For the past week, I've been trying to solve, as a practice homework exercise, Problem 6 of Chapter 11 in Rudin's Functional Analysis, but have not gotten very far it seems. The problem is as follows:
(a) Suppose $A, B$ are commutative Banach algebras, $B$ is semisimple, $\psi: A \rightarrow B$ is a homomorphism with dense range in $B$, and $\alpha: \Delta_{B} \rightarrow \Delta_{A}$ is defined by $$ (\alpha h)(x)= h(\psi(x))$$ for each $x \in A, h \in \Delta_{B}$. Prove that $\alpha$ is a homeomorphism of $\Delta_{B}$ onto a compact subset of $\Delta_{A}$.
(b) Find an example in which $\psi(A)= B$ but $\alpha(\Delta_{B}) \neq \Delta_{A}$.
Here's my attempt: For (a), because $\psi(A)$ is dense in $B$, $\alpha$ is injective such that the topology of $\Delta_{B}$ is the weak topology induced by the Gelfand transforms of $\psi(x)$ for every $x \in A$. In the text, there is a theorem which shows that the maximal ideal space $\Delta_{A}$ is compact Hausdorff, and so a compact subset, say $E$, of $\Delta_{A}$, will be closed. My intuition tells me that if we can show that $\alpha^{-1}(E)= \Delta_{B}$, then we should be done? But after discussing with my instructor of functional analysis, I think we have to show that $\alpha$ is continuous in the Gelfand topology by some sort of sequential argument so that if $h_n \rightarrow h$, where $\{h_n\}$ is a sequence of complex homomorphisms in $\Delta_{B}$ that converges (weakly or in the Gelfand topology?) to $h$, then $\alpha h_n \rightarrow \alpha h$. I'm not sure if this even makes any sense.
For (b), I think I will have to consider the disk algebra compared to some other algebra. The maximal ideal space of the disk algebra is the set of evaluation homomorphisms.
Do you know of any references that could be helpful? Any suggestions would be highly appreciated.