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 $A \hookrightarrow C(\mathbb D)$.