This is from Topology by Munkres:

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.

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.

  • $\begingroup$ It's not good form to ask questions in the imperative; ordering us to "show that" something is true. Also, could you give some context? Is this homework? What have you tried so far? $\endgroup$
    – Josh
    May 20, 2011 at 0:19
  • $\begingroup$ Sorry for that, yes it is a homework.I proved that if $\{V_\alpha\}$ is a slice then each $V_\alpha$ is connected,but I don't know what to do next.thanks. $\endgroup$
    – Jr.
    May 20, 2011 at 0:45
  • $\begingroup$ do not use comments for that: instead of editing the question to observe that you've explained something in a comment, edit it to explain it in the question itself! :) $\endgroup$ May 20, 2011 at 1:38
  • $\begingroup$ By the way, the title of your question is quite unrelated to the question. $\endgroup$ May 20, 2011 at 2:35

3 Answers 3


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.

  • $\begingroup$ While the process of proof is clear, there is a question that comes to mind: Why is the strict inclusion of $V_{\alpha}$ in $W_{\beta}$ not allowed? If it is allowed, then, $V_{\alpha} \setminus W_{\beta}$ can become empty. $\endgroup$ Jan 19, 2021 at 6:45
  • $\begingroup$ I got it! If $V_{\alpha} \subseteq W_{\beta}$, then we can work with $W_{\beta}$ in the same manner to get the exact same contradiction about $W_{\beta}$. $\endgroup$ Mar 25, 2021 at 13:55

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...)

  • 1
    $\begingroup$ How to show that the slices are connected components? Didn't want to start a new question just asking about a part of the proof. $\endgroup$
    – grayQuant
    Nov 27, 2015 at 21:10
  • $\begingroup$ @grayQuant Each slice is of course connected, and since the preimage of $U$ is a disjoint union of the slices, it follows that no two slices are contained in the same component. $\endgroup$
    – Henry
    Nov 8, 2022 at 16:28

I think we could make a simpler argument:

Let $\{U_{\alpha}\}$,$\{V_{\beta}\}$ be partitions of $p^{-1}(U)$.

We know that, if we fix $\alpha_0$, $\forall x\in U_{\alpha_0}$ $\exists! \beta$ s.t $ x \in V_{\beta}$. It follows that if $U_{\alpha_0}$ intersects more than two $V_{\beta}$'s we can write $U_{\alpha_0} = \cup_{\beta} [U_{\alpha_0} \cap V_{\beta}]$ a disjoint open partition; a contradiction to $U_{\alpha_0}$ being connected.


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