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$K$ Theory of Operators II, Higson's notes



The 2019 Stack Overflow Developer Survey Results Are InSpecial elements in the $C^*$ algebra $A otimes mathcalK$.Understanding the map from $K_0(A)$ to homotopy class of maps,The classifying space of a gauge groupThe inverse limit of C$^*$-algebras and whether it commutes with taking the minimal tensor productIn Kunneth Formula for Cohomology, the finitely generated condition is necessary.In what sense is every element of $H_2(G)$ “represented by a free action on some surface”Evaluating Gibbs state in the second quantized formalismExplaining the group structure in homotopy classes of map $[X,K(H)]$Induced $K$-theory maps between $C^*$ algebras.Understanding the map from $K_0(A)$ to homotopy class of maps,Special elements in the $C^*$ algebra $A otimes mathcalK$.$K$-Theory of operators I, Higson notes










4












$begingroup$


What I want to understand is how the map defined in proof of Proposition 3.17 is an inverse of the defined map.




Proposition 3.17: For any (ungraded) $C^*$-algebra $A$, the map
$$Phi:K_0(A) rightarrow [mathcalS, A otimes mathcalK ]$$
defined above* is an isomoprhism. Where $mathcalS=C_0(Bbb R)$, $mathcalK$ compact operators on the graded hilbert space $H=H_0oplus H_1$.





  1. The map (*) is given as mapping an two idempotent matrix $p,q$ into
    $$
    f mapsto
    beginpmatrix
    p f(0) & 0 \
    0 & q f(0)
    endpmatrix
    $$

    where we regard $A otimes mathcalK cong M_2(A otimes mathcalK)$. More details of this map is here.




  2. The details of inverse map can be found in here.



I could not really chase out how this maps are inverse - in particular, the map in 2. defines an element in $K_0(A otimes mathcalK)$, and here we use Morita equivalence to obtain an element,
$$K_0(A) cong K_0(A otimes mathcalK)$$




I would be grateful if someone can spell out more details. I believe, it would be easier for one to first read the original proof.










share|cite|improve this question











$endgroup$
















    4












    $begingroup$


    What I want to understand is how the map defined in proof of Proposition 3.17 is an inverse of the defined map.




    Proposition 3.17: For any (ungraded) $C^*$-algebra $A$, the map
    $$Phi:K_0(A) rightarrow [mathcalS, A otimes mathcalK ]$$
    defined above* is an isomoprhism. Where $mathcalS=C_0(Bbb R)$, $mathcalK$ compact operators on the graded hilbert space $H=H_0oplus H_1$.





    1. The map (*) is given as mapping an two idempotent matrix $p,q$ into
      $$
      f mapsto
      beginpmatrix
      p f(0) & 0 \
      0 & q f(0)
      endpmatrix
      $$

      where we regard $A otimes mathcalK cong M_2(A otimes mathcalK)$. More details of this map is here.




    2. The details of inverse map can be found in here.



    I could not really chase out how this maps are inverse - in particular, the map in 2. defines an element in $K_0(A otimes mathcalK)$, and here we use Morita equivalence to obtain an element,
    $$K_0(A) cong K_0(A otimes mathcalK)$$




    I would be grateful if someone can spell out more details. I believe, it would be easier for one to first read the original proof.










    share|cite|improve this question











    $endgroup$














      4












      4








      4





      $begingroup$


      What I want to understand is how the map defined in proof of Proposition 3.17 is an inverse of the defined map.




      Proposition 3.17: For any (ungraded) $C^*$-algebra $A$, the map
      $$Phi:K_0(A) rightarrow [mathcalS, A otimes mathcalK ]$$
      defined above* is an isomoprhism. Where $mathcalS=C_0(Bbb R)$, $mathcalK$ compact operators on the graded hilbert space $H=H_0oplus H_1$.





      1. The map (*) is given as mapping an two idempotent matrix $p,q$ into
        $$
        f mapsto
        beginpmatrix
        p f(0) & 0 \
        0 & q f(0)
        endpmatrix
        $$

        where we regard $A otimes mathcalK cong M_2(A otimes mathcalK)$. More details of this map is here.




      2. The details of inverse map can be found in here.



      I could not really chase out how this maps are inverse - in particular, the map in 2. defines an element in $K_0(A otimes mathcalK)$, and here we use Morita equivalence to obtain an element,
      $$K_0(A) cong K_0(A otimes mathcalK)$$




      I would be grateful if someone can spell out more details. I believe, it would be easier for one to first read the original proof.










      share|cite|improve this question











      $endgroup$




      What I want to understand is how the map defined in proof of Proposition 3.17 is an inverse of the defined map.




      Proposition 3.17: For any (ungraded) $C^*$-algebra $A$, the map
      $$Phi:K_0(A) rightarrow [mathcalS, A otimes mathcalK ]$$
      defined above* is an isomoprhism. Where $mathcalS=C_0(Bbb R)$, $mathcalK$ compact operators on the graded hilbert space $H=H_0oplus H_1$.





      1. The map (*) is given as mapping an two idempotent matrix $p,q$ into
        $$
        f mapsto
        beginpmatrix
        p f(0) & 0 \
        0 & q f(0)
        endpmatrix
        $$

        where we regard $A otimes mathcalK cong M_2(A otimes mathcalK)$. More details of this map is here.




      2. The details of inverse map can be found in here.



      I could not really chase out how this maps are inverse - in particular, the map in 2. defines an element in $K_0(A otimes mathcalK)$, and here we use Morita equivalence to obtain an element,
      $$K_0(A) cong K_0(A otimes mathcalK)$$




      I would be grateful if someone can spell out more details. I believe, it would be easier for one to first read the original proof.







      algebraic-topology operator-theory operator-algebras k-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 30 at 19:31







      CL.

















      asked Mar 30 at 17:55









      CL.CL.

      2,3222925




      2,3222925




















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