# Covering space induced by properly discontinuous group action

I'm reading from "Galois Groups and Fundamental Groups" of Szamuely about covering spaces.

Given an even, a.k.a. properly discontinuous, continuous action of a group $G$ on $Y$ topological space. Then there is the following:

Lemma: If $Y$ is connected, then the projection $\pi_G : Y \rightarrow Y/G$ is a covering map of $Y$ onto $Y/G$.

Proof: The map is surjective, then given $x \in Y/G$ there exists an open neighbourhood $V \ni x$ such that $V= \pi_G (U)$ and $U$ has the property that $\{ gU \}_{g \in G}$ are pairwise disjoint. So $\pi_G^{-1}(V)= \coprod_\limits{g \in G} gU$, from which the result follows.

Now, my question is: where is the need of connectedness of $Y$? I mean, given a non connected $Y = \coprod Y_\alpha$, where $Y_\alpha$ are the connected components, then because of the action of the group is even, I can repeat the same argument of the proof above and notice that the fiber of $V$ will be distributed all over the connected components, but the fundamental properties of a covering, i.e. fibre which is a disjoint union of opens, everyone homeomorphic to $V$, depends only on the action of the group .

The behavior of the action can change in different connected components, so in general you have no way to guarantee that the fibers of points in distinct components are homeomorphic.

As a simple example consider $Y=\{a,b,c\}$ with the discrete topology and a two element group $G$, where the non-trivial element swaps $a$ and $b$. Then $Y/G$ has two points, and the fibers are the non-homeomorphic sets $\{a,b\}$ and $\{c\}$.

Edit: If in the definition of covering you don't require the fibers to be homeomorphic, your argument does indeed work, i.e. the lemma is valid also for a disconnected $Y$.

Note that some authors may require connectedness of the base space in the definition of covering, since some properties will be only locally constant over the base space (e.g. the degree, as shown in the above example).

• Thank you, but I don't get the point. In general the topology of the quotient is the finest which make the projection map continuous. But then because $Y$ is topologically discrete, then every map from $Y$ to the quotient, whatever the topology is, will be continuous, so the finest topology will be the discrete one. But then, when I take $\pi^{-1}(\{ \bar{a} \}) = \{ a \} \cup \{ b \}$ and both $\{ a \},\{ b \} \sim \{ \bar{a} \}$ – HaroldF Mar 4 '17 at 15:41
• The point is to critique your phrase "disjoint and homeomorphic fiber". Fibers, in this example, are not homeomorphic to each other. One fiber is $\{a,b\}$, the other is $\{c\}$, and a two point space cannot be homeomorphic to a one point space. – Lee Mosher Mar 4 '17 at 17:30
• How the fact that the fibers are not "constant" affect the fact that this is a covering? – HaroldF Mar 4 '17 at 20:15
• For any covering map $f : Y \to X$, fibers over the same component of the base space $X$ will be "constant" in the sense that they have the same cardinality, whereas fibers over different components have no relation to each other. – Lee Mosher Mar 4 '17 at 21:32
• And that's ok, but here the base space is discrete, so totally disconnected. – HaroldF Mar 4 '17 at 22:50