Suppose $p:E\to B$ is a covering map and $B$ is connected. Prove that if $p^{-1}(\{b\})$ has n points $p^{-1}(\{b\})$ has n points for every $b\in B$

Question

My idea is to somehow show that the group $O_n$ is both open closed which will imply $O_n=B$. Then assign to each $n$ the set of points $O_n\subseteq B$ such that $p^{-1}(b)$ has exactly $n$ points. But I don't know how to fill in the details.

Edit: If someone can provide the proof in it's entirety that would be great. I am just doing a self-study, this question has gone unanswered for a while now.

Answer 1

For all $k\in \mathbb N\cup\{\infty\}$, let $O_k=\{x\in B\,:\, |p^{-1}(x)|=k\}$. We will show that each $O_k$ is open. By definition, the sets $O_k$ form a partition of $B$. Suppose that there is more than one nonempty $O_k$, say $O_n$ and $O_m$. Let $U=O_n$ and $V=B\setminus O_n$. Then $U$ is open, while $V$ is a union of the other sets in the partition, which are all open themselves, so it is open too. Furthermore $O_m\subset V$, so $V\neq\emptyset$. The separation $U,V$ then contradicts the connectedness of $B$.

To see each $O_k$ is open, let $x\in O_k$. It has an evenly covered neighborhood $U$. By definition of even covering, $p^{-1}(y)$ will have $k$ elements for all $y\in U$. So $U\subset O_k$. Since we can find a neighborhood $U$ in $O_k$ around every point $x$ in $O_k$, this shows $O_k$ is open.

Answer 2

If $p : E \to B$ is a covering space, $|p^{-1}(b_0)| = n$, then pick an open neighborhood $U_{b_0}$ of $b_0$ which is evenly covered by $p$.

If $b \in B$ is a point with an evenly covered neighborhood $U_b$ such that $U_b \cap U_{b_0} \neq \emptyset$, then $U_b \cap U_{b_0}$ is also evenly covered and hence $p^{-1}(b_1)$ for any $b_1 \in U_b \cap U_{b_0}$ has cardinality $n$. Thus $p^{-1}(U_b)$ must also lift to $n$-many disjoint slices in $E$ by pigeonhole principle. Hence $p^{-1}(b)$ has cardinality the same as $n$.

So the goal is this: for any point $x \in B$, join $x$ and $b_0$ by a chain of evenly covered open sets. If $B$ is path-connected, this is easily done - choose a path $\sigma$ between $x$ and $b_0$ and open cover it by choosing evenly covered neighborhoods at each point.

If no assumption on $B$ is made, one can do the following. For each point $b \in B$ choose an evenly covered open neighborhood $U_b$ around $b$. The collection $\{U_b\}$ open covers $B$. For any $y \in Y$, $U_y$ must always intersect some open set from this cover, otherwise you can obtain a disconnection of $B$ which is impossible since $B$ is connected. That said, you can now easily be able to prove that $x$ and $b_0$ can be joined by a chain of open sets from this cover. Try doing this.