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Thread: Show to be injective *-homomorphism

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

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

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

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

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

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

    Im I doing something wrong here?

    For the adjoint we have
    $\displaystyle \gamma_A(a^*\otimes f^*)=(f^*\circ\gamma)a^*=\left(a(\gamma^*\circ f)\right)^*$
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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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