## A compact Hausdorff space $X$ is homeomorphic to the maximal ideal space of $C(X)$

$\newcommand{\M}{\operatorname{Max}}\newcommand{\C}{\mathbb{C}}$**Theorem**. If $X$ is a compact Hausdorff space, and $\M=\{\text{non-zero algebra homomorphism of }C(X) \text{to } \C\}$ is the maximal ideal space of $C(X)$, then there is a homeomorphism of $X$ onto $\M$.

*Proof*.
For each $x\in X$, define \begin{equation*}
ev_x(f)=f(x), f\in C(X),
\end{equation*}
we have

Claim. \begin{equation*}
H:X\to \M, x\to ev_x
\end{equation*}
is a homeomorphism.

(a)Obviously, $H$ is well-defined.

(b)By Urysohn’s lemma, $H$ is an injection.

(c)If $\{x_\lambda\}$ is a net in $X$ converges to $x\in X$, then $f(x_\lambda)\to f(x)$ for all $f\in C(X)$, i.e., $H(x_\lambda)\xrightarrow{weak*}H(x)$. Hence $H$ is continuous.

(d)Let $I$ be a none-trivial ideal of $C(X)$ and denote all the zero points of $f\in C(X)$ by $Z(f)$. For any finite continuous functions $f_i\in I, i=1,2,\cdots, n$, $$f:=\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n\in I,$$ therefore $f$ is not invertible, and thus $Z(f)\neq\emptyset$.
Since $$Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\supset Z(\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n),$$
$Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\neq\emptyset.$
Hence we have $$\bigcap_{f\in I}Z(f)\neq\emptyset.$$ Note that $Z(f)$ is closed for every continuous function and that $X$ is compact.
In particular, take $I=\ker\tau, \tau\in \M$. Suppose $x\in \bigcap_{f\in I}Z(f)$, then $\ker\tau=I\subset \{f\in C(X)|f(x)=0\}=\ker ev_x$, so $\tau=ev_x$.
Therefore, $H$ is a surjection.

(e) Since $X$ is compact and $\M$ is Hausdorff, $H$ is a homeomorphism.