Let $A$ be an unital $C^\ast$- algebra and let $a \in A$ be normal. If $f \in C(\sigma(a))$ and $g \in C(\sigma(f(a)))$, where $\sigma(a)$ is the spectrum of $a$ in $A$ and $\sigma(f(a))$ is the spectrum of $f(a)$.
How can one proof that $(g \circ f)(a) = g(f(a))$?
My attempt:
I'm using the following facts:
If $X$ is Hausdorff compact and $f \in C(X)$, we have that $\sigma(f) = f(X)$.
Using the Continuous functional calculus, we have an $\ast$-isomorphism $$ f \in C(\sigma (a)) \mapsto f(a) \in C^\ast(a,1) $$ Then, $\sigma(f) = \sigma(f(a))$ and, by 1, $\sigma(f(a)) = f(\sigma(a))$.
By 2, we have that $g \circ f: \sigma(a) \to f(\sigma(a)) = \sigma(f(a)) \to \mathbb C$ is well-defined and is continuous in $\sigma(a)$.
Let's name the continuous functional calculus maps for $a$ and $f(a)$ as $\Phi_1: C(\sigma(a)) \to C^\ast(a,1)$ and $\Phi_2: C(\sigma(f(a))) \to C^\ast(f(a),1)$. We also have:
i. Since $f(a) \in C^\ast(a,1)$, $C^\ast(f(a),1) \subset C^\ast(a,1)$.
ii. $g \circ f \in C(\sigma(a))$ and $\Phi_1 (g\circ f) = (g \circ f) (a)$.
However, I couldn't conclude that $(g \circ f)(a) = g(f(a))$.
Help?
One way is the following: since functional calculus is a $*$-homomorphism, you have $$f(a)^2=f^2(a).$$ So, for $g(t)=t^2$, $g(f(a))=(g\circ f)(a)$. The same reasoning applies to $g(t)=t^n$, and also to polynomials: if $g(f(a))=(g\circ f)(a)$ and $h(f(a))=(h\circ f)(a)$, then $$ (g+h)(f(a))=g(f(a))+h(f(a))=(g\circ f)(a)+(h\circ f)(a)=(g\circ f+h\circ f)(a)=((g+h)\circ f)(a). $$ So $$ p(f(a))=(p\circ f)(a) $$ for all polynomials $p$. Now, given a continuous function $g$ on a compact set that contains $\sigma(a)$ and $\sigma(f(a))$, we can write it as a uniform limit of polynomials $\{p_n\}$. Then $$ g(f(a))=\lim_np_n(f(a))=\lim_n (p_n\circ f)(a)=(g\circ f)(a), $$ since $p_n\circ f\to g\circ f$ uniformly.