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Let $A$ and $B$ be unital $AF$-algebra. By Elliott's theorem we know that if there an order isomorphism $\psi: K_0(A) \rightarrow K_0(B)$ with $\psi([1_{A}]) = [1_{B}]$, then there exists an isomorphism $h : A \rightarrow B$ such that $h_{*} = \psi$.

My question is this is true for homomorphism? In other words, Let $A$ and $B$ be unital $AF$-algebras. Suppose that there is an order homomorphism $\alpha: K_0(A) \rightarrow K_0(B)$ such that $\alpha([1_{A}]) = [1_{B}]$. Is there homomorphism $h: A \rightarrow B$ such that $h_{*} = \alpha$?

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    $\begingroup$ Yes, this is Exercise 7.7 in the book [Rørdam, M.; Larsen, F.; Laustsen, N. An introduction to K-theory for C∗-algebras]. You don't actually need that $B$ is AF, only that $B$ has cancellation of projections (i.e.~if $p,q\in B$ are projections such that $[p]_0=[q]_0\in K_0(B)$, then $p\sim q$). Also, such a homomorphism $h$ is unique up to approximate unitary equivalence. $\endgroup$
    – Jamie Gabe
    Feb 8, 2021 at 15:42
  • $\begingroup$ @JamieGabe Thanks for your help. As it mentioned there, we need a sequence of finite dimensional $C^*$-algebras with unital connecting *-homomorphism. So, we should somehow modify the Elliott's proof? $\endgroup$
    – Peg Leg Jonathan
    Feb 8, 2021 at 15:58
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    $\begingroup$ It's essentially the exact same argument as in Elliott's proof. $\endgroup$
    – Jamie Gabe
    Feb 8, 2021 at 19:21

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