Is composition of covering maps covering map?

Question

In Munkres book, composition of covering maps is covering map when $r^{-1}(z)$ is finite for each $z$ in $Z$ where $q : X\to Y$ , $r:Y\to Z$ are the covering maps. I tried hard to find an example that composition of covering maps is not covering maps but I couldn't find it. Is there such an example?

Answer 1

Yes there is. The following is a classical example (I think it is in Hatcher for example).

Take $H$ to be the hawaiian earings space, that the subspace of $\mathbb C$ which is the union of the circle $C_n\, (n\geq 1)$ of center $\frac 1 {2n}$ and radius $\frac 1 {2n}$. Then you can get a composition of covering maps which is not a covering map as follow :

1. For any $n\geq 1$, find a covering map $p_n \colon E_n \to H$ such that the restriction $p^{-1}(C_n) \to C_n$ is not a trivial covering map.
2. Define $E = \coprod_{n\geq 1} E_n$ and $p \colon E \to H$ the map induced by all the $p_n$'s. Show that $p$ is not a covering map.
3. Show that there is a trivial covering map $t \colon T \to H$ and a (non trivial) covering map $q \colon E \to T$ such that $p = t \circ q$.

I let you fill the details, but the idea is the following : $p \colon E \to H$ should be a map such that for any open neighbourhood $U$ of $0$, $p^{-1}(U)$ is almost a (countable) disjoint union of copy of $U$ except one of the summand screws things up.

Answer 2

Here is a simple visualization of the example that Pece is talking about. The top map is a 2-fold covering but the bottom map has infinite fibers. The composition isn't locally trivial and is therefore not a covering map. However, it is still a semicovering map in the sense that it is a local homeomorphism which has the unique path lifting property.