Proving $K_1(\mathcal{T})=0$ (is trivial)

Question

Let $\mathcal{T}$ the toeplitz algebra and we define the short exact sequence? where $C(\mathbb{T})=\{z\in \mathbb{C}/ |z|\leq 1\}$: $$ 0 \rightarrow \mathcal{K} \rightarrow \mathcal{T }\rightarrow C(\mathbb{T}) \rightarrow0 $$

and we want to prove that $K_1(\mathcal{T})=0$ (is trivial)

NB: I have proved that $\psi : K_1(\mathcal{T}) \rightarrow K_1(C(\mathbb{T}))$ is a morphism such $Rank(\psi)={0}.$

Answer 1

Since $K_1(\mathcal K)=0$, we know the map $K_1(\mathcal T)\to K_1(C(\mathbb T))$ is injective. But the index map $K_1(C(\mathbb T))\to K_0(\mathcal K)$ is an isomorphism (it takes the class of the identity function on $C(\mathbb T)$ to minus the class of a rank-one projection in $\mathcal K$), so the range of the map $K_1(\mathcal T)\to K_1(C(\mathbb T))$ is trivial, and therefore $K_1(\mathcal T)=0$.