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Math Help - Show to be injective *-homomorphism

  1. #1
    Member Mauritzvdworm's Avatar
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    Show to be injective *-homomorphism

    \gamma : \left[0, 1\right]\rightarrow \mathbb{T},t\mapsto e^{i2\pi t}
    Let S be the closed ideal consisting of functions \{f\in C_{0}(\left[0,1\right],\mathbb{T}):f(1)=0\}
    Where C_{0}(\left[0,1\right],\mathbb{T}) denotes the continuous functions vanishing at infinity
    from \left[0,1\right] to the unit circle, \mathbb{T}

    Show that the map \gamma_A:A\otimes_{*}S\rightarrow S(A)
    where S(A)=\{f\in C_{0}(\left[0,1\right],\mathbb{T}):f(1)=0=f(0)\}
    is an injective *-homomorphism

    Linearity is easy, so is multiplicativity, but how will I show that the adjoint is preserved and that the map is injective?
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  2. #2
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    Wouldn't you just use the standard thing where you map elements a and b to the same value in the image and then show that a and b must be the same? That is f(a) = f(b) implies a = b.
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  3. #3
    Member Mauritzvdworm's Avatar
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    I see I forgot to put in how \gamma_A is defined

    \gamma_A(a\otimes f)=(f\circ \gamma)a

    so if \gamma_A(a\otimes f)=\gamma_A(b\otimes f) we must have
    (f\circ \gamma)a=(g\circ\gamma)b so
    (f\circ\gamma)a-(g\circ\gamma)b=0 then
    (fa\circ\gamma)-(gb\circ\gamma)=((fa)-(gb))\circ\gamma=0
    but \gamma\neq 0 for any t\in\left[0,1\right]

    Now I can say fa=gb, but what I would like to show is a\otimes f=b\otimes g

    Im I doing something wrong here?

    For the adjoint we have
    \gamma_A(a^*\otimes f^*)=(f^*\circ\gamma)a^*=\left(a(\gamma^*\circ f)\right)^*
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  4. #4
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    Where did g come from?
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  5. #5
    Member Mauritzvdworm's Avatar
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