori* connections between the $\phi_n$s.