I am trying to prove the equivalence of categories between compact Hausdorff spaces and unital $C^*$ algebras.

The maps \begin{align*} f: X \mapsto C(X) \\ g: A \mapsto \hat{A} \end{align*} is a contravariant category equivalence between the category of compact Hausdorff spaces and continuous maps and the category of commutative unital $C^*$ algebras and unital $*$-homomoprhisms.

I know one direction $fg \simeq id:A \rightarrow C(\hat{A})$ is Gelfand Naimark.

It then suffices to prove $gf \simeq id$, i.e. $$ X \mapsto \widehat{C(X)}, \quad x \mapsto \phi_x:f \mapsto f(x)$$ is a homeomorphism?


Yes. And the only nontrivial part is that the map $x\longmapsto \phi_x$ is surjective. Below is the argument that I know.

Assume that $\varphi:C(X)\to\mathbb C$ is a character. We prove a few things

  • if $\varphi(f)=0$, then there exists $x_f\in X$ such that $f(x_f)=0$. Indeed, if $f(x)\ne0$ for all $x$, then by compactness there exists $c>0$ with $|f(x)|\geq c$ for all $x$. Then $1/f\in C(X)$ and $1=\varphi(f/f)=\varphi(f)\varphi(1/f)$ and $\varphi(f)\ne0$.

  • for each $f\in C(X)$, there exists $x_f\in X$ with $\varphi(f)=f(t_x)$. Apply the above to the function $f-\varphi(f)$.

  • For each $f\in C(X)$, consider the set $T_f=\{x:\ f(x)=\varphi(f)\}$. By the above, $T_f\ne\varnothing$. As $T_f=f^{-1}(\{\varphi(f)\})$, it is compact.

  • Given $f_1,\ldots,f_n\in C(X)$, we have $\bigcap_{j=1}^n T_{f_j}\ne\varnothing$. Indeed, let $$ g=\sum_{j=1}^n|f_j-\varphi(f_j)|^2. $$ As $\varphi$ is linear and multiplicative, $$ \varphi(g)=\sum_{j=1}^n \varphi[(f_j-\varphi(f_j))\overline{(f_j-\varphi(f_j))}] =\sum_{j=1}^n \varphi[f_j-\varphi(f_j)]\varphi[\overline{f_j-\varphi(f_j)}]=0 $$ As we showed above, there exists $x\in X$ with $g(x)=0$. It follows that $f_j(x)=\varphi(f_j)$ for $j=1,\ldots,n$ so $x\in\bigcap_{j=1}^n T_{f_j}$.

  • The family $\{T_f\}_{f\in C(X)}$ has the finite intersection property. As $X$ is compact, it follows that $\bigcap_{f\in C(X)}T_f\ne\varnothing$. That is, there exists $x\in X$ such that $\varphi(f)=f(x)$ for all $f\in C(X)$.

  • $\begingroup$ Thanks a lot Martin, I will take some time to digest this, mean while, if you have time, I hope you could have a look at the problems I have regarding Higson's works. $\endgroup$ – Cy L Shih Apr 6 at 20:50

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