Denote by $S^2$ the sphere. By identifying $C(S^2)$ with functions on $D^2=\{z\in\mathbb C:|z|\leq 1\} $ which are constant on the boundary, one can define a projection in $M_2(C(S^2))$:
$$f(z)=\left(\begin{array}{}1-zz^*&-z(1-z^*z)^{1/2}\\-z^*(1-z^*z)^{1/2}&z^*z\end{array}\right)$$
Note that $1_2-f(z)$ is a projection that is Murray-von Neumann equivalent to $1_1$ (in $M_2(C(D^2))$), and the equivalence is given by the partial isometry $$u(z)=\left(\begin{array}{}z&0\\(1-z^*z)^{1/2}&0\end{array}\right)$$ (pay attention $u\notin C(S^2)$). My question is, how do I show $[f]\neq [1]$ in $K_0(C(S^2))$?
$[f]$ is in fact one of the two generators of $K_0(C(S^2))\simeq \mathbb Z^2$, which is obtained by computing the image of the generator $v\in K_1(C(S^1))$ in $K_0(C_0(\mathbb R^2))$ by the index map $\delta_1:K_1(C(S^1))\to K_0(C_0(\mathbb R^2))$ (given by the short exact sequence $C_0(\mathbb R^2)\to C(D^2)\to C(S^1)$).
However, is there a easier and more understandable way to show the proposition?