Question: This is from *Topology* by Munkres:

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> Let $p:E \to B$ be a covering map. Suppose $U$ is a 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.
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### What I've tried so far

I proved that if $\{V\_\alpha\}$ is a slice then each $V\_\alpha$ is connected, but I don't know what to do next.

Answer: 
Let $\{V\_\alpha\}$, $\{W\_\beta\}$ be partitions of $p^{-1}(U)$. 

Pick $\alpha$ and consider $V\_\alpha$, choose $e \in V\_\alpha$. Since $e \in p^{-1}(U) = \bigsqcup\_\beta W\_\beta$, there exists a unique $\beta$ such that $e \in W\_\beta$.

We want to show that $V\_\alpha = W\_\beta$. Suppose by way of contradiction that $V\_\alpha \neq W\_\beta$. Then $V\_\alpha \setminus W\_\beta \neq \emptyset$. We know $V\_\alpha \cap W\_\beta \neq \emptyset$ since each contain $e$. Notice that $V\_\alpha \cap W\_\beta$ is open since it is the intersection of two open sets. Moreover $V\_\alpha = \big( V\_\alpha \cap W\_\beta \big) \cup \big( V\_\alpha \setminus W\_\beta \big)$. And of course $\big( V\_\alpha \cap W\_\beta \big) \cap \big( V\_\alpha \setminus W\_\beta \big) = \emptyset$. If we can show that $V\_\alpha \setminus W\_\beta$ is open, then we have a separation of $V\_\alpha$. This is a contradiction to our hypothesis that $U$ is connected since $V\_\alpha$ is homeomorphic to $U$ and is this connected as well. Hence, we can conclude that $V\_\alpha = W\_\beta$ and thus the partition is unique.

Now why is $V\_\alpha \setminus W\_\beta$ open? If $e' \in V\_\alpha \setminus W\_\beta$, there exists $\beta\_{e'} \neq \beta$ such that $e' \in W\_{\beta\_{e'}}$. Notice that $V\_\alpha \setminus W\_\beta = V\_\alpha \cap \bigcup\_{e' \in V\_\alpha \setminus W\_\beta} W\_{\beta\_{e'}}$ which is open because $\bigcup\_{e' \in V\_\alpha \setminus W\_\beta} W\_{\beta\_{e'}}$ is open since it is the union of open sets, and the intersection of two open sets is open.
Show that the slices are the connected components of $p^{-1}(U)$. It follows immediately from this that the partition of $p^{-1}(U)$ into slices is unique.

(You should consider an example where $U$ is *not* connected, to see *why* the partition is not unique in that case, by the way...)