Maximal analytic continuation gives rise to a covering

Suppose that $a$ is a point on a connected Riemann surface $X$ and $\varphi \in \mathcal{O}_a$ admits an analytic continuation along every curve in $X$ starting at $a$. Let $(Y, p, f, b)$ be the maximal analytic continuation of $\varphi$.

Problem: Prove that $p : Y \to X$ is a covering map.

My try: View $\operatorname{id} : X \to X$ as the analytic continuation of itself, then we must have a map $F : X \to Y$ such that $F(a) \in p^{-1}(a)$ and $f \circ F = \operatorname{id}$. I suspect $\# p^{-1}(a)$ gives the number of sheets of the covvering. I don't see yet how we can get a homeomorphism between an open $U \subset X$ and some $V_i$ in each of the sheets.