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Let $p:E\to B$ be a covering map. 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 have no idea what can I do here. Well ... let's take a representation of the preimage into the disjoint slices $$ \coprod\limits_{\alpha \in A} {V_\alpha } = p^{ - 1} \left( U \right) $$ since $ V_{\alpha} \cong U $ each $ V_{\alpha}$ is in particular connected. Maybe I have to consider the connected components of $ \coprod\limits_{\alpha \in A} {V_\alpha } = p^{ - 1} \left( U \right) $ (well... I think that the components are just $V_{\alpha}$ but I'm not completely sure) maybe that would help

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Consider the covering space $\widetilde X= \mathbb R$ of $X=S^1$ with $p:\mathbb R \to S^1,r\mapsto(\cos(2\pi r),\sin(2\pi r))$ as the covering map. The fibre of $(1,0)$ is $\mathbb Z$. Let $U$ be the open disconnected neighborhood of $(1,0)$ which is the disjoint union of two open arcs $U_1=\{(x_1,x_2)\in S^1\mid x_1>0\}$ and $U_2=\{(x_1,x_2)\in S^1 \mid x_1<0\}$. Then the preimage is the disjoint union of open intervals of length $\frac12$ centered at the points of $\frac12\mathbb Z$, say $J_n=(\frac n2-\frac14\ ,\ \frac n2+\frac14)$. For an open set $V_\alpha$ which is supposed to be mapped to $U$, you can take $V_\alpha=J_{2n}\cup J_{2n+1}$ but you could also take $V_\alpha=J_{2n}\cup J_{2n-1}$ or even $V_\alpha=J_{2n}\cup J_{2n-k}$ for an odd integer $k$.

On the other hand, if $U$ is connected, then the partition of its preimage is unique, since the $V_\alpha$ are the connected components.

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