For $\phi^*:C(\Delta)\rightarrow C(\mathbb{T})$ defined by $(\phi^* f)(\lambda) = f(\omega_{\lambda})$, where $\omega_{\lambda}((a_n)_n) = \sum_{n\in\mathbb{Z}} a_n\lambda^n$, and $\Gamma$ the Gelfand transform $\Gamma(a) = \hat{a}$. $\mathcal{C}$ is the subset of $C(\mathbb{T})$ with summable Fourier coefficients. $$F:\mathcal{C}\rightarrow \mathcal{A}$$ given by $(F f)_n = \frac{1}{2\pi}\int_{0}^{2\pi} \exp(-i n\theta)f(\exp(i\theta)) d\theta$.
I have shown that $F^{-1}=\phi^*\circ \Gamma$ and I think I showed that $\mathcal{C}$ is a complex unital subalgebra of $C(\mathbb{T})$ (since $\mathcal{A}$ is a commutative, unital Banach algebra).
I also showed that $(\phi^*)^{-1}(f) = \hat{a}$ for some $a\in\mathcal{A}$.
I have a theorem stating that $a$ is invertible if and only if $\omega(a)\neq 0$ for all $\omega\in\Delta$ (the characterspace of $\mathcal{A}$) (I also have other theorems, but I don't know what to use and what to do from here).
How do I show that $a$ is invertible? I looked at what would happen if $0=\omega(a)=((\phi^*)^{-1}(f))(\omega)$ (where we assume $f(z)\neq 0$ for all $z$), but I don't think it helped.
And what does that have to do with invertibility of $f$ in $\mathcal{C}$?