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Why, in the definition of covering spaces, are evenly covered neighborhoods called this instead of the more obvious covered neighborhoods? (which also corresponds better to the root/ancestral terms "covering space" and "covering map")

Let $p: C \to X$ be a continuous surjective map between two topological spaces. Then a neighborhood $U \subset X$ is called an evenly covered neighborhood if $p^{-1}(U)$ is the disjoint union of open subsets of $U$, call them $C_i$, such that $p|_{C_i}$, $p$ restricted to any one of the $C_i$, is a homeomorphism between that $C_i$ and $U$.

In other words, if the preimage of $U$ under $p$ is the disjoint union of sets homeomorphic to $U$ under $p$.

Is there some type of regularity or niceness or naturalness inherent to this definition that is lacking from some more general definition? Or would I not be abusing terminology if I called "evenly covered neighborhoods" just "covered neighborhoods"?

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    $\begingroup$ I'd say that the "evenly" comes because all the $C_i$ are homeomorphic (because all are homeomorphic to $U$), so they're all "equal" in a topological sense. I'm just thinking out loud, though. $\endgroup$ – Ivo Terek Jul 31 '16 at 21:00
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The example to have in mind is where $X$ is the unit circle $S^1\subseteq\Bbb{C}$ and where $p : \Bbb{R} \to S^1$ is defined by $p(z) \mapsto e^{2\pi i z}$. This $p$ is a covering projection: every $x \in X$ has a neighbourhood that is evenly covered by $p$. The pre-image $p^{-1}(U)$ of a small open arc $U$ symmetric around $1 \in S^1$ is the disjoint union $\ldots (-1 -\epsilon, -1 + \epsilon) \cup (-\epsilon, \epsilon) \cup (1-\epsilon, 1+\epsilon) \cup \ldots$ of open sets $(i-\epsilon, i+\epsilon)$, $i \in \Bbb{Z}$, each of which is mapped homeomorphically onto $U$.

Now let $q : \Bbb{R}_{\ge0} \to S^1$ be the restriction of $p$ to the non-negative reals and consider the pre-image $q^{-1}(U)$ of our small open arc $U$: $q^{-1}(U)$ is the disjoint union $[0, \epsilon) \cup (1-\epsilon, 1+\epsilon) \cup (2-\epsilon, 2+\epsilon) \cup \ldots$. So $q$ does not evenly cover $U$ because the component $[0, \epsilon)$ of $q^{-1}(U)$ is not mapped homeomorphically to $U$. Important properties likes path-lifting that hold for $p$ will fail for $q$, because the local properties of $\Bbb{R}_{\ge0}$ at $0$ are different from the local properties at other points of $q^{-1}(1)$. The word "evenly" emphasises the requirement that each point in $p^{-1}(x)$ has the same local properties.

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Because $p:C\to X$ is surjective and continuous, every open set $U$ in $X$ is covered by its preimage $p^{-1}(U)$ (which is open): here covered means that $p(V)=U$. The adjective evenly is used to mean that $p^{-1}(U)$ is a disjoint union of open sets $U_i$, all homeomorphic to $U$ under $p$; that is, each $p:U_i \longrightarrow U$ an homeomorphism.

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