# How do I prove this map is a covering Projection

Suppose $$X$$ is a Riemann surface and $$a\in\ X$$ suppose $$\phi\in\mathcal O_a$$ is a holomorphic function germ at $$a.$$ According to the theorem 7.8 of Forster's book Lectures on Riemann surfaces on page 47 $$\phi$$ has a maximal analytic continuation $$(Y,p,f,b).$$

Now suppose $$\phi\in\mathcal O_a$$ admits analytical continuation along every curve in $$X$$ which starts at $$a$$.Is it always true that $$p:Y\rightarrow X$$ has path lifting property.

If I assume another extra condition that , $$X$$ is simply connected then Monodormy theorem imply that $$p:Y\rightarrow X$$ has path lifting property.

Another question is that whether $$p:Y\rightarrow X$$ is covering or not?

Obviously path lifting property of $$p:Y\rightarrow X$$ implies that $$p:Y\rightarrow X$$ is a covering since $$p:Y\rightarrow X$$ is unbranched holomorphic according to the definition. Conversely any covering map has path lifting property.

Definition of path lifting property of a map given in the book of Otto Forster's book on page 25 is that - A continuous map $$p:Y\rightarrow X$$ is said to have path lifting property if for every path $$u:[0,1]\rightarrow X$$ and for every point $$y_0 \in Y$$ with $$p(y_0)=u(0)$$ there exists a lifting $$\tilde u :[0,1]\rightarrow Y$$ of $$u$$ such that $$\tilde u(0)=y_0$$.