Half exactness of $K_0$ functor

Theorem. Let $\require{amscd}$ $\begin{CD} 0 @> >> A @>i>> [email protected]>\pi>>[email protected]>>>0 \end{CD}$ be an exact sequence of $\mathbb{C}$-algebras, then $\begin{CD}K_0(A)@>{i_*}>>K_0(B)@>{\pi_*}>>K_0(C)\end{CD}$ is exact in the middle.

Key point. For any invertible element $u$ in $M_m(B)$, $$\begin{pmatrix}u & \\ & -u^{-1}\end{pmatrix}=\begin{pmatrix}& 1 \\ 1& \end{pmatrix}\begin{pmatrix}1 & \\u & 1\end{pmatrix}\begin{pmatrix}1 & -u^{-1}\\ & 1\end{pmatrix}\begin{pmatrix}1 & \\u & 1\end{pmatrix}.$$ Since each factor on the right hand side lifts to a invertible element in $M_{2m}(B^+)$, $\begin{pmatrix}u & \\ & -u^{-1}\end{pmatrix}$ lifts to an invertible element in $M_{2m}(B^+)$.

Proof. Let $[e']-[e'_0]$ be an element in the kernel of $\pi_*$, where $e'\in M_{n'}(B^+)$ and $e'_0\in M_{n'}(\mathbb{C})$ satisfies $e'- e'_0\in M_{n'}(B)$. Then $[\pi_* e']=[\pi_* e'_0]$ and thus there is a unitary $w$ in $M_{n}(C)$ such that $w (\pi_* e) w^{-1}=\pi_* e_0=e_0$, where $n=2n', e= \begin{pmatrix}e & \\ & 0\end{pmatrix}, e_0=\begin{pmatrix}e_0 & \\ & 0\end{pmatrix}.$

Pick a lift $v\in M_{2n}(B^+)$ of $\begin{pmatrix}w& \\ & -w^{-1}\end{pmatrix}$, we have $$\pi_*(v\begin{pmatrix}e & \\ & 0\end{pmatrix}v^{-1}) =\begin{pmatrix}w& \\ & -w^{-1}\end{pmatrix} \begin{pmatrix}\pi_*e& \\ & 0\end{pmatrix} \begin{pmatrix}w^{-1}& \\ & -w\end{pmatrix} =\begin{pmatrix}e_0& \\ & 0\end{pmatrix}.$$ Since $\begin{CD} A @>i>> [email protected]>\pi>>C \end{CD}$ is exact, there is some $d\in M_{2n}(A^+)$ such that $i_* d=v\begin{pmatrix}e & \\ & 0\end{pmatrix}v^{-1}.$ Moreover, $d=d^2$ due to that $i$ is an injection. A direct computation $$i_*([d]-[e_0])=[v\begin{pmatrix}e & \\ & 0\end{pmatrix}v^{-1}]-[e_0]=[e]-[e_0]$$ yields that $\begin{CD}K_0(A)@>{i_*}>>K_0(B)@>{\pi_*}>>K_0(C)\end{CD}$ is exact in the middle.

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