I have been trying to do exercise 2 in section 53 of Munkres' Topology for quite some time. I looked at solutions online, and I encountered two. However, neither seem to make much sense to me. First the exercise:

Let p : E → 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.

Now the solutions:

(1) http://www.math.cornell.edu/~erin/topology/munkres.pdf

This one doesn't make sense to me because he claims that $E = \cup V_\beta$. I do not see why that has to be true. The open set $U$ in the exercise is not said to be the whole codomain. Also, I don't see how the bijection will be a solution to the exercise anyhow.



Here, I dont see why A and B have to be both open and closed. Sure, in their respective subspaces they have to be open and closed, but not in $p^{-1}(U)$.

If I am understanding the argument correctly, since A is open and closed for $p_{|A} : A \rightarrow U$, it must be closed in E as well since $p_{|A} $ is a homeomorphism. Shouldn't this, in that case, always be true for covering spaces? Every open set that is evenly covered is also a closed set?

  • $\begingroup$ I think you are missing the theorem from Chapter 23 of Munkres that states that a set $X$ is connected iff the only sets that are both open and closed in $X$ are $X$ itself and the empty set. $\endgroup$
    – Alex
    Dec 13 '15 at 21:26
  • $\begingroup$ @Alex Actually, no, I am not. That was not the problem. Yeldarbskich cleared that up (I forgot to see $p^{−1}(U)$ as a subspace as well) $\endgroup$
    – Avatrin
    Dec 13 '15 at 23:03
  • $\begingroup$ OK great. My misunderstanding. Glad he cleared it up! It was tough to tell what the sticking point was from your question. For the record, I think proof (1) offers the best write-up, but I think you're right that the proof should read $p^{-1}(U) = \bigcup V_{\beta}$. $\endgroup$
    – Alex
    Dec 14 '15 at 4:19

To your concerns for (2), we give $p^{-1}(U) \subset E$ the subspace topology. Then $A$ and $B$ are open by $U$ being evenly covered, thus they are open in the subspace topology. To see that they are closed in $p^{-1}(U)$ as well note that $p^{-1}(U)-A$ is just the preimage minus a disjoint slice; since all the other slices are open in $E$ then their union intersected with $p^{-1}(U)$ is open in the subspace topology. Therefore $A$ is also closed in $p^{-1}(U)$, and similarly for $B$.

As to your last question, certainly not. Consider $\mathbb{R}$ as a trivial cover of itself, i.e. the identity $\mathbb{R} \to \mathbb{R}$, or more generally any homeomorphism of a space.


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