Why is the restriction of a character here non zero?

Question

Let $A$ be a unital $C^\ast$-algebra, let $a$ be normal, $B$ the $\ast$-subalgebra generated by $1$ and $a$ and $f\in C(\sigma (a))$. Let $C$ be the $\ast$-algebra generated by $1$ and $f(a)$.

If $\tau \in \Omega (B)$ why is $\tau\mid_C$ non-zero?

Answer 1

Because if $\tau|_C=0$, then $\tau(1)=0$. So for any $x\in A$, $$ \tau(a)=\tau(1a)=\tau(1)\tau(a)=0. $$