AF-algebras and K-theory

Question

Suppose that $A$ and $B$ are AF-algebras and $\varphi, \psi \colon A \to B$ be $*$-homomorphisms with $K_0(\varphi) = K_0(\psi)$. Since $A$ is an AF-algebra, we may write $A = \bigcup_{n \in \mathbb{N}} A_n$ where each $A_n$ is finite dimensional and such that the sequence of $A_n$'s is increasing. Now let for each $n \in \mathbb{N}$ the maps $\varphi_n$ and $\psi_n$ be the restrictions of $\varphi$ and $\psi$ to $A_n$, respectively. Is it now true that $K_0(\varphi_n) = K_0(\psi_n)$ for each $n \in \mathbb{N}$? And if it is the case, why is it true?

Answer 1

Yes this is true. If $A$ is the limit of $(A_n,i_n)$ denote by $i_{n,\infty}$ the resulting maps from $A_n$ into $A$. Now the restrictions are given by $\varphi \circ i_{n,\infty}$ resp. $\psi \circ i_{n,\infty}$. But by functoriality of $K_0$ we get $$ K_0(\varphi\circ i_{n,\infty})= K_0(\varphi) \circ K_0(i_{i,\infty}) = K_0(\psi) \circ K_0(i_{n,\infty}) = K_0(\psi \circ i_{n,\infty}). $$ This proves your statement. Another remark is:

Since $B$ has cancellation and $A_n$ is finite dimensional, the restrictions are even unitarily equivalent.