why for every $ f\in C(\sigma(x))$ we have $ \Phi (f(x))= f(\Phi(x))$?

Question

In a book about $ C^* $-algebra, in the section of continuous functional calculus says that:

Suppose $ x $ is a normal element of $ C^*$-algebra $ A $, then the continuous functional calculus has this property that If $ \Phi: A \to B $ is a $ C^*$-homorphism ($B$ is an arbitrary $C^*$-algebra) then for every $ f\in C(\sigma(x))$ we have $ \Phi (f(x))= f(\Phi(x))$. Please help me about the proof of this statement. thanks

Answer 1

This is clear if $f$ is a polynomial in $t,\overline{t}$. But these polynomials are dense in $C(\sigma(x))$ (and actually this is how one constructs the functional calculus) and both sides of the equation are continuous in $f$. This proves the equation in general.