Proposition 6.2.10 (Spectrum Evaluation Homomorphism).label Using the same notations as Definition 6.2.9. Let $\holo{\cali{N}\parens{\spec{A}{x}}}{A}$ be the algebra of all functions holomorphic in a neighborhood of $\spec{A}{x}$ into $A$ then
\begin{align*}\Phi_{x}:\holoa\to A&&f\mto f(x)\end{align*}
is a homomorphism such that
\begin{align*}\Phi_{x}(1)=1_{A}&&\Phi_{x}(\indi{\com})=x\end{align*}
where $1$ is the constant function and $\indi{\com}:\lam\in\com\mto \lam\in\com$
Proof. Let $f,g$ be functions holomoprhic in neighborhoods $U_{f},U_{g}$ of $\spec{A}{x}$, respectively. Let $U=U_{f}\cap U_{g}$, and $C_{i}$ ($i=1,2$) be smooth simple closed curve in $U$ enclosing $\spec{A}{x}$ such that $C_{2}$ lies inside $C_{1}$ then
\begin{align*}f(x)g(x)&=\parens{\frac{1}{2\pi i}\integral{C_1}{}f(\lam)(\lam-x)\inv d\lam}\parens{\frac{1}{2\pi i}\integral{C_2}{}g(\mu)(\mu-x)\inv d\mu}\\&=-\frac{1}{4\pi^{2}}\iintegral{C_1\ti C_2}{}f(\lam)g(\mu)(\lam-x)\inv(\mu-x)\inv d\lam d\mu\\&=-\frac{1}{4\pi^{2}}\iintegral{C_1\ti C_2}{}\frac{f(\lam) g(\mu)}{\lam-\mu}(\mu-x)\inv d\lam d\mu+\frac{1}{4\pi^{2}}\iintegral{C_1\ti C_2}{}\frac{f(\lam) g(\mu)}{\lam-\mu}(\lam-x)\inv d\lam d\mu\end{align*}
where the last equality is by partial fractions. Since $\frac{g(\mu)}{\lam-\mu}$ is holomorphic inside the curve $C_{2}$ if $\lam$ lies on $C_{1}$
\begin{align*}\frac{1}{4\pi^{2}}\integral{C_1}{}\parens{\integral{C_2}{}\frac{g(\mu)}{\lam-\mu}d\mu}f(\lam)(\lam-x)\inv d\lam=0\end{align*}
Hence we get the map $f\mto f(x)$ is a homomorphism
\begin{align*}f(x)g(x)&=\frac{1}{2\pi i}\parens{\integral{C_2}{}\frac{1}{2\pi i}\integral{C_1}{}\frac{f(\lam)}{\lam-\mu}d\lam}g(\mu)(\mu-x)\inv d\mu\\&=\frac{1}{2\pi i}\integral{C_2}{}f(\mu)g(\mu)(\mu-x)\inv d\mu\\&=(fg)(x)\end{align*}
Suppose now $f(\lam)\equiv 1,\lam\in\com$ then
\begin{align*}f(x)=\frac{1}{2\pi i}\integral{C}{}(\lam-x)\inv d\lam\end{align*}
where we pick $C=\curl{\lam:\abs{\lam}=\norm{x}{}+\e}$ for some $\e>0$. We have then
\begin{align*}(\lam-x)\inv=\sums{\infty}{n=0}\frac{x^{n}}{\lam^{n+1}}\end{align*}
uniformly for every $\lam\in C$. Hence by residue theorem
\begin{align*}f(x)=\frac{1}{2\pi i}\integral{C}{}\sums{\infty}{n=0}\frac{x^{n}}{\lam^{n+1}}d\lam=\sums{\infty}{n=0}\frac{1}{2\pi i}\parens{\integral{C}{}\frac{1}{\lam^{n+1}}d\lam}x^{n}=1\end{align*}
Similarly if $f(\lam)=\lam,\fall \lam\in \com$ then $f(x)=x$.$\square$