If $p:Y\to X$ is a covering map and $U \subset X$ is evenly covered, then $p^{-1}(U)$ as a partition into slides doesn't have to be unique. There's a excercie problem in Munkres's topology book like this, but it has a hypothesis over $U$ which is $U$ is a connected set. In these conditions, it says $p^{-1}(U)$ as a partition into slides is unique. But I'd like to see an example where $U$ is not connected and how the partition in that case isn't unique.
 A: Take $X = D_1 \cup D_2 \subset \mathbb R^2$ with the induced topology from $\mathbb R^2$ where $D_1$ is the disk centered on $(-1,0)$ with radius equal to $1/2$ and $D_2$ is the disk centered on $(1,0)$ with the same radius.
Take $Y = (D_1 \times \mathbb Z\}) \cup (D_2 \times \mathbb Z\}) \subset \mathbb R^3$ endowed with the topology induced from $\mathbb R^3$ and $p(x,y,z) = (x,y)$. $p$ is a covering map.
Then $P_1 = \{(D_1 \times \{n\})\cup(D_2 \times \{n\}) \mid n \in \mathbb Z\}$ and $P_2 = \{(D_1 \times \{n\})\cup(D_2 \times \{n+1\}) \mid n \in \mathbb Z\}$ are two different partitions of $p^{-1}(X)$ into sheets.
In a more general way, suppose that $ X = X_1 \cup X_2$ is the union of two distinct connected components. Take $I = \{1,2\}$ endowed with the discrete topology and $Y= X \times I$ endowed with the product topology. $p(x,i) = x$ is a covering map $Y \to X$.
$Q_1 = \{X_1 \times \{1\}, X_2 \times \{2\}\}$ and $Q_2 = \{X_1 \times \{2\}, X_2 \times \{1\}\}$ are two different partitions of $p^{-1}(X)$ into sheets.
A: If $U$ is evenly covered and not connected, then the partition of $p^{-1}(U)$ into slides is never unique unless there is only one slide.
$U$ not connected means that $U = U_1 \cup U_2$ with disjoint open $U_i \subset U$. Let $p^{-1} (U) = \bigcup_{\alpha \in  A} V_\alpha$ be  a decomposition into slides. Then each $p_\alpha : V_\alpha \stackrel{p}{\to} U$ is a homeomorphism and the $V_\alpha^i = (p_\alpha)^{-1}(U_i)$ are open and disjoint subsets of $V_\alpha$ such that $V_\alpha = V_\alpha^1 \cup V_\alpha^2$. If $A$ has more than one element, consider a nontrivial bijection $f : A \to A$. Define $V_\alpha^f = V_\alpha^1 \cup V_{f(\alpha)}^2$. The $V_\alpha^f$ are pairwise disjoint open subsets of $p^{-1} (U)$ whose union is $p^{-1} (U)$. Each $V_\alpha^f$ is mapped by $p$ homeomorphically onto $U$, thus $\{V_\alpha^f\}$ is partition of $p^{-1} (U)$ into slides which is different from $\{V_\alpha\}$.
