Consider inductive sequence of $C^\ast$-algebra $A_1 \overset{φ_1}→A_2\overset{φ_2}→···$ inductive limit $(A,(μ_n)_n∈ℕ)$. How is $K_{0}$ continous?

Question

Consider an inductive sequence of $C^\ast$-algebra $A_1 \overset{\varphi_1}\to A_2\overset{\varphi_2}\to···$with inductive limit $(A,(\mu_n)_{n\in \Bbb N})$. Explain how continuity of $K_{0}$ follows from:

$K_{0}(A)= \bigcup^\infty_{n=1} K_0(\mu_n)(K_0(A_n))$

1. For question can I use, $K_0(A)$ is isomorphic to the colimit of the $K_0$ groups of the individual algebras?

I have answered as follows: $$ K_0(A) = \bigcup_{n=1} K_0(\mu_n)(K_0(A_n)). $$ The $K_0$ functor is a contravariant functor from the category of $C^\ast$-algebras to the category of abelian groups. Given an inductive sequence of $^\ast$-algebras and the inductive limit $A$ and the connecting maps $(\mu_n)_{n=1}$ define a new $C^\ast$-algebra, which is the colimit of this sequence. The $K_0$ functor respects colimits, meaning that $K_0(A)$ is isomorphic to the colimit of the $K_0$ groups of the individual algebras, i.e., $K_0(A) \simeq \operatorname{colim}_{n=1} K_0(A_n)$.

The union $\bigcup_{n=1} K_0(\mu_n)(K_0(A_n))$ is a natural way to express this colimit. It is essentially the union of the $K_0$ groups of the algebras $A_n$, where each $K_0(A_n)$ is mapped to $K_0(A)$ through the connecting map $K_0(\mu_n)$.