why this sequence is exact?

Question

let $A$ denotes the Toeplitz algebra, $u$ is the unilateral shift, let $\tau$ be the unique *-homomorphism such that $\tau (u)=1$, let $A_{0}$ be the kernel of $\tau$. And $S$ is the closed ideal of $C(T)$ such that $f(1)=0$.then we have exact sequence : $$0 \rightarrow K \rightarrow A_{0} \rightarrow S \rightarrow 0$$ I dont know why this is true, what I know is that we have exact sequence: $$0 \rightarrow K \rightarrow A \rightarrow C(T) \rightarrow 0$$ since the Toeplitz operator $T_{\phi}$ is compact if and only if $\phi =0$. so how can I prove the first sequence is exact? Thanks for any hints.

Answer 1

Since $K$ admits no nonzero $*$-homomorphism, you have that $\tau|_K=0$. Thus $K\subset A_0$. As $A=\{T_f+ k:\ f\in C(\mathbb T),\ k\in K\}$, we have that $\tau(T_f)=f(1)$. Then $\tau|_{A_0}$ is the map $T_f+k\longmapsto f(1)=0$.

So we map $K\to A_0$ with the identity map, and $A_0\to S$ with the map $T_f+k\longmapsto f$.