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Let $X$ be a connected topological space, and $\pi : Y \rightarrow X$ a surjective covering space map. Suppose that the group of deck transformations of $\pi$ contains a subgroup $\mathbb Z_p$, where $p$ is a prime number, such that $\mathbb Z_p$ acts freely and transitively on fibers of $\pi$. If $Y$ is not connected, is it necessarily that $Y$ is homeomorphic to a disjoint union of $p$ copies of $X$? (Note: Here we do not assume that $X$ is locally connected.)

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By "a subgroup $\mathbb{Z}_p$", do you mean "a subgroup which is isomorphic to $\mathbb{Z}_p$"? And I assume in this context $\mathbb{Z}_p$ means the cyclic group of order p? –  Harry Altman Aug 17 '12 at 8:31
    
@HarryAltman Yes. I should write more clearly. –  Hezudao Aug 17 '12 at 13:29
    
Does anyone have examples of pathological covering spaces, perhaps with X not locally connected? That might be instructive. –  Hew Wolff Aug 24 '12 at 5:36

1 Answer 1

Here's a partial result: the original claim is true if we do assume that X is locally connected.

Claim. The original statement above is true if $X$ is locally connected.

Proof. Let $F$ be a fiber, and $Y'$ be the set of connected components of $Y$. There's an obvious map $f: F \rightarrow Y'$ taking a point to the connected component containing that point. The group $G = \mathbb{Z}_p$ acts on $F$, and also (because it's a group of homeomorphisms of $Y$) it acts on $Y'$ in the obvious way. These actions are compatible in the sense that for any $g \in G$, the corresponding permutations of $F$ and $Y'$ commute with $f$.

Because the action of $G$ is both free and transitive, the size of $F$ is $p$.

By Claim 1 below, $f$ is onto.

I claim that $f$ is also one-to-one. Suppose not. Then the size of $Y'$ is less than $p$. Because $p$ is prime, the action of $G$ on $Y'$ is trivial. By assumption, the size of $Y'$ is at least two, so pick two distinct elements $y'_0$, $y'_1$. $f$ is onto, so the subsets $F_i = f^{-1}(y'_i)$ of $F$ are nonempty. Each $F_i$ is fixed setwise under the action by $G$. But then $G$ does not act transitively on $F$, which is against our assumption.

So $f$ is an isomorphism of $G$-sets. $Y$ has exactly $p$ connected components which are cyclically permuted by the deck transformation group. Because $Y$ is locally the same as $X$, $Y$ is locally connected. By Claim 2 below, $Y$ is homeomorphic to the disjoint union of its connected components. QED.

Claim 1. If $p: E \rightarrow B$ is a covering map and $B$ is connected and locally connected, then every fiber of $p$ meets every connected component of $E$.

Proof. Take $E'$ to be the set of connected components of $E$, and suppose $G \subseteq E'$. Define $B_G$ to be the set of $b \in B$ such that $p^{-1}(b)$ meets the components in $G$, and does not meet the other connected components of $E$. Suppose $b \in B_G$. We can choose a neighborhood $U$ of $b$ which is evenly covered. Because $B$ is locally connected, we can choose $U$ to be connected. The sheets over $U$ are also connected, so $U \subseteq B_G$. Since this is true for any $b$, $B_G$ is open.

Every point in $B$ is in exactly one $B_G$, so $\{B_G: G \subseteq E'\}$ forms a partition of $B$. Since each $B_G$ is open, each $B_G$ must be a union of connected components of $B$. But $B$ is connected, so there is exactly one nonempty $B_G$.

So every fiber meets the same set of connected components of $E$. Suppose $W \in E'$. Clearly at least one fiber meets $W$. Therefore all fibers meet $W$. QED.

Claim 2. If $X$ is locally connected, then it's homeomorphic to the disjoint union of its connected components.

Proof. Suppose the connected components are $X_i$. There's an obvious map $q: \amalg_i X_i \rightarrow X$, which is a continuous bijection. We need only prove that $q$ is open. So suppose $U \subseteq \amalg_i X_i$ is open. Each $X_i$ is locally connected, so $\amalg_i X_i$ is also locally connected. So $U$ is a union of sets $U_j$ which are open and connected in $\amalg_i X_i$. Since each $U_j$ is connected, it must lie in a single component of $X$, and therefore must be open in $X$. So $U$ is also open in $X$, and $q$ is open. QED.

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Could you explain why $f$ is onto? –  Hezudao Aug 18 '12 at 5:17
    
Yea, what is this obvious map? –  Chris Gerig Aug 19 '12 at 7:40
    
$f$ takes a point in $Y$ to the connected component containing that point. I'll try to edit to make that clear. But I don't know how to show that it's onto, sorry. I'll think some more. –  Hew Wolff Aug 21 '12 at 0:21

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