# Generators of $K_0(C(\mathbb{C}P^1))$

$\newcommand{\C}{\mathbb{C}}\newcommand{\R}{\mathbb{R}}$ I know that $K_0(C(\mathbb{C}P^1))\simeq K^0(\mathbb{C}P^1))\simeq K^0(S^2)\simeq \mathbb{Z}^2$ (we're talking about complex K-theory). This means that $K_0(C(\mathbb{C}P^1))$ has two generators, I'd like to know how do they look in $K_0(C(\mathbb{C}P^1))$ i.e. I'd like get two functions $f_1,f_2:\C P^1\rightarrow M_n(\C P^1)$ such that $f_i(x)$ is always a projection and $\{f_i\}$ generates $K_0(C(\mathbb{C}P^1))$.

I know that to calculate $K_0(C(\mathbb{C}P^1))$ one considers the following split-exact sequence: $\require{AMScd}$ \begin{CD} 0 @>>> C_0(\mathbb{C}) @>>> C(\mathbb{C}P^1)@>>> \mathbb{C} @>>> 0 \end{CD} And uses Bott periodicity to show that $K_0(C_0(\mathbb{C}))\simeq \mathbb{Z}$. According to Wegge-Oslen (in his book K-theory and C$^\ast$-Algebras, a friendly approach ) one can use the fact that $S^2\C\simeq C_0(\R^2)\simeq C_0(\C)$ to show this. How does he obtain that? Moreover, is there any way to obtain that $K_0(C_0(\mathbb{C}))\simeq \mathbb{Z}$ without using the Bott map? I'm trying to get the calculations as explicit and simple as possible.

• This has been answered here – Zorngo Jun 9 '18 at 1:52