# What does the “evenly” in the term “evenly covered neighborhood” signify?

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"?

• 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. – Ivo Terek Jul 31 '16 at 21:00

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.

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.