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I recall my definition of a covering map.

A continuous and surjective map $p:E\to X$ between topological space, where $X$ is connected and locally arcwise-connected, is called a covering map if for every $x\in X$ there exists an open and connected neighbourhood $V\subseteq X$ of $x$ such that, for every connected component $U$ of $p^{-1} (V) $, the restriction $p:U \to V$ is a homeomorphism.

It is true that also $E$ is locally connected? It would be if the connected components of $p^{-} (V) $ in the previous definition had been open in $E$, but my definition (that is, the definition given in my textbook) doesn't makes that request.

Can you help me?

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  • $\begingroup$ $U$ being a connected component means it is open in $p^{-1}(V)$, which in turn means it is the intersection of $p^{-1}(V)$ with an open subset of $E$. By continuity of $p$, $p^{-1}(V)$, hence also $U$, is open in $E$. $\endgroup$ – doetoe Feb 13 '15 at 13:11
  • $\begingroup$ not true @doetoe $\endgroup$ – Anubhav Mukherjee Feb 13 '15 at 16:29
  • $\begingroup$ @AnubhaV what goes wrong in my argument? $\endgroup$ – doetoe Feb 13 '15 at 17:13
  • $\begingroup$ Ah, I see. It is the definition of connected component (which is a maximal connected subset, but isn't required to be open). $\endgroup$ – doetoe Feb 13 '15 at 17:17
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One should require the components of $p^{-1}(V)$ to be open or, equivalently, that $p^{-1}(V)$ is the topological sum of its components. Alternatively, one could require $E$ to be locally connected, as that makes the components of open subsets of $E$ open.

For a counterexample, let $p:\Bbb Q\to X$ be a map where $X$ has only one point $x$. Then the neighborhood of $x$ is $X$, and its preimage is $\Bbb Q$, whose components are singletons, thus they are homeomorphic to $X$.

The same holds with "connected" replaced by "path connected". In the theory of covering spaces, one usually requires all spaces to be locally path connected, as that leads to a smoother theory, and many theorems would fail without that local assumption.

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