# Regarding two different definitions of Covering spaces and their equivalence

I got two different definitions of 'Covering spaces' from two different books while reading. I would like to know if they are equivalent, or is it the case that the authors are simply following different conventions. Here they are:

All spaces are assumed to be path-connected and locally path-connected.

Definition 1 (following Hatcher's book) : $$(\tilde X, p : \tilde X \rightarrow X)$$ is called a covering space of $$X$$ if for each $$x \in X$$, there exists an open neighbourhood $$U$$ of $$x \in X$$ such that $$p^{-1}(U) = \bigcup\limits_{\alpha \in \cal(A)} U_{\alpha}$$, where $$U_{\alpha}$$ are disjoint, open subsets of $$\tilde X$$ and $$p: U_{\alpha} \rightarrow U$$ is a homeomorphism.

Definition 2 (following Massey): $$(\tilde X, p : \tilde X \rightarrow X)$$ is called a covering space of $$X$$ if for each $$x \in X$$, there exists an open path-connected neighbourhood $$U$$ such that each path-component of $$p^{-1}(U)$$ is homeomorphic to $$U$$ under the map $$p$$.

My attempt at showing equivalence:

First, $$(2) \implies (1)$$ : Recall that if $$X$$ is locally path-connected space, any open subset of $$X$$ is also locally path-connected, and path components of $$X$$ are open. Now, let $$V$$ be the open connected nbd. obtained from (2). $$p^{-1}(U)$$ is open in $$\tilde X$$, hence locally path-connected. So, path components of $$p^{-1}(U)$$ are open. Also, $$p^{-1}(U)$$ is union of disjoint path components of $$p^{-1}(U)$$, hence we get (1).

Now, $$(1) \implies (2)$$ : Suppose (1) holds with $$U$$ as the obtained nbd. Since, $$X$$ is locally path-connected, there exists $$U'$$ open, path-connected in $$X$$ such that $$x \in U' \subset U$$. Now, $$p^{-1}(U') = \bigcup\limits_{\alpha \in \cal(A)} U'_{\alpha}$$, where $$U'_{\alpha} = U_{\alpha} \cap p^{-1}(U)$$. It can be shown that $$p : U'_{\alpha} \rightarrow U'$$ is a homeomorphism. I don't understand how to show that path components are homeomorphic to some open connected nbd. of $$X$$ after this.

You simply have to check that the $$U'_\alpha$$ are the path components of $$p^{-1}(U')$$.
First of all, they are disjoint and open by (1) and the choices made. Moreover, $$U'$$ is path-connected and $$U'_\alpha$$ is homeomorphic to it under $$p$$, so $$U_\alpha'$$ is also path-connected.