Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?

Question

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.

---

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

Answer 1

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 $C^*(G)$.