Let $ \varphi :C\to D $ be a $*$-homomorphism of $C^*$-algebras and suppose that $g \in\ker(K_0(\varphi))$. Show that there is a natural number $n$ and a projection $ p\in P_n(\tilde C)$ such that $ g= [p]_0−[s(p)]_0 $ and $\tilde\varphi(p) \sim_h s((\tilde\varphi p))$.
Is there any other way for this question? I am not sure about this so.
Since $g\in\ker(K_0(\varphi))$, we know that $[g]_0 = 0$ in $K_0(C)$, where $[g]_0$ is the $K_0 $-class of $g$.
By definition, $[g]_0 = [p]_0 - [s(p)]_0$, where $p$ and $s(p)$ are projections in $\tilde C$, the unitization of $C$.
This implies that $[p]_0 = [s(p)]_0$ in $K_0(C)$.
Since $K_0$ is an abelian group, $[p]_0 - [s(p)]_0 = 0$ implies that $[p]_0 = [s(p)]_0$.
Thus, we have $[g]_0 = [p]_0 - [s(p)]_0 = 0$, which means $[p]_0 = [s(p)]_0$.
Since $[p]_0 = [s(p)]_0$, there exists a unitary $u\in\tilde C $ such that $p = u^*s(p)u$.
Now, consider the projection $q = uu^* ∈ \tilde C$. Since $u$ is unitary, we have $q = uu^* = u^*u = s(p)$.
We can define $n$ to be the rank of $p$, which is equal to the rank of $s(p)$ (since $p$ and $s(p)$ have the same $K_0$ -class).
Therefore, we have found a natural number $n$ and a projection $p \in P_n(\tilde C)$ such that $g = [p]_0 - [s(p)]_0$ and $p = s(p)$.
Finally, let's consider $\tilde\varphi(p)$ and $s(\tilde\varphi(p))$. Since $p = s(p)$, it follows that $\tilde\varphi(p) =\tilde\varphi(s(p))$.
Thus, there exists a natural number $n$ and a projection $p ∈ P_n(\tilde C)$ such that $g = [p]_0 - [s(p)]_0$ and $\tilde\varphi(p) ∼_h s( \tilde\varphi (p))$.