If $A$ is a $C^\ast$-algebra and $u\in A$ is unitary and $\tau_u : A \to \mathbb C$ is the evaluation map why can $\tau_u$ no map to any number in $(-\infty, 0]$?
This is used in the proof here:
Claim: Let $A$ be a $C^\ast$-algebra and $u$ be unitary. If $\sigma (u) \neq \mathbf T$ then there exists $a \in A$ self-adjoint s.t. $u=e^{ia}$.
Proof: We may assume that $-1 \notin \sigma (u)$ and that $A$ is commutative. Let $\varphi : A \to C_0 (\Omega (A))$ be the Gelfand representation and let $f = \varphi (u)$ and $\ln : \mathbb C \setminus (-\infty , 0]\to \mathbb C$ the principal branch of the logarithm. Then $g = \ln \circ f$ is a well-defined element of $C_0 (\Omega)$....
Why is $g$ well-defined?