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 $h_{*} = \alpha$?
The result is true: See this paper https://www.researchgate.net/publication/264425120_The_Category_of_Bratteli_Diagrams by Elliot, Golestani and Amini, theorem 6.1. What you are asking is whether the $K_0$ functor is full when restricted to the subcategory of unital AF algebras and is co-restricted to the category of ordered abelian groups with unital, positive group homomorphisms as morphisms. This is exactly what the above paper shows in theorem 6.1. I think that it can also be proved by more elementary means, but I need to think a bit more about this.