Suppose that $A$ is a unital C$^*$-algebra. An element $a \in A$ is said to be normal if $a^*a=a^*a$. Further, let $\sigma(a)$ be the spectrum of $a$. I keep seeing these references to $f(a)$ where $f \in C(\sigma(a))$, the set of continuous functions $f: \sigma(a)\to \mathbb{C}$. But what does it mean? $f$ is a $\mathbb{C}-$function on $\sigma(a) \subset \mathbb{C}$! I think there is some sort of identification going on, but I don't know what.
I guess I should be using the condition that $A$ is unital. In that case, let $C^*(a,1)$ denote the $C^*$-algebra generated by $a$, given by $C^*(a,1)= \overline{\mathrm{span}\{(a^*)^ma^n:m,n\ge 0\}}$, which by the Gelfand-Naimark theorem is $*-$isomorphic to $\Sigma(C^*(a,1))$, the set of characters on $C^*(a,1)$ via the Gelfand transform $\Gamma$.
Question: What does $f(a)$ mean for $f \in C(\sigma(a))$? Why should $f(a)$ belong to $A$? I'll be very grateful if someone could explicity write down the map. Thanks!
As you observed, $C^*(a,1)$ is isomorphic to continous functions on $\mathrm{Spec}(C^*(a,1))$. The thing is this is homeomorphic to $\mathrm{sp}(a)$, we get this by sending a character $\gamma$ to $\gamma(a)$ (check that the codomain is correct and this is infact a homeomorphism). After we gather all the information, we end up with an isomorphism
$$C(\mathrm{sp}(a)) \cong C^*(a,1) $$
This isomorphism sends the function $1\mapsto 1_A$ (i.e function of all 1s to the identity element of $A$) and sends $\mathrm{id}_{\mathrm{sp}(a)} \mapsto a$. The image of a function $f$ on $\mathrm{sp}(a)$ under this isomorphism is called $f(a)$.
In general this is known as the functional calculus or the spectral theorem.
As an example if $a$ is a positive element of $A$, i.e self adjoint and $\mathrm{sp}(a) \subset [0,\infty)$, then $f(x) = \sqrt{x}$ is a continuous function on the spectrum and so you can use this isomorphism to get $\sqrt{a}$.