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This question is similar to my question here, but not the same question.

Let $(A_{i},\alpha_{i})$ be directed system of C* algebras and *-homomorphisms. Let the $\beta_{i}:A_{i}\rightarrow A$ are the canonical *-homomorphisms. Consider some subalgebra $B$ belongs to direct limit $A$. Does it always exists a $j\in I$ such that $B\subseteq \beta_{j}(A_{j})$?

My guess is that is true. Since $A$ is quotient algebra of disjoint union of algebras, we can write every elements $x\in A$ in the form $x=(x_{1},x_{2},...)+N$, with $x_{i}\in A_{i}$, $N$ is the ideal form by equivalence relation. If $B$ is a sub-algebra, then every component of $B$ form a sub-algebra. Is my thought correct?

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  • $\begingroup$ The inverse image of a subalgebra under a homomorphism is always a subalgebra. $\endgroup$ – Ruy Feb 22 at 20:47
  • $\begingroup$ Yes, I am asking if we do not assume $\beta_{i}$ are surjective, does all subalgebra in $A$ lies in some image of $\beta_{i}$ $\endgroup$ – Ken.Wong Feb 22 at 22:12
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    $\begingroup$ Well, this is quite different from asking if $\beta^{-1}_j(B)$ is a subalgebra. So I guess you should have asked whether $B\subseteq \beta_j(A_j)$, for some $j$. If this is what you want to know then the answer is no: just take $B$ to be the algebra generated by some element not in the union of the ranges of the $\beta_j$. $\endgroup$ – Ruy Feb 23 at 0:14
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    $\begingroup$ The question you have written is actually weaker than your preveious question. Of course there is a $j$ such that the preimage of $B$ under $\beta_j$ is a subalgebra - this is true for every $j$! If you meant to ask a different question, you should edit to clarify. $\endgroup$ – Alex Kruckman Feb 23 at 2:53
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    $\begingroup$ Maybe the confusion is this: the notation $f^{-1}(X)$ makes sense even when $X$ is not contained in the image of $f$. $\endgroup$ – Alex Kruckman Feb 23 at 2:55
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The inductive limit of C$^*$-algebras is often not the union of the ranges of the $\beta _j$.

For instance, consider the C$^*$-algebra $K$, formed by all compact operators on $\ell ^2$. Also, for each $n$, consider the subset $K_n\subseteq K$, formed by all operators whose matrix $(a_{i, j})_{i, j}$, relative to the canonical basis of $\ell ^2$, have nonzero entries only in the top left $n\times n$ block.

Then

  • each $K_n$ is a closed $^*$-subalgebra of $K$ (isomorphic to $M_n(\mathbb C)$),

  • $K_n\subseteq K_{n+1}$, and

  • the union $\bigcup_nK_n$ is dense in $K$.

With this much information you are able to deduce that $K$ is the inductive limit of the $K_n$, with the connecting maps being the inclusions $K_n\hookrightarrow K_{n+1}$.

Observing that every operator in $\bigcup_nK_n$ has finite rank, we see that $\bigcup_nK_n$ is not equal to $K$. It is only a dense $^*$-subalgebra (and hence can't be the inductive limit in the category of C$^*$-algebras since we want the inductive limit to be a C$^*$-algebra).

Any infinite rank compact operator will therefore generate a subalgebra that is not contained in the union of the ranges of the $\beta _j$.

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  • $\begingroup$ Thank you for the great answer. I would like to ask if we require $B$ to be dense also, will the answer changes? $\endgroup$ – Ken.Wong Feb 23 at 13:04
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    $\begingroup$ No, the answer will still be negative. In fact the worst case is when $B$ is the whole inductive limit! $\endgroup$ – Ruy Feb 23 at 13:06

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