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Let $A$ be a $C^*$-algebra and let $(A_n)$ be an increasing sequence of $C^*$-subalgebras of $A$ whole union is dense in $A$.Let $\phi_n:A_n\to A_{n+1}$ be the inclusion map.Then $A$ is $*$ isomorphic to the direct limit of the direct sequence $(A_n,\phi_n)$.

Suppose $B=lim(A_n,\phi_n)$,how to show that $A$ is $*$ isomorphic to $B$?

math112358
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  • There is a map from $B$ to $A$ using that $B$ is the inductive limit. You can show that this map is a $*$-isomorphism. – Epsilon Dec 10 '19 at 21:40

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Maps out of both $A$ and $B$ are uniquely determined by their restrictions to $\cup_n A_n$, via density. So there are unique maps $A\to B$ and $B\to A$ restricting to the identity on $\cup_n A_n$; the composites give morphisms $A\to A$ and $B\to B$ which fix the dense subspaces $\cup_n A_n$, which are thus the respective identities.

Kevin Carlson
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  • I can prove that there exist a unique map from $A$ to $B$ since $B$ is the inductive limit,but how to show that there also exists a unique map from $B$ to $A$? – math112358 Dec 08 '19 at 18:32
  • @math112358 The existence of these maps follows from basic properties of complete metric spaces. See the answers here: https://math.stackexchange.com/questions/76854/extending-a-function-by-continuity-from-a-dense-subset-of-a-space – Kevin Carlson Dec 08 '19 at 19:22
  • But it does not use the properties of inductive limit. – math112358 Dec 08 '19 at 20:47
  • @math112358 Well, one way of phrasing the general results there about metric spaces is that a complete metric space is the inductive limit (in the category whose morphisms are uniformly continuous) of any sequence of subspaces with dense union. – Kevin Carlson Dec 08 '19 at 20:53