Consider unital (complex) abelian Banach algebras $A$ and $B$ with corresponding ideal spaces $\Sigma_A$ and $\Sigma_B$. Suppose $\rho:A\to B$ is a continuous algebra homomorphism that maps $1_A$ to $1_B$ and posseses an adjoint $\rho^*:B^*\to A^*$.
It is easy to see that for $h_B\in\Sigma_B$, $\rho^*(h_B)\in\Sigma_A$.
How would be the cleanest way of arguing that $\rho^*|_{\Sigma_B}:\Sigma_B\to\Sigma_A$ is indeed continuous?
My current thoughts: I need to show that $\Sigma_B\ni h_B\mapsto\rho^*(h_B)\in\Sigma_A$ is continuous, which is by definition (of adjoint equivalent) of $h_B\mapsto h_B\circ\rho$ being continuous. This sort of reminds me of the Gelfand transform. Is this the way to go? Is it as simple as arguing that both $h_B\mapsto h_B(b)$ and $\rho$ are continuous?
A fairly clean way to do this would be an argument involving nets. Let $(h_i)_{i\in I}$ be a net in $\Sigma_B$ that converges to $h$ in the relative weak$^*$-topology. For any $x\in A$ we have
\begin{align*} \rho^*(h)x=h(\rho(x))=\lim_{i\in I}h_i(\rho(x))=\lim_{i\in I}\rho^*(h_i)x. \end{align*}
Thus $(\rho^*(h_i))_{i\in I}$ is weak$^*$-convergent to $\rho^*(h)$, which establishes continuity.