Let $A, B$ be commutative unital Banach algebras and let $\varphi: A \rightarrow B$ be a continuous unital map such that $$\overline{\varphi(A)} = B$$
Let $$\varphi^{*}: \text{Max}(B) \rightarrow \text{Max}(A)$$ $$\varphi^{*}(m) = m(\varphi)$$ be the map from the space of maximal ideals of $B$ to the space of maximal ideals of $A$ induced by $\varphi$.
How to prove that $\varphi^{*}$ is a topologically injective map?
(Recall that an operator $T: X \rightarrow Y$ is called topologically injective if $T: X \rightarrow \text{im}(T)$ is a homeomorphism)
My progress on the problem is the following: first of all, the space of maximal ideals of a commutative Banach algebra $A$ can be identified with the space of continuous functionals of the form $m: A \rightarrow \mathbb{C}$. Clearly the map above is continuous, since the pointwise convergence of a net $(n_{i})$ in $\text{Max}(B)$ implies the convergence of the net $ m_{i} \circ \varphi) $
(the space of continuous linear functional is endowed with the weak* topology)
If we assume for the moment that the map is bijective, then the fact that a continuous bijective map between compact Hausdorff spaces is a homeomorphism, yields the result.
(here the space of maximal ideals is compact in weak* topology since the algebra is unital)
The suggested proposition looks like a relaxation of the aformentioned reasoning above though i cannot figure out an easy way to modify it to make it work. Are there any hints?
Use that all characters are of norm one.
Indeed, they are bounded since their kernel is a maximal ideal and therefore closed. The fact that they are contractive follows from spectral theory. If $a - \lambda 1$ is invertible so is $\chi(a) - \lambda$. The counter-reciprocal gives that when $\lambda$ is in the image of $a$ under $\chi$ then $a - \lambda 1$ is not invertible and so $$ |\chi(a)| \leq \sup \{ |\lambda| : \lambda \in \mathrm{sp}(a) \} \leq \| a \| $$
Using the fact that $\varphi[A]$ is dense in $B$ you have that if two continuous functional $\varphi_1$ and $\varphi_2$ agree on $\varphi[A]$, then they are equal. This gives the injectivity.
For topological injectivity you need to see that if $\varphi^\ast(\chi_n) = \chi_n \circ \varphi \in \mathrm{im}(\varphi^*)\subset \mathrm{Max}(A)$ converge to $\chi \circ \varphi$ then $\chi_n \to \chi$ in $\mathrm{Max}(B)$. Let $b \in B$, we can find $a_\epsilon$ with $\|\varphi(a_\epsilon) - b\| < \epsilon$ and \begin{eqnarray*} |\chi_n(b) - \chi(b) | & \leq &\| \chi_n(b) - \chi_n(\varphi(a_\epsilon))\| + |\chi_n(\varphi(a_\epsilon)) - \chi(\varphi(a_\epsilon))| + | \chi(\varphi(a_\epsilon)) - \chi(b)|\\ & \leq & \epsilon + |\chi_n(\varphi(a_\epsilon)) - \chi(\varphi(a_\epsilon))| + \epsilon. \end{eqnarray*} But that implies that the limit of $|\chi_n(b) - \chi(b)|$ is smaller or equal than $\epsilon$ for every $\epsilon$ and therefore $0$.