# Compact Hausdorff space and its double dual: Gelfand Naimark.

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 $$f\in C(X)$$.

• 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. – Cy L Shih Apr 6 at 20:50