functional calculus under the $*$ homomorphism











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If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$



My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?






operator-theory operator-algebras c-star-algebras von-neumann-algebras







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  • This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
    – André S.
    18 hours ago












  • I have reedited the question
    – mathrookie
    12 hours ago















up vote
0
down vote

favorite












If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$



My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?






operator-theory operator-algebras c-star-algebras von-neumann-algebras







share|cite|improve this question
























  • This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
    – André S.
    18 hours ago












  • I have reedited the question
    – mathrookie
    12 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$



My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?






operator-theory operator-algebras c-star-algebras von-neumann-algebras







share|cite|improve this question















If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$



My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?






operator-theory operator-algebras c-star-algebras von-neumann-algebras




operator-theory operator-algebras c-star-algebras von-neumann-algebras






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago

























asked 20 hours ago









mathrookie

686512




686512












  • This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
    – André S.
    18 hours ago












  • I have reedited the question
    – mathrookie
    12 hours ago


















  • This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
    – André S.
    18 hours ago












  • I have reedited the question
    – mathrookie
    12 hours ago
















This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
18 hours ago






This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
18 hours ago














I have reedited the question
– mathrookie
12 hours ago




I have reedited the question
– mathrookie
12 hours ago










1 Answer
1






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This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
$$
psi(x,y)=(x,0), b=(0,1).
$$
Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.






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    1 Answer
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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

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    up vote
    1
    down vote



    accepted










    This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
    $$
    psi(x,y)=(x,0), b=(0,1).
    $$
    Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
      $$
      psi(x,y)=(x,0), b=(0,1).
      $$
      Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
        $$
        psi(x,y)=(x,0), b=(0,1).
        $$
        Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.






        share|cite|improve this answer












        This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
        $$
        psi(x,y)=(x,0), b=(0,1).
        $$
        Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 12 hours ago









        Martin Argerami

        121k1073172




        121k1073172






























             

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