Let $A,B$ be commutative unital Banach algebras and let $\phi :A\to B$ be a continuous unital map such that

$\overline{\phi (A)}=B$

Let

${\phi}^{\ast}:\text{Max}(B)\to \text{Max}(A)$

${\phi}^{\ast}(m)=m(\phi )$

be the map from the space of maximal ideals of $B$ to the space of maximal ideals of $A$ induced by $\phi $.

How to prove that ${\phi}^{\ast}$ is a topologically injective map?

(Recall that an operator $T:X\to Y$ is called topologically injective if $T:X\to \text{im}(T)$ 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 $m:A\to \mathbb{C}$. 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 ${m}_{i}\circ \phi )$)

(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?

$\overline{\phi (A)}=B$

Let

${\phi}^{\ast}:\text{Max}(B)\to \text{Max}(A)$

${\phi}^{\ast}(m)=m(\phi )$

be the map from the space of maximal ideals of $B$ to the space of maximal ideals of $A$ induced by $\phi $.

How to prove that ${\phi}^{\ast}$ is a topologically injective map?

(Recall that an operator $T:X\to Y$ is called topologically injective if $T:X\to \text{im}(T)$ 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 $m:A\to \mathbb{C}$. 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 ${m}_{i}\circ \phi )$)

(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?