Confusion on analytic continuation.

The definition of analytic continuation of holomorphic function is stated as follows:

Let $f_{1}$ and $f_{2}$ be two analytic functions on two domains (open and connected) $Omega_{1}$ and $Omega_{2}$ such that $Omega_{1}capOmega_{2}neqvarnothing$. If $f_{1}$ and $f_{2}$ agree on $Omega_{1}cap Omega_{2}$, we say $f_{2}$ is the analytic continuation of $f_{1}$ on $Omega_{2}$, and vice versa.

A smaller version of it is that:

If $f$ is analytic on a domain $Dsubsetmathbb{C}$ and $F$ is analytic on a bigger domain $Esubsetmathbb{C}$ such that $f=F$ on $Dsubset E,$ then $F$ is the analytic continuation of $f$ on $E$.

From what I read, this kind of technique allows us to define a function in a smaller domain and extend it analytically to a larger domain. But I don’t understand why this definition allows us to do so.

What confuses me is that the definition only guarantees $f=F$ on the intersection $Omega_{1}capOmega_{2}$, so perhaps $fneq F$ on $Omega_{2}$, then how do I know $f$ is analytic on $Omega_{2}setminusOmega_{1}$?

I tried to use the identity theorem as follows:

Let $f$ and $g$ be two holomorphic functions on a domain $D$ such that $f=g$ on a subset $Ssubset D$ that contains a limit point, then $f=g$ on the whole $D$.

But this seems backward. By the hypothesis of analytic continuation, we only have $f=g$ on $S$, and $g$ is analytic on $D$, we do not really know if $f$ is analytic on the whole $D$ (this is the purpose of analytic continuation, right? to extend $f$ analytically to the whole $D$.)

Am I overthinking this and confusing myself?? I guess we should have, say $f_{1}=f_{2}$ on the whole $Omega_{1}cupOmega_{2}$, but I don’t know how to prove it.

Edit 1: (Some Clarification, Possible Answer and Reference)

I am sorry if I am asking a confusion (bad) question. My confusion is that, even though the analytic continuation exists, I don’t think that means anything helpful. It only gives us an analytic function $F$ on a bigger domain $Omega_{2}$ such that $F|_{Omega_{1}}=f$ for $Omega_{1}subsetOmega_{2}$. But it does not say anything about $f$, $f$ is still in $Omega_{1}$. So I do not understand why analytic continuation can extend the domain on which $f$ is analytic.

The book "Complex Analysis and Applications" by Hemant Kumar Pathak, has a chapter about analytic continuation.

As Jose suggested, it does not make sense to say $f=F$ on $Omega_{2}$, because $f$ is on $Omega_{1}$.

The book explains that if we have an analytic continuation of $f_{1}$ from $Omega_{1}$ into $Omega_{2}$ via $Omega_{1}capOmega_{2}$, then the aggregate value of $f_{1}$ in $Omega_{1}$ and $f_{2}$ in $Omega_{2}$ can be regarded as a single function $f(z)$ analytic in $D_{1}cup D_{2}$ such that $$f(z)=left{
begin{array}{ll}
f_{1}(z), zin D_{1}\
f_{2}(z), zin D_{2}
end{array}
right.$$

This actually clarifies things out. This is like what we did when we want to remove the singularity: if $f_{1}$ has a removable singularity at $z_{0}$, then we actually extend $f_{1}$ to $f$ by defining $$f(z)=f_{1}(z), zneq z_{0} text{and} f(z_{0})=lim_{zrightarrow z_{0}}f_{1}(z).$$

Thus, we are actually extending $f_{1}(z)$ to $f(z)$, not to $f_{2}(z)$. We sort of complete $f_{1}(z)$ into $Omega_{2}$ by defining $f(z)$.

I hope my explanation can help other people who study complex analysis and find analytic continuation confusing.

Feel free to add anythings more!