If $p:E \to B$ is a covering map, and if $E$ is compact, prove that $p^{-1}(b) $ is finite for all $b \in B$.

I need to verify correctness of my proof and ask if there is a more straight-forward proof for this problem.

Each $b \in B$ has an evenly covered neighborhood $U_b$, i.e. $p^{-1}(U_b)=\bigsqcup_{\alpha \in J} M_\alpha$, where $M_{\alpha}$ is an open set in $E$, which is homeopmorphic to $U_b$ under $p$, and $M_{\alpha} \cap M_{\beta} = \emptyset$ if $\alpha \neq \beta$.

$K=E \setminus (\bigsqcup_{\alpha \in J} M_\alpha)$ is a closed set in $E$, since $E$ is compact, $K$ is also compact. It is obvious that $\displaystyle\bigcup_{a \in B \ \& \ a\neq b}p^{-1}(U_a)$ is an open cover for $K$, and hence has a finite subcover $\mathcal{A}=\{A_1, \ldots,A_n\}$. Let $\mathcal{M}=\{M_{\alpha}\}_{\alpha \in J}$. Therefore $\mathcal{M} \cup \mathcal{A}$ is an open cover for $E$, hence has a finite subcover $\mathcal{A'}=\{A_1, \ldots,A_n , M_1,\ldots, M_k\}$. Now I need to show that $J=\{1,\ldots,k\}$ This has to be the case, because otherwise there exists $M_{\alpha} \in \mathcal{M}$ such that $M_{\alpha} \notin \{M_1,\ldots, M_k\}$ Obviously $M_{\alpha} \notin \{A_1,\ldots, A_n\}$ but $M_{\alpha} \in E$, contradicts the fact that $\mathcal{A'}$ covers $E$.


Your proof is certainly correct, and in full fleshed out detail there is not any way to make it more straightforward.

But one thing you can do with a proof like this is to use standard terminology to boil it down to a few pithy sentences, for example: Since $b$ has an evenly covered open neighborhood it follows that $p^{-1}(b)$ is a discrete closed subset of $E$. Every closed subset of a compact space is compact. Also, every discrete compact space is finite. It follows that $p^{-1}(b)$ is finite.

Notice what's going on here: all of your details have been moved out of your proof and into proofs of well-known, standard results about subspace topologies and compact spaces. Feel free to quote those kind of standard results.

  • $\begingroup$ I was thinking @the8thone's proof has a flaw in the last sentence ($M_\alpha\in E$). $\endgroup$ – Quang Hoang Jul 26 '14 at 17:44
  • $\begingroup$ This assumes that singletons are closed, no? $\endgroup$ – Ayman Hourieh Jul 26 '14 at 18:14
  • $\begingroup$ @QuangHoang $M_{\alpha} \subset p^{-1}(U_b) \subset E$ $\endgroup$ – the8thone Jul 26 '14 at 19:09
  • $\begingroup$ @the8thone: I don't get it, why $M_\alpha\subset E$ contradicts $M_\alpha\not\in\{A_1,\dots, A_n\}$? Also, I'm not sure why $M_\alpha\not\in\{A_1,\dots,A_n\}$? $\endgroup$ – Quang Hoang Jul 26 '14 at 19:24
  • $\begingroup$ It is obvious that $$\displaystyle\bigcup_{a \in B \ \& \ a\neq b}p^{-1}(U_a)$$ is an open cover for $K$, note this is a union over all $a\in B$ and "$a \neq b$". And $\{A_1, \ldots , A_n\}$ are open sets in $\displaystyle\bigcup_{a \in B \ \& \ a\neq b}p^{-1}(U_a)$, Furthermore $M_{\alpha} \subset p^{-1}U_b$. Is it clear why $M_{\alpha} \notin \{A_1, \ldots , A_n\}$ ? $\endgroup$ – the8thone Jul 26 '14 at 20:32

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