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Let $A$ be a C* algebra, $J$ an ideal, $\pi: A \to A/J$ the quotient map. Recall that the relative K theory group $K_0(A, A/J)$ consists of equivalence classes of triples $(p,q,x)$ where $p$ and $q$ are projections over $A$ and $\pi(x)$ implements a Murray-Von-Neumann equivalence between $\pi(p)$ and $\pi(q)$. One has the standard equivalence relations involving homotopies and direct sums, together with the relation $[p,q,x] = 0$ if $x$ implements a Murray-Von-Neumann equivalence between $p$ and $q$.

One can check that the sequence $K(A, A/J) \to K(A) \to K(A/J)$ is exact in the middle, where the first map is given by $[p,q,x] \mapsto [p] - [q]$. I am trying to understand how the excision theorem for K-theory of C* algebras works, and it would help if I had a direct proof of the following lemma: if the exact sequence $0 \to J \to A \to A/J \to 0$ splits then the map $K(A, A/J) \to K(A)$ is injective. It would even be helpful to understand what is going on in the simple case where $A$ is the unitalization of a nonunital C* algebra $J$ (where the relevant short exact sequence necessarily splits).

Can anyone help?

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If the sequence $0 \rightarrow J \rightarrow A \rightarrow A/J \rightarrow 0$ splits and $A$ is unital, then $K_0(A,A/J) \cong \ker \left(K_0(A) \rightarrow K_0(A/J) \right)$. To show this w/o excision is exercise 4.10.7 on p. 114 here: books.google.com/books?id=r-Icer50QoIC –  Steve Huntsman May 20 '10 at 16:02
Arg... you're right, but unfortunately they don't give any hints. I'll poke through that chapter and see if there are any useful tools, but otherwise I'm just as stuck. –  Paul Siegel May 21 '10 at 19:42

2 Answers 2

For the case where $A = C(X)$ for some compact Hausdorff space and $J$ is an ideal in $A$, the lemma you state is Theorem 2.2.10 in my book "Complex Topological K-Theory." As the title suggests, I only consider K-theory of topological spaces in my book, but it would not be too hard for you to take the material in Section 2.2 and adapt it to the K-theory of $C^*$-algebras.

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The unitalization case $A=J^+$ is treated in Blackadar's $K$-theory for Operator Algebras. In Proposition 5.4.1 he directly showes that $$ K_0(J^+,J)\cong\ker(K_0(J^+)\to K_0({J^+}/J)). $$

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