Definition of transpose map in Murphy's C* algebras

Question

Given a continuous map $\theta : \Omega \rightarrow \Omega'$, the author defines its transpose $\theta^t : C(\Omega) \rightarrow C(\Omega')$ by $f \mapsto f\circ \theta$. Then he states that it is a unital *-homomorphsim. Why is it unital though? The transpose map does not map the identity to the identity in $C(\Omega)$ (unless $\theta$ is the identity map)

Answer 1

The unit in $C(\Omega)$ is not the identity function, but rather the constant function $f=1$, as multiplication is pointwise and not composition (composition makes no sense if you think about it). So, if $g=1$, then $g=g\circ\theta$ which is to say that the $*$-homomorphism satisfies $\tilde\theta(1)=1$.