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Thank you for answering my question.

I have another question about the $K_1$ group. As you may know, some books define the $K_1$ group like below:

enter image description here

Also, it defines the $K_0$ group for an arbitrary C*-algebra like below:

enter image description here

Where $K_{00}(A)$ is the Grothendieck group for the abelian semigroup $V(A)$.

My question is how can I define $K_1(A)$ like the way of $K_0(A)$ more precisely how can I define $K_1(A)$ like below:

enter image description here

Improve this question
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    $\begingroup$ It seems when we define $K_0$ like above, we add some 0s to the matrices and when we calculate the $Ker$, we nicely remove those 0s. $\endgroup$
    – Peg Leg Jonathan
    Jul 11, 2020 at 7:41
  • $\begingroup$ anyone can help? $\endgroup$
    – Peg Leg Jonathan
    Jul 11, 2020 at 10:16
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    $\begingroup$ This is how $K_1$ is already defined. Since $K_1(\mathbb C) = 0$ you have $\ker (K_1(A^+) \to K_1(\mathbb C)) = K_1(A^+) = K_1(A)$. $\endgroup$
    – Jamie Gabe
    Jul 11, 2020 at 10:51

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