Analytic continuation - How is it that $f_1(z)=f_2(z)$ in some part of the $\Omega_1 \cap \Omega_2$ but not for the rest? I am struggling with Complex Analysis and really need some help to understand my lecture notes:

Definition: Let $f(z)$ be holomorphic in a domain $\Omega$, and let $g(z)$ be defined on a subset S of $\Omega$ that has a point of accumulation in $\Omega$. If $f(z) = g(z)$ for all $z \in S$, we call $f(z)$ the analytic continuation of g(z) to the domain $\Omega$.

*

*If $f_1(z)$ and $f_2(z)$ are both analytic continuations of $g(z)$ to the same domain $\Omega$, then $f_1(z) = f_2(z),\ \forall z \in \Omega$

*If $f_1(z)$ is an analytic continuation of $g(z)$ to the domain $\Omega_1$ and $f_2(z)$ is an analytic continuation of $g(z)$ to a different domain $\Omega_2$ we may have
$\qquad f_1(z) \neq f_2(z)$ at some points in $\Omega_1 \cap \Omega_2$.
We are only certain that $f_1(z) = f_2(z), \forall z \in T$ where $T$ is a domain (we emphasise connectedness of $T$) such that $\qquad S \subset T \subset \Omega_1 \cap \Omega_2$.


I get the first bullet point but not quite for the second. How is it that $f_1(z)=f_2(z)$ in some part of the $\Omega_1 \cap \Omega_2$ but not for the rest? Can someone shed some light on this? Really appreciate the help!
 A: How about $\Omega_1$ being the complex plane with the origin and negative real axis removed, and $\Omega_2$ being the complex plane with the origin and positive real axis removed? Let $f_1$ and $f_2$ be branches of the complex
square root. Precisely, $f_1$ on $\Omega_1$ is defined by
$f_1(re^{it})=\sqrt r e^{it/2}$ for $r>0$ and $-\pi<t<\pi$ and $f_2$ on $\Omega_1$ is defined by
$f_2(re^{it})=\sqrt r e^{it/2}$ for $r>0$ and $0<t<2\pi$. Then $f_1=f_2$
on the upper half-plane, but $f_1=-f_2$ on the lower half-plane.
A: Consider 
$$
\Omega_1=\mathbb C\setminus\{ir: r\ge 0 \}, \qquad
\Omega_2=\mathbb C\setminus\{-ir: r\ge 0 \}.
$$
Let $f_1,f_2$ be the logarithms on $\Omega_1,\Omega$, respectively, so that
$$
f_1(x)=f_2(x)\in\mathbb R, \quad x\in\mathbb R^+.
$$
Then
$$
f_1(z)=f_2(z), \quad\text{if Re$\,z$>0}.
$$
But,
$$
f_1(x)-f_2(x)=-2\pi i, \quad\text{if Re$\,z$<0}.
$$
Note. Two analytic continuations define the same function ONLY in the connected component of the intersection of their domains which intersects the original domain.
