K theory of the wedge of circles

Question

I am interested in finding the $K_0, K_1$ groups of $C(S^1 \vee S^1)$.

We know that $K_0 (C(S^1)) = K_1(C(S^1)) = \mathbb Z$, but is this directly helpful?

Answer 1

If $X$ is a compact space written as $X=A\cup B$, where $A$ and $B$ are closed subsets, with $I=A\cap B$, then the Mayer-Vietoris sequence is an exact sequence of topological $K$-groups $\require{AMScd}$ \begin{CD} K_0(C(X)) @>({\rho_A}_*,{\rho_B}_*)>> K_0(C(A))\oplus K_0(C(B)) @>{\pi_A}_*-{\pi_B}_*>> K_0(C(I))\\ @A \partial AA @. @VV\text{ind}V\\ K_1(C(I)) @<<{\pi_A}_*-{\pi_B}_*< K_1(C(A))\oplus K_1(C(B)) @<<({\rho_A}_*,{\rho_B}_*)< K_1(C(X)) \end{CD}

where

• $\rho_A:C(X)\to C(A)$,

• $\rho_B:C(X)\to C(B)$,

• $\pi_A:C(A)\to C(I)$,

• $\pi_B:C(B)\to C(I)$,

are the restriction maps. Regarding the case in point, where $A=B=S^1$, and $I=\{*\}$, we have \begin{CD} K_0(C(X))@>({\rho_A}_*,{\rho_B}_*)>> \mathbb Z\oplus \mathbb Z @>{\pi_A}_*-{\pi_B}_*>> \mathbb Z\\ @A \partial AA @. @VV\text{ind}V\\ 0 @<<{\pi_A}_*-{\pi_B}_*< \mathbb Z\oplus \mathbb Z @<<({\rho_A}_*,{\rho_B}_*)< K_1(C(X)) \end{CD} so we must have that $K_0(C(X))=\mathbb Z$, and $K_1(C(X))=\mathbb Z\oplus \mathbb Z$.