Let A , B be commutative unital Banach algebras and let &#x03C6;<!-- φ --> : A &#x2

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?