# If $p:E\to X$ is a covering map ($X$ connected and locally arcwise connected) then is $E$ locally connected?

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?

• $U$ 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 $E$. – doetoe Feb 13 '15 at 13:11
• not true @doetoe – Anubhav Mukherjee Feb 13 '15 at 16:29
• @AnubhaV what goes wrong in my argument? – doetoe Feb 13 '15 at 17:13
• Ah, I see. It is the definition of connected component (which is a maximal connected subset, but isn't required to be open). – doetoe Feb 13 '15 at 17:17

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 $X$.