imagine $f_1$ is analytic over $S_1$.
$f_2$ is analytic over $S_2$.
$f_3$ is analytic over $S_3$.
if in $S_1 \cap_{}^{} S_2$ , $$f_1=f_2$$ and in if in $S_2 \cap_{}^{} S_3$ , $$f_2=f_3$$
is it true that in if in $S_1 \cap_{}^{} S_3$ , $$f_1=f_3$$
for checking this I thought there could be two states in the first state where $S_1 \cap_{}^{} S_2 \cap_{}^{} S_3 \neq \emptyset$ then we can use the equalities mentioned to conclude that all three functions are equal in $S_1 \cap_{}^{} S_2 \cap_{}^{} S_3$. then with using the theoream "The behavior of a function that is analytic in a region $S ⊂ C$ is completely determined by its behavior in a (small) neighborhood of an arbitrary point in that region" we can understand that $f_1=f_3$ is true in $S1 \cap_{}^{} S_3$ . but I don't know what to do if $S_1 \cap_{}^{} S_2 \cap_{}^{} S_3 = \emptyset$
Let \begin{align*} f_{1}(z)&=\sqrt{r}e^{i\theta/2}\quad r>0,0<\theta<\pi\\ f_{2}(z)&=\sqrt{r}e^{i\theta/2}\quad r<0,\pi/2<\theta<2\pi\\ f_{3}(z)&=\sqrt{r}e^{i\theta/2}\quad r<0,3\pi/2<\theta<5\pi/2\\ \end{align*}
All three functions are analytic in their domains, $f_ {1} = f_ {2}$ at the intersection of their domains $ \frac{\pi}{2} <\theta <\pi $. So $ f_ {2}$ is an analytic extension of $ f_ {1}$.
Similarly, $ f_ {2} = f_ {3} $ in $ \frac{3\pi}{2} <\theta <2 \pi $, so $ f_ {3} $ is an analytic extension of $ f_ {2} $, however $ f_ {3} (z) = -f_ {1} (z) $, for any $ z $ in the first quadrant,
for example:
$z = 1 + i $, $ f_ {3} (1 + i ) = f_ {2} (\sqrt {2} e^{i (\frac{\pi}{4} + 2 \pi}) = - 2 ^ {1/4} e^{i \frac{\pi}{8}} $, but $ f_ {1} (1 + i) = f (\sqrt {2} e^ {i\frac{\pi}{4}}) = 2 ^ {1/4} e ^ {\frac{\pi}{8}} $, therefore $ f_ {3} $ is not an analytic extension of $ f_ {2} $.