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Consider the short exact sequence

$$0 \longrightarrow C_0(\mathbb{R}^2) \overset{\varphi}\longrightarrow C(\mathbb{D}) \overset{\psi}\longrightarrow C(\mathbb{T}) \longrightarrow 0$$

I need to show that

1) $K_1(C(\mathbb{D}))=0$

2) $K_0(C(\mathbb{D}))\cong \mathbb{Z}$

3) $K_o(\psi)$ is injective

And finally I need to conclude that $\delta_1: K_1(C(\mathbb{T})) \to K_0(C_0(\mathbb{R}^2))$ is an isomorphism and that $K_0(C_0(\mathbb{R}^2))$ is non zero.

Here $\mathbb{D}= \lbrace z \in \mathbb{C}: \vert z \vert \leq 1 \rbrace$ and $\mathbb{T}= \lbrace z \in \mathbb{C}: \vert z \vert = 1 \rbrace$

I have already proved 1) and 2) but I am having some trouble with 3). I am using the book "Introduction to k-theory" by M. Rørdam.

By proposition 9.3.3 (Rørdam) the short exact sequence induce an exact sequence:

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And by exactness we obtain that Im($K_0(\varphi)$)=Ker($K_0(\psi))$ so I want to show that that Im($K_0(\varphi)$)=Ker($K_0(\psi))=\lbrace 0 \rbrace$ as it the injectivity then follows. But I can't seem to prove this, so I was wondering if there is another way.

I can see how 1, 2 and 3 allows me to conclude that $\delta_1: K_1(C(\mathbb{T})) \to K_0(C_0(\mathbb{R}^2))$ is an isomorphism by using exactness, but how do I know that $K_0(C_0(\mathbb{R}^2))$ is non zero?

Miep
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1 Answers1

1

Note that $K_0(C\mathbb(\mathbb D))\cong\mathbb Z$, with generator the identity element $1\in C(\mathbb D)$. Thus to show that $K_0(\psi)$ is injective, it is enough to show that $K_0(\varphi)[1]_0\neq0$. But $\varphi$ takes the identity of $C(\mathbb D)$ to the identity of $C(\mathbb T)$, so $K_0(\varphi)[1]_0=[1]_0\neq 0$.

Upon knowing that $\delta_1$ is an isomorphism, we can conclude that $K_0(C_0(\mathbb R^2))$ is nonzero by showing that $K_1(C(\mathbb T))$ is nonzero, and the computation of $K_1(C(\mathbb T))$ is handled in your previous question

Aweygan
  • 23,232