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anginih86 2022-06-19 Answered
Let A , B be commutative unital Banach algebras and let φ : A B be a continuous unital map such that
φ ( A ) ¯ = B
Let
φ : Max ( B ) Max ( A )
φ ( m ) = m ( φ )
be the map from the space of maximal ideals of B to the space of maximal ideals of A induced by φ.
How to prove that φ is a topologically injective map?
(Recall that an operator T : X Y is called topologically injective if T : X 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 C. Clearly the map above is continuous, since the pointwise convergence of a net ( n i ) in Max ( B ) implies the convergence of the net m i φ ))
(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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Answers (1)

kpgt1z
Answered 2022-06-20 Author has 23 answers
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 λ 1 is invertible so is χ ( a ) λ. The counter-reciprocal gives that when λ is in the image of a under χ then a λ 1 is not invertible and so
| χ ( a ) | sup { | λ | : λ s p ( a ) } a
Using the fact that φ [ A ] is dense in B you have that if two continuous functional φ 1 and φ 2 agree on φ [ A ], then they are equal. This gives the injectivity.
For topological injectivity you need to see that if φ ( χ n ) = χ n φ i m ( φ ) M a x ( A ) converge to χ φ then χ n χ in M a x ( B ). Let b B, we can find a ϵ with φ ( a ϵ ) b < ϵ and
| χ n ( b ) χ ( b ) | χ n ( b ) χ n ( φ ( a ϵ ) ) + | χ n ( φ ( a ϵ ) ) χ ( φ ( a ϵ ) ) | + | χ ( φ ( a ϵ ) ) χ ( b ) | ϵ + | χ n ( φ ( a ϵ ) ) χ ( φ ( a ϵ ) ) | + ϵ .
But that implies that the limit of | χ n ( b ) χ ( b ) | is smaller or equal than ϵ for every ϵ and therefore 0.
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