Banach-Stone

$T^{-1}$ is linear and isometric, because $\mathrm{\parallel }{T}^{-1}g\mathrm{\parallel }=\mathrm{\parallel }T(T^{-1}g)\mathrm{\parallel }=\mathrm{\parallel }g\mathrm{\parallel }$ for all $g$; hence by [L2] the transpose $T^{\ast }$ is a bounded linear bijection with $(T^{\ast }{)}^{-1}=(T^{-1}{)}^{\ast }$, and $\mathrm{\parallel }{T}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }(T^{\ast }{)}^{-1}\mathrm{\parallel }=1$.

For every $y\in L$ the point evaluation ${\delta }_y$ is an extreme point of $B_{C(L{)}^{\ast }}$ of the form $1\cdot {\delta }_y$, so by [L1] and [L3] its image $T^{\ast }{\delta }_y$ is an extreme point of $B_{C(K{)}^{\ast }}$; by [L1] there are unique $h(y)\in K$ and $u(y)\in \mathbb{K}$ with $\mathrm{\mid }u(y)\mathrm{\mid }=1$ and $T^{\ast }{\delta }_y=u(y){\delta }_{h(y)}$. This defines functions $h:L\to K$ and $u:L\to \mathbb{K}$.

Conversely, let $h:L\to K$ be a homeomorphism and $u:L\to \mathbb{K}$ continuous with $\mathrm{\mid }u\mathrm{\mid }=1$, and set $Tf:=u\cdot (f\circ h)$. Then $T$ is $\mathbb{K}$-linear, $\mathrm{\parallel }Tf{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_y\mathrm{\mid }u(y)f(h(y))\mathrm{\mid }={\sup }_x\mathrm{\mid }f(x)\mathrm{\mid }=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ because $h$ is surjective, and $T$ is surjective with inverse $Sg:=(g\circ h^{-1})\mathrm{/}u(h^{-1})$.

Evaluating $T^{\ast }{\delta }_y=u(y){\delta }_{h(y)}$ at the constant function $\mathbf{1}$ gives $u(y)=(T^{\ast }{\delta }_y)(\mathbf{1})={\delta }_y(T\mathbf{1})=(T\mathbf{1})(y)$, so $u=T\mathbf{1}$ is continuous, and $\mathrm{\mid }u(y)\mathrm{\mid }=1$ for all $y$ by [step 1.2].

For $f\in C(K)$ and $y\in L$: $Tf(y)={\delta }_y(Tf)=(T^{\ast }{\delta }_y)(f)=u(y)f(h(y))$, using the definition of the transpose in [L2] and [step 1.2].

The map $h$ is continuous: for every $f\in C(K)$ the function $y\mapsto f(h(y))=Tf(y)\mathrm{/}u(y)$ is continuous, since $Tf$ is continuous and $u$ is continuous with $\mathrm{\mid }u\mathrm{\mid }=1$ so $1\mathrm{/}u=\bar{u}$ is continuous; the family $\{f\circ h:f\in C(K)\}$ therefore consists of continuous functions and the evaluation embedding $e_K$ of $K$ into the product over $C(K,[0,1])$ has continuous composition $e_K\circ h$, whence $h$ is continuous because $e_K$ is an embedding by [L4].

Applying [step 1.2] and [step 2.2] to the surjective isometry $T^{-1}:C(L)\to C(K)$ (which is a surjective linear isometry by [step 1.1]) produces $h^{\mathrm{\prime }}:K\to L$ continuous and $v:K\to \mathbb{K}$ with $\mathrm{\mid }v\mathrm{\mid }=1$ such that $T^{-1}g(x)=v(x)g(h^{\mathrm{\prime }}(x))$ for all $g,x$.

From $T{T}^{-1}={\mathrm{i}\mathrm{d}}_{C(L)}$: for $g\in C(L)$ and $y\in L$ one computes $g(y)=T(T^{-1}g)(y)=u(y)(T^{-1}g)(h(y))=u(y)v(h(y))g(h^{\mathrm{\prime }}(h(y)))$ by [step 2.2] and [step 3.2], with $\mathrm{\mid }u(y)v(h(y))\mathrm{\mid }=1$; if $h^{\mathrm{\prime }}(h(y))\mathrm{\ne }y$ then [L5] gives $g$ with $g(y)\mathrm{\ne }g(h^{\mathrm{\prime }}(h(y)))$, contradicting the displayed identity; so $h^{\mathrm{\prime }}\circ h={\mathrm{i}\mathrm{d}}_L$, and the same argument with the roles reversed gives $h\circ h^{\mathrm{\prime }}={\mathrm{i}\mathrm{d}}_K$. Hence $h$ is a bijection with continuous inverse $h^{\mathrm{\prime }}=h^{-1}$, that is, a homeomorphism.

By [step 2.2] and [step 4.1] every surjective linear isometry has the asserted form with $h$ a homeomorphism and $\mathrm{\mid }u\mathrm{\mid }=1$; by [step 1.3] every pair $(h,u)$ of that form defines a surjective linear isometry; and the pair is unique since $u=T\mathbf{1}$ by [step 2.1] and then $h$ is recovered from $T$ by the formula.