If $a$ is normal in a $C^*$-algebra $A$, then $\tau(a^*)=\overline{\tau(a)},\forall \tau\in\Omega(A)$? I'm reading Murphy's text. I don't understand that why the underlined map preserves adjoints. ($\Omega(A)$ is the character space of $A$.) Can someone help me? Thanks a lot.

(Is it because $f(a)$ is normal? But why $\tau(a^*)=\overline{\tau(a)} $ for any normal element $a$?)
 A: $\tau$ is a unital algebra homomorphism from $B$ to $\mathbb C$.  For even Banach algebras this implies $\|\tau\|=1$ because $\tau(b)\in\sigma(b)$ and the spectral radius is bounded above by the norm.
Suppose that $u\in B$ is unitary.  Then $\tau(u)$ and $\tau(u^*)$ are reciprocals and each has modulus at most one.   Hence $\tau(u^*)=\overline{\tau(u)}$.  Each $b\in B$ is a linear combination of unitaries in $B$, so $\tau(b^*)=\overline{\tau(b)}$ follows.

That might not be useful.  After writing the above I opened Murphy thinking references there might be more helpful, in part because the logical order of results would make more sense.  (E.g., you would need to know that $\tau\in \Omega(B)$ is $*$-preserving in order for the Gelfand map be to a $*$-hom.)
What you are asking about is covered in Theorem $2.1.9.$
That is proved before showing that $B$ is spanned by its unitary elements (I don't know precisely where that happens, but it is an application of theory developed in that chapter including somewhere even the result in question), and using the latter to prove $2.1.9$ would be circular reasoning. 
