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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.

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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.

The rest follows simply.

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  • $\begingroup$ Thank you, I somehow overlooked this. $\endgroup$ – P-addict Oct 21 '19 at 6:07

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