This is an extension of the following question from Munkres. Section 53-problem 2.
Let $p: E \to B$ be continuous and surjective. Suppose that $U$ is an open set of $B$
that is evenly covered by $p$. Show that if $U$ is connected, then the partition of $p^{-1}(U)$ into slices is unique.
I could solve this problem but one question that I could not answer:
Can we find an example where $U$ is not connected and the partition of $p^{-1}(U)$ into slices is not unique?
I tried simple examples of the covering map $f : \mathbb{R} \to S^1$ and the non-covering maps $g : \mathbb{R}^{+} \to S^1$ without any success.
Best Answer
Let $p : E \to B$ be a continuous surjection and let $U \subset B$ be a nonempty open set which is evenly covered, with slices $V_\alpha, \alpha \in A$. Let $U_1, U_2$ be two disjoint nonempty open subsets of $U$. As an example take $p : \mathbb R \to S^1$ , $U = S^1 \setminus \{1\}$ and $U_i$ any two disjoint nonempty open subsets of $U$.
Obviously the $U_i$ are evenly covered with slices $V^i_\alpha = V_\alpha \cap p^{-1}(U_i)$. The set $W = U_1 \cup U_2$ is not connected but of course also evenly covered. However, the partition of $p^{-1}(W)$ into slices is not unique. In fact, for any bijection $f : A \to A$ we get the partition $$W^f_\alpha = V^1_\alpha \cup V^2_{f(\alpha)}$$ of $p^{-1}(W)$ into slices over $W$.