The covering space of connected space 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.)
 A: 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.
