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Let $\mathcal{A}$ be a commutative unital C*-algebra and $a\in\mathcal{A}$ an element such that the polynomials in $a$, $a^*$ and $1$ are dense in $\mathcal{A}$. There is an isometric *-isomorphism $\Psi:C(\sigma(a)\rightarrow\mathcal{A}$ such that \begin{align} \Psi(id_{\sigma(a)})=a,\quad\Psi(1_{\sigma(a)})=1. \end{align} It is the unique continuous *-homomorphism from $C(\sigma(a))$ to $\mathcal{A}$.

So, I am having trouble proving both the existence part and the uniqueness part. For the existence part I used the theorem:

Let $\mathcal{A}$ be a commutative unital C*-algebra and $a\in\mathcal{A}$ an element such that the polynomials in $a$, $a^*$ and $1$ are dense in $\mathcal{A}$. Let $\Delta$ denote the character space of $\mathcal{A}$. Then the map $\Phi:C(\sigma(a))\rightarrow C(\Delta)$ given by $\Phi(f)=f(\phi)$ is an isometric *-isomorphism obeying $\Phi(id_{\sigma(a)})=â$.

â is the function $â:\Delta\rightarrow \mathbb{C}$ with $â(\omega)=\omega(a)$ from the Gelfand transform and $\sigma(a)$ the spectrum of $a$.

I used $\Phi$ from this theorem to obtain $\Psi(id_{\sigma(a)})=a$ by putting $\Psi=\Gamma^{-1}(\Phi)$ and computing $\Gamma^{-1}(\Phi(id_{\sigma(a)}))$, where $\Gamma^{-1}$ is the inverse of the Gelfand transform but I am not sure how to derive the other equation from this construction of $\Psi$.

The uniqueness part I neither am sure how to prove.

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I'll try to use your notation, hopefully I managed to do so. Concerning $\Psi$ being unital, we will need the explicit identification $\mu\colon \sigma(a) \cong \Delta$ given by

$$\mu(\omega) = \omega(a), \quad a\in A, \ \omega \in \Delta.$$

Set $$\pi \colon C(\sigma(a)) \rightarrow C(\Delta); \quad \pi(f)(\omega) = (f \circ \mu)(\omega), \quad \omega \in\Delta.$$

The isomorphism you're looking for, is the composition $$\Phi = \Gamma^{-1} \circ \pi$$

First of all, $\Gamma^{-1}$ is unit-preserving due to inverse of unital $\ast$-homomorphisms being unital again: If $\rho\colon E \rightarrow D$ is a unital $\ast$-isomorphism, then $$\rho^{-1}(1_D) = \rho^{-1}(\rho(1_E)) = 1_E.$$

Now, for every character $\omega$ on $A$, $1_{\sigma(a)}$ maps $\mu(\omega)=\omega(a)$ into the real number $1$ (by definition), so

$$ \Phi(1_{\sigma(a)})(\omega) = \Gamma^{-1}((1_{\sigma(a)}\circ \mu)(\omega))=\Gamma^{-1}(1)=1_A $$

I hope this helps with the second equation.

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  • $\begingroup$ Thank you for your answer! I think I can work with it, but I am no quiet sure why we have to introduce the function $\mu $ ? $\endgroup$ – Smissi Jun 2 '18 at 16:38
  • $\begingroup$ You don't really need it, it's just to be explicit about everything. $\endgroup$ – Munk Jun 23 '18 at 22:08

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