# If Banach Algebras are isomorphic, then their corresponding maximal ideal spaces are homeomorphic

Let $A$ ans $B$ be two unital commutative Banach algebras which are isomorphic to each other. Then I need to prove that their maximal ideal spaces are homeomorphic, where by maximal ideal space of $A$, I mean the set of multiplicative linear functionals on the Banach algebra $A$.

I can see this as an immediate consequence of Banach Stone Theorem when A and B are $C^{*}$ algebras by using the fact that Gelfand map becomes a isometric isomorphism when the underlying Banach algebra is a $C^{*}$ algebra.

But how to prove this is general?

• To be clear, your question is about what happens if $A$ and $B$ are isomorphic but not isometrically isomorphic? – Eric Wofsey Apr 18 '18 at 16:36
• Yes and also $A$ and $B$ need not be $C^{*}$ algebras – sachin garg Apr 18 '18 at 16:53

Let $\phi:A\to B$ an isomorphism. Write $c(A), c(B)$ for the spaces of characters, and let $\tilde\phi:c(A)\to c(B)$ be given by $\tilde\phi(f)=f\circ\phi^{-1}$. It is easy to check that $\tilde\phi$ is a bijection. And it is continuous: the topology on the characters is the weak$^*$-topology, that is pointwise convergence. If $f_n\to f$ in $c(A)$, then for all $b\in B$ we have $$\tilde\phi(f_n)=f_n(\phi^{-1}(b))\xrightarrow[n\to\infty]{} f(\phi^{-1}(b))=\tilde\phi(f).$$ So $\tilde\phi$ is continuous. An entirely similar computation shows that $\tilde\phi^{-1}$ is also continuous, so $\tilde\phi$ is a homeomorphism.