# The fiber of a covering space over a connected space has constant cardinality [duplicate]

Let $p: E\to B$ be a covering map; let $B$ connected. Show that if $p^{-1}(b_0)$ has $k$ elements for some $b_0 \in B$, then $p^{-1}(b)$ has $k$ elements for every $b \in B$.

I know that $E$ has a unique slice because $B$ is connected, but I don't know what to do next.

For the sake of providing some context, this is Section 53, Exercise 3 of Munkres' Topology.

• When we talk about $p$ being a covering map, it has some special properties. Perhaps one of these properties says something nice about local things which you can apply here!
– user2959
May 20, 2011 at 20:06

## 2 Answers

Let $p : E \to B$ be a covering map with $B$ connected. Show that if $p^{-1}(b_0)$ has $k$ elements for some $b_0 \in B$, then $p^{-1}(b)$ has $k$ elements for every $b \in B$.

Proof: Suppose there exists $b_0 \in B$ with $p^{-1}(b_0)$ having $k$ elements. Let $A := \{b \in B : p^{-1}(b)$ has $k$ elements $\}$. Observe that

1. $A \neq \emptyset$ since $b_0 \in A$.
2. $A \cap (B \setminus A) = \emptyset$.
3. $A \cup (B \setminus A) = B$.
4. Claim 1: $A$ is open.
5. Claim 2: $B \setminus A$ is open.

Conclude that $B \setminus A = \emptyset$, otherwise we would have a separation of $B$, a contradiction that $B$ is connected. Since $B \setminus A = \emptyset$, we must have $A = B$ which means every fiber of $p$ (ie, every set $p^{-1}(b)$) has $k$ elements.

Proof of Claim 1: $A$ is open: Let $a \in A$, then since $p$ is a covering map, there exists $U \ni a$ open such that $p^{-1}(U) = \bigsqcup_{\alpha \in \Lambda} V_\alpha$ where each $V_\alpha$ is homeomorphic to $U$ via $p \vert_{V_\alpha}$. Notice that $\Lambda$ has $k$ elements since $p^{-1}(a)$ intersects with each $V_\alpha$ at exactly one point and there are $k$ such points. Hence we can write $p^{-1}(U) = \bigsqcup_{n = 1}^k V_n$. Now let $x \in U$ be arbitrary and notice that $p^{-1}(x) \subseteq p^{-1}(U) = \bigsqcup_{n = 1}^k V_n$ and $p \vert_{V_n}$ is a homeomorphism and hence $p^{-1}(x)$ has $k$ elements as well, so $x \in A$. Conclude that $A$ is open since for each $a \in A$, $a \in U \subseteq A$.

Proof of Claim 2: $B \setminus A$ is open: Same reasoning as above.

Let $m\in \mathbb N \cup \{\infty\}$ and $B_m\subset B$ be the set of points $b\in B$ such that the cardinality of $p^{-1}(b)$ is $m$.
From the definition of "covering map" it is clear that each $B_m$ is open in $B$ and from the definition of "set-theoretic map" it follows that $B$ is the disjoint union $B=\sqcup_{m=0}^\infty B_m$.
The definition of connected applied to $B$ forces every $B_m$ save one, necessarily $B_k$, to be empty.
We have then proved what we wanted: that $B=B_k$, i.e. that all fibers of $p$ have $k$ elements.

• why can't the cardinality of $p^{-1}(b)$ be uncountable? Aug 25, 2021 at 10:40