Is there a covering map with uncountably many slices?

Question

The following is from the Topology by Munkres:

Let $E$ and $B$ be two topological spaces and $p:E\to B$ a continuous surjective map. The open set $U\subset B$ is said to be evenly covered by $p$ if the i nverse image $p^{-1}(U)$ can be written as the union of disjoint open sets $V_\alpha$ in $E$ such that for each $\alpha$, the restriction of $P$ to $V_\alpha$ is a homeomorphisam of $V_\alpha$ onto $U$. The collection $\{V_\alpha\}$ is called a partition of $p^{-1}(U)$ into slices. If every $b\in B$ has a neighborhood $U$ that is evenly covered by $p$, then $p$ is called a covering map.

Here is my question: Is there an example that $p^{-1}(U)$ contains uncountably many slices?

A bit better example would be connected: Take the universal cover of an uncountable wedge of circles. — Moishe Kohan, Nov 21 '15 at 19:14

score 8 · Accepted Answer · answered Nov 21 '15 at 17:14

Sure, why not? If $X$ is an uncountable discrete set and $Y$ is any space, the projection $X \times Y \to Y$ is a covering map. If you don't like this, then you could instead construct a CW complex (which always have universal covers) $X$ with $\pi_1(X)$ uncountable. Then the universal covering map $\tilde X \to X$ has uncountably many slices.

It just so happens that when doing topology the spaces and covering maps we prefer to think about almost always have countable fibers.

score 3 · Answer 2 · answered Nov 21 '15 at 17:15

You can take the trivial covering space $E = B \times F$ where $F$ is an uncountable set with the discrete topology with $\pi \colon B \times F \rightarrow B$ the projection. Then $B$ is evenly covered by the disjoint union $\bigcup_{f \in F} B \times \{ f \}$ of open subsets of $E$ homeomorphic to $B$.

Is there a covering map with uncountably many slices?

2 Answers2