Apply holomorphic functional calculus to the following functions which are defined on a dis connected open set in the plane containing $\Gamma_0 , \Gamma_1$

$f(z)\begin{cases} 1& \text{Around 0}\\0& \text{around 1}\end{cases}$

$g(z)\begin{cases} 0& \text{Around 0}\\z& \text{around 1}\end{cases}$

Then $p_0=f(a), a_1=g(a)$ but $fg=0$.

Apply holomorphic functional calculus to the following functions which are defined on a disconnected open set in the plane containing $\Gamma_0 , \Gamma_1$

$f(z)=\begin{cases} 1& \text{Around 0}\\0& \text{around 1}\end{cases}$

$g(z)=\begin{cases} 0& \text{Around 0}\\z& \text{around 1}\end{cases}$

Then $p_0=f(a), a_1=g(a)$ but $fg=0$.

By the same argument one can shows that $p_0a=a_0$ where

$$a_0 = \frac{1}{2\pi i} \int_{\Gamma_0}\lambda(\lambda -a)^{-1}d\lambda.$$

But the holomorphic functional calculus implies that $p_0 a_1=a$ not zero.

Apply holomorphic functional calculus to the following functions which are defined on a dis connected open set in the plane containing $\Gamma_0 , \Gamma_1$

$f(z)\begin{cases} 1& \text{Around 0}\\0& \text{around 1}\end{cases}$

$g(z)\begin{cases} 0& \text{Around 0}\\z& \text{around 1}\end{cases}$

Then $p_0=f(a), a_1=g(a)$ but $fg=0$.