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I tried to prove that if $A$ is the disk algebra then the Gelfand transform is the identity map. The statement can be found here in Theorem 4.4 but it is given without proof. Please can someone check my proof?

Proof:

$\Omega (A)=$ character space of $A$

$\sigma (a) = $ spectrum of $a$

If $A$ is the disk algebra then it is generated by $1$ and $z$ and $\widehat{z}$ is a homeomorphism $\Omega (A) \to \sigma (z)$. We know that $\sigma (z) =\mathbb D$ hence it follows that $\Omega (A) \cong \mathbb D$, that is, every character corresponds to exactly one $\lambda \in \mathbb D$. We now claim that the Gelfand transform which we will denote by $\Gamma: A \to C_0(\Omega (A))$ is the identity map. First of all note that $\Omega (A) $ is compact so that $C_0 (\Omega (A)) = C(\Omega (A))$. Since $\Omega (A) = \mathbb D$, the Gelfand transform is a map $\Gamma : A \to C(\mathbb D)$. It now only remains to be shown that $\Gamma (1) = 1$ and $\Gamma (z) = z$. To this end, note that characters are unital and hence $\Gamma (1) = 1$ is clear. If $f \in A$ denotes the map $f(z) = z$ then for a character $\chi$ corresponding to $\lambda \in \mathbb D$ it then is clear that $\chi (f) = f(\lambda) = \lambda$, that is, $\Gamma (z) = z$ for all $z \in \mathbb D$. Since $1$ and $z$ generate $A$ it follows that $\Gamma$ is the inclusion map $A \hookrightarrow C(\mathbb D)$.

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First things first: Your proof is correct.

Depending on the target audience it may be necessary to elaborate a few points. What it means that the disk algebra is generated by $1$ and $z$, and how that together with $\Gamma(1) = 1$ and $\Gamma(z) = z$ implies that $\Gamma$ is the inclusion map might not be evident to everybody. Or it may not be evident to an examiner/grader that you know that, if you are tasked with proving it in an exam or homework.

On the other hand, things like the unitality of characters need not be mentioned if it is clear that they are well-known to the audience (and the presenter of the proof).

But if no special circumstances demand extending or shortening it, it's a good proof.

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  • $\begingroup$ Thank you! I am writing some notes just for myself and in these notes, a few pages before this proof, I prove both that the disk algebra is generated by $1$ and $z$ and also why two homomorphisms are equal if they agree on a generating set. Thank you very much for your help and your time! $\endgroup$ – Student Apr 18 '14 at 7:58

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