I am trying to show that for a normal invertible element $b$ of a unital $C^*$-algebra $A$, there exists $f \in C(\text{sp}(b))$ such that $f(b) = b^{-1}$.
My attempt is as follows: Since $0$ is not in the spectrum of an invertible element, we can consider the identity function $\text{id} \in C(\text{sp}(b))$, and note that the function $\text{id}^{-1}$ is well - defined. Then, if $\Phi : C(\text{sp}(b)) \rightarrow C^*(b, 1)$, then $\Phi(\text{id}^{-1}) = b^{-1}$. Thus $\text{id}^{-1}$ is the desired function.
This, to me, seems to be the correct idea. However, it seems that it is not generally the case that $\text{id}^{-1}$ is an element of $C(\text{sp}(b))$, as the image of an element of the spectrum of $b$ under this map may not be an element of the spectrum of $b$. Is this the case?
The space $C(\operatorname{sp}(b))$ is the space of all continuous functions $f: \operatorname{sp}(b) \rightarrow \mathbb{C}$. There is no need for $\operatorname{rng}(f)$ to be a subset of $\operatorname{sp}(b)$.
The problem that could occur is that $\operatorname{id}^{-1}$ is not defined at $0$ (and can not be continuously extended to $0$). Hence for $a \in A$ we have $\operatorname{id}^{-1} \in C(\operatorname{sp}(a)) \iff 0 \notin \operatorname{sp}(a)$. This is of course not the case for $b$ as it is invertible and $0 \notin \operatorname{sp}(b)$.