K Theory of $C^*$ algebras I, Higson's notes The 2019 Stack Overflow Developer Survey Results Are InK-theory for non-separable C*-algebrasWhy must trivial extension of C*-algebra be split Short Exact Sequence?Exact sequence of tensor productDirect limit of certain $C^*$ algebras is simpleTopological K-Theory-Spectrum for $C^*$-Algebrasinductive limit of nuclear c*algebras is nuclearInterpretation of the $K$-groups of a $C^*$-algebra.Elliott-Natsume-Nest proof of Bott periodicity for $K$-theory of $C^ast$-algebrasShort exact sequence in $K_0$ of non unital rings.$K$-Theory of operators I, Higson notes
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K Theory of $C^*$ algebras I, Higson's notes
The 2019 Stack Overflow Developer Survey Results Are InK-theory for non-separable C*-algebrasWhy must trivial extension of C*-algebra be split Short Exact Sequence?Exact sequence of tensor productDirect limit of certain $C^*$ algebras is simpleTopological K-Theory-Spectrum for $C^*$-Algebrasinductive limit of nuclear c*algebras is nuclearInterpretation of the $K$-groups of a $C^*$-algebra.Elliott-Natsume-Nest proof of Bott periodicity for $K$-theory of $C^ast$-algebrasShort exact sequence in $K_0$ of non unital rings.$K$-Theory of operators I, Higson notes
$begingroup$
Let $B$ be a $C^*$ algebra, and $A(B)$ denote the algebra of bounded continuous functions from $[1,infty)$ to $B$. Let $I(B)$ be the ideal of functions which vanish at infinity. Let $Q(B)$ be the quotient $C^*$ algebra.
We have an exact sequence
$$0 rightarrow I(B) rightarrow A(B) rightarrow Q(B) rightarrow 0$$
Claim:
$$K_0(A(B)) cong K_0(Q(B))$$
I know: from algebraic $K$ theory of rings we have a middle exact sequence, and from homotopy invaraince, that $K_0(I(B)) = 0$.
How does one deduce surjectivity of $K_0(A(B)) rightarrow K_0(Q(B))$?
I suppose this follows from six-term exact sequence of $C^*$ algebras. I wonder if there is an elementary way to see this.
Reference: Page 47 of Higson's notes.
operator-algebras k-theory topological-k-theory algebraic-k-theory
asked Apr 5 at 16:08
CL.CL.
2,3222925
$endgroup$
add a comment |
$begingroup$
Let $B$ be a $C^*$ algebra, and $A(B)$ denote the algebra of bounded continuous functions from $[1,infty)$ to $B$. Let $I(B)$ be the ideal of functions which vanish at infinity. Let $Q(B)$ be the quotient $C^*$ algebra.
We have an exact sequence
$$0 rightarrow I(B) rightarrow A(B) rightarrow Q(B) rightarrow 0$$
Claim:
$$K_0(A(B)) cong K_0(Q(B))$$
I know: from algebraic $K$ theory of rings we have a middle exact sequence, and from homotopy invaraince, that $K_0(I(B)) = 0$.
How does one deduce surjectivity of $K_0(A(B)) rightarrow K_0(Q(B))$?
I suppose this follows from six-term exact sequence of $C^*$ algebras. I wonder if there is an elementary way to see this.
Reference: Page 47 of Higson's notes.
operator-algebras k-theory topological-k-theory algebraic-k-theory
asked Apr 5 at 16:08
CL.CL.
2,3222925
$endgroup$
add a comment |
$begingroup$
Let $B$ be a $C^*$ algebra, and $A(B)$ denote the algebra of bounded continuous functions from $[1,infty)$ to $B$. Let $I(B)$ be the ideal of functions which vanish at infinity. Let $Q(B)$ be the quotient $C^*$ algebra.
We have an exact sequence
$$0 rightarrow I(B) rightarrow A(B) rightarrow Q(B) rightarrow 0$$
Claim:
$$K_0(A(B)) cong K_0(Q(B))$$
I know: from algebraic $K$ theory of rings we have a middle exact sequence, and from homotopy invaraince, that $K_0(I(B)) = 0$.
How does one deduce surjectivity of $K_0(A(B)) rightarrow K_0(Q(B))$?
I suppose this follows from six-term exact sequence of $C^*$ algebras. I wonder if there is an elementary way to see this.
Reference: Page 47 of Higson's notes.
operator-algebras k-theory topological-k-theory algebraic-k-theory
asked Apr 5 at 16:08
CL.CL.
2,3222925
$endgroup$
Let $B$ be a $C^*$ algebra, and $A(B)$ denote the algebra of bounded continuous functions from $[1,infty)$ to $B$. Let $I(B)$ be the ideal of functions which vanish at infinity. Let $Q(B)$ be the quotient $C^*$ algebra.
We have an exact sequence
$$0 rightarrow I(B) rightarrow A(B) rightarrow Q(B) rightarrow 0$$
Claim:
$$K_0(A(B)) cong K_0(Q(B))$$
I know: from algebraic $K$ theory of rings we have a middle exact sequence, and from homotopy invaraince, that $K_0(I(B)) = 0$.
How does one deduce surjectivity of $K_0(A(B)) rightarrow K_0(Q(B))$?
I suppose this follows from six-term exact sequence of $C^*$ algebras. I wonder if there is an elementary way to see this.
Reference: Page 47 of Higson's notes.
operator-algebras k-theory topological-k-theory algebraic-k-theory
operator-algebras k-theory topological-k-theory algebraic-k-theory
asked Apr 5 at 16:08
CL.CL.
2,3222925
asked Apr 5 at 16:08
CL.CL.
2,3222925
edited Apr 6 at 17:37
CL.
asked Apr 5 at 16:08
CL.CL.
2,3222925
asked Apr 5 at 16:08
CL.CL.
2,3222925
asked Apr 5 at 16:08
CL.CL.
2,3222925
2,3222925
add a comment |
add a comment |
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