Consider the short exact sequence $$0\xrightarrow{\ \ \ }C_0(ℝ^2)\xrightarrow{\ \varphi \ }C(ⅅ)\xrightarrow{\ \psi\ }C()\xrightarrow{\ \ \ }0,$$ where $\psi$ is the restriction mapping and $\varphi$ is obtained by identifying $\mathbb D\setminus\mathbb T$ with $ℝ^2$. We make the following identification for the unitisation $C_0(ℝ^2)^\sim$ of $C_0(ℝ^2)$ $$C_0(ℝ^2)^\sim = \{f∈C(ⅅ):\ f|_ \ \text{ is constant}\}.$$
Let $u∈C()$ be the element given by $u(z) =z$. Show that $δ_1([u]_1) = [e]_0−[f]_0$, where $e,f∈M_2(C_0(ℝ^2)^∼)$ are given at $z∈\mathbb D$ by $$e(z) =\begin{bmatrix}\lvert z\rvert^2& z(1-\lvert z\rvert^2)^\frac{1}{2} \\\overline z(1-\lvert z\rvert^2)^\frac{1}{2}& 1-\lvert z\rvert^2\end{bmatrix},\qquad\qquad f(z)=\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}.$$
I am trying to show $δ_1([u]_1) = [e]_0−[f]_0$. Hhow can i do that? Any hint?
This is my try:
Applying the boundary map $δ_1$ to the class of $[]_1$. The boundary map $δ_1: K_1(C_0(ℝ^2)^∼)$$→ K_0(C()) $is defined as follows: $δ_1([]_1)$ $= [ ∘ ]_0 - []_0$
Now, let's compute $ ∘ $ and substitute it into the formula.
$ ∘ $$: → C()$ $( ∘ )()$$ =\psi(()) = () = $
Therefore, we have $( ∘ )()$$ = $ for all $ ∈ $.
Substituting the results back into the formula: $δ_1([]_1)= []_0 - []_0$ To compute $[]_0$ and $[]_0$, we need to express them as matrices in $_2(C_0(ℝ^2)^∼)$. $$[]_0 =\begin{bmatrix} & 0\\0 & 0 \end{bmatrix}$$ $$[]_0 =\begin{bmatrix}0 & 0\\0 & 1 \end{bmatrix}$$ Finally, we can compute $δ_1([]_1)$: $δ_1([]_1)= []_0 - []_0$ =$\begin{bmatrix}& 0\\0 & 0\end{bmatrix}$ - $\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}$ =$\begin{bmatrix}& 0\\0& -1\end{bmatrix}$ Therefore, $δ_1([]_1) = \begin{bmatrix}& 0\\0 & -1\end{bmatrix}$ In conclusion, we have shown that $δ_1([]_1) =\begin{bmatrix}& 0\\0 & -1\end{bmatrix}$, which matches the expression for $[]_0 - []_0$ given in the problem statement.