Theorem: Spaces of Continuous Functions on Homeomorphic Compact Hausdorff Spaces are Isometrically *-Algebra Isomorphic

Let $K, L$ be homeomorphic compact Hausdorff spaces.
Let $\map \CC K$ and $\map \CC L$ be the spaces of continuous functions on $K$ and $L$ respectively.
Then $\map \CC K$ and $\map \CC L$ are isometrically $\ast$-algebra isomorphic.

Proof:
Let $f : K \to L$ be a homeomorphism.
Let $\phi \in \map \CC L$.
From Composite of Continuous Mappings is Continuous, $\phi \circ f \in \map \CC K$.
Hence we can define $T : \map \CC L \to \map \CC K$ by: 
:$T \phi = \phi \circ f$
for each $\phi \in \map \CC L$.
Let $\phi, \psi \in \map \CC L$, $\lambda \in \C$ and $x \in K$. 
Then we have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {T \phi + \lambda T \psi} } x
		| r = \map {\paren {T \phi} } x + \lambda \map {\paren {T \psi} } x
}}
{{eqn	| r = \map \phi {\map f x} + \lambda \map \psi {\map f x} 
}}
{{eqn	| r = \map {\paren {\phi + \lambda \psi} } x
}}
{{eqn	| r = \map {\paren {\map T {\phi + \lambda \psi} } } x
}}
{{end-eqn}}
hence:
:$T \phi + \lambda T \psi = \map T {\phi + \lambda \psi}$
So $T$ is linear. 
We also have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {\paren {T \phi} \paren {T \psi} } } x
		| r = \map {\paren {T \phi} } x \map {\paren {T \psi} } x
}}
{{eqn	| r = \map \phi {\map f x} \map \psi {\map f x}
}}
{{eqn	| r = \map {\phi \psi} {\map f x} 
}}
{{eqn	| r = \map {\paren {\map T {\phi \psi} } } x
}}
{{end-eqn}}
Hence we obtain:
:$\paren {T \phi} \paren {T \psi} = \map T {\phi \psi}$
Hence $T$ is an algebra homomorphism.
Additionally we have:
{{begin-eqn}}
{{eqn	| l = \map {\overline {T \phi} } x
		| r = \overline {\map {\paren {T \phi} } x}
}}
{{eqn	| r = \overline {\map \phi {\map f x} }
}}
{{eqn	| r = \map {\overline \phi} {\map f x}
}}
{{eqn	| r = \map {T \overline \phi} x
}}
{{end-eqn}}
Hence:
:$\overline {T \phi} = T \overline \phi$
So $T$ is a $\ast$-algebra homomorphism.
We now need to establish that $T$ is an algebra isomorphism.
Note that since $f$ is a bijection, if $\phi, \psi \in \map \CC L$ are such that:
:$\phi \circ f = \psi \circ f$
we have that:
:$\phi = \psi$ 
Hence $T$ is an injection.
Let $u \in \map \CC K$.
Further, since $f^{-1}$ is continuous, we obtain that $u \circ f^{-1} \in \map \CC L$ from Composite of Continuous Mappings is Continuous.
Hence we obtain:
:$\map T {u \circ f^{-1} } = \paren {u \circ f^{-1} } \circ f = u$ 
So $T$ is surjective.
So $T$ is indeed an algebra isomorphism.
We finally show that $T$ is an isometry. 
By the definition of the supremum norm, we wish to show that:
:$\ds \sup_{y \mathop \in L} \cmod {\map \phi y} = \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} }$
We note that since $\map f x \in L$ for each $x \in K$:
:$\ds \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} } \le \sup_{y \mathop \in L} \cmod {\map \phi y}$
From the extreme value theorem for normed vector spaces, since $L$ is compact, there exists $y_\ast \in L$ such that:
:$\ds \cmod {\map \phi {y_\ast} } = \sup_{y \mathop \in L} \cmod {\map \phi y}$
Since $f$ is a bijection, there exists $x_\ast \in K$ such that $\map f {x_\ast} = y_\ast$ and:
:$\ds \sup_{y \mathop \in L} \cmod {\map \phi y} = \cmod {\map \phi {\map f {x_\ast} } } \le \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} }$
So we obtain:
:$\ds \sup_{y \mathop \in L} \cmod {\map \phi y} = \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} }$
Hence $T$ is an isometric $\ast$-algebra isomorphism.
{{qed}}
Category:Space of Continuous Functions on Compact Hausdorff Space
Category:*-Algebras
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