For the sake of this question I will assume the following


**Definition** A pre-C\*-algebra $A$ is *local* in the sense of [1], i.e. if there is a family of C\*-subalgebras $\{A\_i\}$ of $A$ with the property that for any $i,j$ there is $k$ such that $A\_i,A\_j\subset A\_k$ and $A=\bigcup A\_i$.


It is clear that $M\_\infty=\bigcup M\_n(\mathbb C)$ is a local C\*-algebra in the sense above. So, for a fixed positive element $k\in M\_\infty$, consider the map $\psi:M\_\infty\to M\_\infty$ given by
$$\psi(m):=m\otimes k,\qquad k\in M\_\infty.$$
As there is $\nu\in\mathbb N$ such that $k\in M\_\nu(\mathbb C)$, it follows that for any $n\in\mathbb N$ there exists $m\in\mathbb N$ such that $\psi(M\_n(\mathbb C))\subset M\_m(\mathbb C)$.


Let $A=\bigcup A\_i,B=\bigcup B\_i$ be local C\*-algebras and let $\phi:A\to B$ be a c.p.c. order zero map. Does the statement
$$\forall n\in\mathbb N\qquad\exists m\in\mathbb N\quad|\quad\phi(A\_n)\subset B\_m,$$
hold in general? It seems unlikely to me but I haven't been able to craft a counterexample so far.




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[1] Antoine, R., Perera, F., & Thiel, H. *Tensor products and regularity properties of Cuntz semigroups*. ([arxiv:1410.0483](http://arxiv.org/abs/1410.0483))



Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
The answer is `No'.


**Example:**


Let $A$ be any C\*-algebra that is not finitely generated as a C\*-algebra. For example, let $A=C(X)$ for a compact, metric space $X$ that has infinite covering dimension, such as the Hilbert cube.


Since $A$ is a C\*-algebra, the family $\mathfrak{A}=\{A\}$, consisting just of $A$ itself, is a trivial exhausting, directed family of complete subalgebras. However, there is another such family: For a subset $F$ of $A$, let $C^\*(F)$ denote the sub-C\*-algebra generated by $F$. Consider the family
$$
\mathfrak{B} = \{ C^\*(F) \colon F \text{ finite subset of } A\}.
$$
It is easy to check that $\mathfrak{B}$ is an exhausting, upward directed family of complete subalgebras of $A$.
However, since $A$ itself is not finitely generated, the subalgebra $A$ does not belong to the family $\mathfrak{B}$.


Now, consider the identity map $A\to A$, which is clearly a c.p.c. order zero map. When the source is equipped with the family $\mathfrak{A}$ and the target is equipped with the family $\mathfrak{B}$, then the OP's question has a negative answer.


**Explanation**


A local C\*-algebra is only assumed to have an exhausting, directed family of complete subalgebras. But this family is not part of the structure of the local C\*-algebra. Indeed, there may be different (inequivalent) such families for the same local C\*-algebra, as the above example shows.


A different (and maybe better) way to think of local C\*-algebras is as follows. Given a pre-C\*-algebra $A$, consider the embedding into its completion $\overline{A}$. Then $A$ is a local C\*-algebra if and only if for each finite subset $F$ of $A$, the C\*-algebra $C^\*(F)$ (generated in $\overline{A}$) is contained in $A$. This shows that every local C\*-algebra has a canonical exhausting, directed family of complete subalgebras, namely the collection of all finitely generated sub-C\*-algebras.


It seems plausible that the structure theory for c.p.c. order zero maps between C\*-algebra (as developed by Winter and Zacharias), can be extended to the context of local C\*-algebras to show the following:


Let $A$ and $B$ be local C\*-algebras, and let $\varphi\colon A\to B$ be a c.p.c. order zero map. Let $F$ be a finite subset of $A$. Then there exists a finite subset $G$ of $B$ (presumably of cardinality at most one more than that of $F$) such that the image of the C\*-algebra $C^\*(F)$ under $\varphi$ is contained in the C\*-algebra $C^\*(G)$.


