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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)

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  • $\begingroup$ By local C*-algebra, do you mean a local pre-C*-algebra that is complete, i.e. a directed colimit in the category of C*-algebras and $*$-homomorphisms? Wouldn't any C*-algebra satisfy that (taking $\{A_i\}=\{A\}$)? $\endgroup$ – Chris Heunen Mar 11 '15 at 12:55
  • $\begingroup$ @ChrisHeunen The definition of local C*-algebra I would like to consider for this question is given in the OP. As a special example you can consider inductive limits of C*-algebras (where for simplicity you can take, say, injective connecting maps, although this shouldn't be strictly necessary), but you omit the completion w.r.t to the norm in order to remain with a pre-C*-algebra. In the above example, $M_\infty$ can be constructed this way from the sequence of matrix algebras $M_n(\mathbb C)$ with obvious inclusions, whose C*-limit is the C*-algebra of compact operators. $\endgroup$ – Phoenix87 Mar 11 '15 at 13:14
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    $\begingroup$ The closed subspaces $\phi^{-1}(B_1)\cap A_n$, $\phi^{-1}(B_2)\cap A_n$... would cover $A_n$ so one of them would have to agree with it by Baire's theorem. $\endgroup$ – Leonel Robert Mar 11 '15 at 20:57
  • $\begingroup$ @Phoenix87 What is the acronym c.p.c in the question. $\endgroup$ – Zbigniew Mar 13 '15 at 6:08
  • $\begingroup$ Completely positive contractive $\endgroup$ – Phoenix87 Mar 13 '15 at 9:57
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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 $C^*(G)$.

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  • $\begingroup$ Hi Hannes. Many thanks for your answer. In your example I can see how you can get $A$ from the completion of the union of all the finitely generated C*-subalgebras of $A$, but it seems to me that you are somehow implying that one doesn't need to take the norm-completion, and this is not clear to me at the moment. Perhaps what's misleading me is that I'm thinking of $C(X)$ as the infinite tensor product $C(I)^{\otimes\infty}$ with $I=[0,1]$, seen as an inductive limit with the obvious connecting maps. $\endgroup$ – Phoenix87 May 4 '15 at 16:52
  • $\begingroup$ Yes, this might seem counter-intuitive at first. But every element in $C(X)$ is contained in a finitely generated sub-C*-algebra of $C(X)$, and therefore $C(X)$ is the union (no closure needed) of all its finitely generated sub-C*-algebras. $\endgroup$ – Hannes Thiel May 4 '15 at 20:55
  • $\begingroup$ Thanks for the clarification, I think I can now see why this is the case (just considering real functions on $X$). However your example shows that there might be a suitable choice of exhaustive families for which the property in the OP holds. Of course, for any c.p.c. order zero map $\phi:A\to B$ between C*-algebras one has that $\phi(A)$ is contained in a sub-C*-algebra of $B$ (which is actually the property I'm trying to get to when $A$ and $B$ are local C*-algebras). $\endgroup$ – Phoenix87 May 4 '15 at 21:14
  • $\begingroup$ Also, about the second part of your answer, is there something already known in that direction for c.p.c. order zero maps between C*-algebras? For example, if $a\in A^+$, then $\phi(C^*(a))\subset C^*(h_\phi,\pi_\phi(a))\cap B$, but are there elements $g_1,\ldots, g_n\in B$ such that $\phi(C^*(a))\subset C^*(g_1,\ldots,g_n)$? $\endgroup$ – Phoenix87 May 5 '15 at 15:15
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    $\begingroup$ Yes, a c.p.c. order-zero map $\varphi\colon A\to B$ between two C*-algebras corresponds to a *-homomorphism $\alpha\colon C_0((0,1])\otimes A \to B$ such that $\varphi(a)=\alpha(t\otimes a)$. Assume $B$ is $\sigma$-unital and $b$ is a strictly positive element in $B$. Given a set $F$ in $A$, consider $G=\{\alpha(t\otimes b)\}\cup \varphi(F)$. Then $\varphi(C^*(F))\subset \alpha(C_0((0,1])\otimes C^*(F)) = C^*(G)$. $\endgroup$ – Hannes Thiel May 6 '15 at 14:08

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