 # Let A , B be commutative unital Banach algebras and let &#x03C6;<!-- φ --> : A &#x2 anginih86 2022-06-19 Answered
Let $A,B$ be commutative unital Banach algebras and let $\phi :A\to B$ be a continuous unital map such that
$\overline{\phi \left(A\right)}=B$
Let
${\phi }^{\ast }:\text{Max}\left(B\right)\to \text{Max}\left(A\right)$
${\phi }^{\ast }\left(m\right)=m\left(\phi \right)$
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 $T:X\to \text{im}\left(T\right)$ 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 $\left({n}_{i}\right)$ in $\text{Max}\left(B\right)$ implies the convergence of the net ${m}_{i}\circ \phi \right)$)
(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?
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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 $\chi \left(a\right)-\lambda$. The counter-reciprocal gives that when $\lambda$ is in the image of $a$ under $\chi$ then $a-\lambda 1$ is not invertible and so
$|\chi \left(a\right)|\le sup\left\{|\lambda |:\lambda \in \mathrm{s}\mathrm{p}\left(a\right)\right\}\le ‖a‖$
Using the fact that $\phi \left[A\right]$ is dense in $B$ you have that if two continuous functional ${\phi }_{1}$ and ${\phi }_{2}$ agree on $\phi \left[A\right]$, then they are equal. This gives the injectivity.
For topological injectivity you need to see that if ${\phi }^{\ast }\left({\chi }_{n}\right)={\chi }_{n}\circ \phi \in \mathrm{i}\mathrm{m}\left({\phi }^{\ast }\right)\subset \mathrm{M}\mathrm{a}\mathrm{x}\left(A\right)$ converge to $\chi \circ \phi$ then ${\chi }_{n}\to \chi$ in $\mathrm{M}\mathrm{a}\mathrm{x}\left(B\right)$. Let $b\in B$, we can find ${a}_{ϵ}$ with $‖\phi \left({a}_{ϵ}\right)-b‖<ϵ$ and
$\begin{array}{rcl}|{\chi }_{n}\left(b\right)-\chi \left(b\right)|& \le & ‖{\chi }_{n}\left(b\right)-{\chi }_{n}\left(\phi \left({a}_{ϵ}\right)\right)‖+|{\chi }_{n}\left(\phi \left({a}_{ϵ}\right)\right)-\chi \left(\phi \left({a}_{ϵ}\right)\right)|+|\chi \left(\phi \left({a}_{ϵ}\right)\right)-\chi \left(b\right)|\\ & \le & ϵ+|{\chi }_{n}\left(\phi \left({a}_{ϵ}\right)\right)-\chi \left(\phi \left({a}_{ϵ}\right)\right)|+ϵ.\end{array}$
But that implies that the limit of $|{\chi }_{n}\left(b\right)-\chi \left(b\right)|$ is smaller or equal than $ϵ$ for every $ϵ$ and therefore $0$.