# 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 $$D$$!

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).