I'm trying to understand the following statement.
Let $C = C(T)$ the collection of all continuous maps from the unit circle $T \subset \mathbb{C}$ into $\mathbb{C}$, and let $A$ be the algebra of all continuous maps on the closed unit disk $D \subset \mathbb{C}$ into $\mathbb{C}$ which are analytic on $\text{Int} \: D$.
Define the map $\varphi: A \longrightarrow C$ by $\varphi(f) = f|_{T}$. Then $\varphi$ is a unital homomorphism onto the closed subalgebra $B$ of $C$ generated by the set $\{1,z\} \subset C$ for $1$ the unit map and $z$ the inclusion function.
My question, why is $B \neq C$? $B$ is the linear span of $1$ and $z$ right? So it seems to me by the Stone Weirstrass theorem that since $B$ is the closure of the collection of polynomials of the form $\sum_{k=0}^n \alpha_k z^k$ ($n \in \mathbb{N}$) should give you $C=C(T)$.
Is $B \subset C$ properly? If so, why? Thank you.