I am trying to prove the following:
Let $B=\{f:\mathbb R \rightarrow \mathbb C | \;\; f\; \mbox{ is continuous and } 2\pi - \mbox{periodic}\}$. Remember that $\widehat B = \{\varphi:B\longrightarrow \mathbb C|\;\; \varphi \;\mbox{ is an homomorphism} \}\setminus \{0\}$. Show that $\Psi: \mathbb S^1 \longrightarrow \widehat B$, given by: $$\Psi (e^{it})(f):= f(t) \qquad \forall f\in B $$ is a homeomorphism.
What I have done so far:
I have already proved that $\Psi$ is injective and continuous using standard arguments involving nets, existence of angle functions defined on connected open sets of $\mathbb S^1$, and the topology of pointwise convergence in $\widehat B$, i.e.: $$ \varphi_\alpha \rightarrow \varphi \iff \forall f\in B:\quad \varphi_\alpha(f) \rightarrow \varphi(f).$$
Since $\mathbb S^1$ is compact and $\widehat B$ is Hausdorff, it follows that $\Psi$ is a homeomorphism onto its image.
My problem:
I wasn't able to prove that $\Psi$ is surjective, since I don't know a natural way to find $z_0=e^{it_0}\in \mathbb S^1$, such that, for a given $\varphi \in \widehat B$: $$\Psi(e^{it_0})(f) = f(t_0) = \varphi(f), \forall f \in B.$$
Assume there is some $\varphi \in \widehat B$ such that $\varphi\neq\Psi(z_0)$ for all $z_0\in S_1$. Then there is a maximal ideal $M$($=\ker\varphi$) in $B$ such that for each $t\in\mathbb R$ (in particular, each $t\in[0,2\pi]$) there is some $f\in M$ such that $f(t)\neq0$. By compactness of $[0,2\pi]$, there exist $f_1,\ldots,f_n\in M$ such that for each $t\in[0,2\pi]$ $f_j(t)\neq0$ for some $j$. But then $f=|f_1|^2+\ldots+|f_n|^2\in M$ and $f(t)\neq0$ for all $t\in[0,2\pi]$, hence all $t\in\mathbb R$. Thus $f$ is invertible, and $M$ cannot contain an invertible element, a contradiction.