# Can we find an example where $U$ is not connected and the partition of $p^{-1}(U)$ into slices is not unique?

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.

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 $$W$$.
Take $$B = \{a,b\}$$ and $$E = B \times \{0,1\}$$ with $$p$$ being the map that forgets the second coordinate. Now let $$U = B$$. We could take as the two slices, either $$U_1 = \{(a,0), (b,0)\} \mbox{ and } U_2 = \{(a,1), (b,1)\}$$ or $$U'_1 = \{(a,0) , (b,1) \}\mbox{ and } U'_2 = \{(a,1), (b,0)\}.$$