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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.

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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 $W$.

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Simplest example:

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 $$U'_1 = \{(a,0) , (b,1) \}\mbox{ and } U'_2 = \{(a,1), (b,0)\}.$$

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