# If a connected open set is evenly covered, then its preimage is uniquely partitioned into slices

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.

• 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? – Josh May 20 '11 at 0:19
• 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. – Jr. May 20 '11 at 0:45
• 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! :) – Mariano Suárez-Álvarez May 20 '11 at 1:38
• By the way, the title of your question is quite unrelated to the question. – Mariano Suárez-Álvarez May 20 '11 at 2:35

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...)

• 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. – grayQuant Nov 27 '15 at 21:10

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.