Analytic continuation and two versions of Monodromy theorem So in complex analysis, the Monodromy Theorem says that:

Let $\gamma_0,\gamma_1$ be two paths in $\mathbb C$ s.t $\gamma_0(0)=\gamma_1(0)=a$ and $\gamma_0(1)=\gamma_1(1)=b$. Let $\{\gamma_s\}_{s\in[0,1]}$ be a homotopy between $\gamma_0$ and $\gamma_1$ fixing the end points.
Let germ $[f]_a\in \mathcal O_a$ where $\mathcal O_a$ is the stalk at $a$. Suppose that $[f]_a$ can be continued analytically along $\gamma_s$ for all $s\in [0,1]$. Then analytic continuation of $[f]_a$ along $\gamma_0$ and along $\gamma_1$ result the same germ at $b$

Now, I don't see how to conclude the "classical" Monodromy theorem from this,
I mean how to show that: 

If a complex function $f$ is analytic in a disk contained in a simply connected domain $D$ and $f$ can be analytically continued along every polygonal arc in $D$, then $f$ can be analytically continued to a single-valued analytic function on all of $D$! 

 A: I think now I can answer my question. 
In general, if $\Omega\subset\mathbb C$ is a simply-connected region and $f$ is an analytic function on disc $a\in D\subset \Omega$ such that the germ $[f]_a$ at $a$ of $f$ admits an analytic continuation along any curve $\gamma$ in $\Omega$ from $a$ to $z$. Then, there exists an analytic function $F:\Omega\rightarrow \mathbb C$ with $F{|_D}=f$. 
For the proof, let $\gamma$ be a curve in $\Omega$ from $a$ to $z$, and let $\tilde{\gamma}$ be the analytic continuation of $[f]_a$  (i.e the lifting to $\mathcal O$).
Define, $F(a):=\tilde\gamma(1)(z)$ i.e, the value of the germ $\tilde\gamma(1)$ at $z$ (recall that $\tilde\gamma(1)$ is the germ of some function at $z$)
Now, since any two curves $\gamma_1,\gamma_2$ from $a$ to $z$ are homotopic, then $\tilde\gamma_1(1)=\tilde\gamma_2(1)$ and hence the definition of $F$ is well-defined. 
For continuity; We note that along any curve $\gamma$ in $\Omega$ and for all $z\in \gamma([0,1])$ we have $F(z)=\tilde\gamma(t)(\gamma(t))$. i.e, $F$ is continuous along all curves in $\Omega$ thus continuous. 
It remains to show that $F$ is analytic. Well, at any $z_0\in \Omega$, let the function element $(V,h)$ be a representative of the germ $\tilde \gamma(1)$ where $z_0\in V$. Hence, at a small neighborhood $z_0\in W\subset V$ , $F(z)=h(z)$. (i.e, at each point $z_0$, $F$ admits power series expansion).
