Suppose $A$ is a unital $C^{*}$-algebra and $a \in A$ is normal, with spectrum $\sigma(a)$. A common analysis in textbooks is the following. Let $B = C^{*}(a)$ the unital $C^{*}$-algebra generated by $1$ and $a$. This is an abelian algebra, so by Gelfand Isomorphism Theorem, we get $B \cong C(\hat{B})$, where $\hat{B}$ is the space of all nonzero $*$-homomorphisms $\chi: B \to \mathbb{C}$. Moreover, the function $\hat{a}: \hat{B} \to \mathbb{C}$ defined by $\hat{a}(\chi) := \chi(a)$ is a homeomorphism between $\hat{B}$ and $\sigma(a)$.
My question is: if $A$ is not only unital but also abelian, does the same analysis hold for $B$ replaced by $A$? In other words, is $\hat{a}$ a homeomorphism between $\hat{A}$ and $\sigma(a)$. I believe the usual analysis hold in this case too, since: $$\operatorname{Im}(\hat{a}) = \{\chi(a): \chi \in \hat{A}\} = \sigma(a)$$ and this map is a homeomorphism if I am not mistaken. In short, do we still get $\hat{A} \cong C(\sigma(a))$ for $A$ abelian?
No, you don't. When you do the Gelfand transform business with an algebra that is bigger than $C^*(a)$, there is the possibility of having characters that are zero on $a$, so they don't exist as characters in $C^*(a)$. Consider $A=C[0,1]$ and $a(x)=\frac12\,x$. Then $$ \hat A=[0,1],\qquad\qquad\text{ while }\qquad\qquad \sigma(a)=\big[0,\frac12\big] $$ You can make this as extreme as you want, in the sense that it is possible to have a fixed $\sigma(a)$ and $\hat A$ any compact subset of the complex plane that contains $\sigma(a)$. Indeed, you can always form $A=B\oplus C(K)$. Then $\sigma(a,a)=\sigma(a)$, while $\hat A=K$.