# Relative K-theory and split exact sequences of C* algebras

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

## 3 Answers

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

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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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Although this question was posted and answered a very long time ago, I just stumbled upon it and I thought it might be worthwhile if an answer is provided near the post (it actually seems like a full answer is not yet available).

First I will assume $A$ is untital. Second I will use the fact that given $[(p',q',x')]\in K_0(A,A/J)$ I can write it as $[(\ell_k,q,\ell_k)]$ where $\pi(q)=\pi(\ell_k)=\ell_k$ where $\ell_k$ is the standard (bigger than $k\times k$) matrix over the complex numbers with ones on the first $k$ entries of the diagonal and zeros elsewhere. A sketch of the proof is:

-note that $$[(p',q',x')]=[(p'\oplus 1-p',q'\oplus 1-p', x'\oplus 1-p')]$$ -note that $$[(p'\oplus 1-p',q'\oplus 1-p', x'\oplus 1-p')]=[(\ell_k,q'',x'')]$$ - Use the fact that given a partial isometry $y$ we can find a path of unitaries in matrices of quadruple the size from the identity to $V$ such that $V^*yy^*V=y^*y$ and $yy^*Vy^*y=y$ (where we include partial isometries and projections in the upper left hand corner in bigger matrices).

-So let $u_t$ this path of unitaries for the partial isometry $\pi(x)$ and let $v_t$ a lift starting at the identity.

-Note that $$[(\ell_k,q'',x'')]=[(\ell_k, q, x)],$$ where $\ell_k=\pi(\ell_k)=\pi(q)=\pi(x)$ (use $(\ell_k,v_t^*q''v_t,v_t^*x'')$).

-Note that $$[(\ell_k, q, x)]=[(\ell_k,q,\ell_k)]$$ by a linear path $t\ell_k+(1-t)x$ since $\pi(x)=\pi(\ell_k)$.

Denote by $\sigma:A/J\rightarrow A$ the section which exists by assumption. Now suppose $r=[(\ell_k, q, \ell_k)]$ is a general element in $K_0(A,A/J)$ such that $[\ell_k]-[q]=0$ in $K_0(A)$. Then there must exist a path of unitaries $u_t$ starting at the identity such that $u_1^*qu_1=\ell_k$. So we have $r=[(\ell_k,\ell_k, u_1^*\ell_k)]$. Now denote $U:=\sigma(\pi(u_1))$ then $\pi(U)=\pi(u_1)$ and thus by a linear path $r=[(\ell_k,\ell_k, U^*\ell_k)]$. Now note that $U^*\ell_k U=\sigma\pi(u_1^*\ell_ku_1)=\ell_k$ which shows that $(\ell_k,\ell_k,U^*\ell_k)$ is degenerate thus $r=0$. So we find that the kernel of $K_0(A,A/J)$ is trivial. Which was to be shown.

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