Let $C$ be a finite sheeted covering space of compact space $X$.

How do I prove that $C$ is compact?

Someone please give me a proof sketch..

Let $p:C\rightarrow X$ be a covering map.

Let $\mathscr{A}$ be an open cover of $C$.

Since $p$ is open, $\{p(V)\}_{V\in\mathscr{A}}$ is an open cover of $X$.

Hence, there exists a finite subcover $\{p(V_1),...,p(V_n)\}$ of $X$.

However, there is no gurantee that $V_i = p^{-1}(p(V_i))$.

How do I tackle this problem?


Given an open cover of the covering space, refine the cover to obtain a new open cover where every open set in the new cover maps homeomorphically onto an open set in the base space, and the pullback of the neighborhood is a disjoint union of $k$ homeomorphic neighborhoods, where the cover has $k$ sheets, and each of the disjoint open neighborhoods is contained in some element in the original cover (refining means each open set in the new cover will be contained in some open set of the old cover, so if we can find a fine subcover of the refinement we are done. We can ensure that all of the disjoint neighborhoods are contained in some set in the cover because open sets are closed under finite intersection).

Since the base space is compact, the projection of the refinement, which is an open cover, has a finite subcover, say with $n$ sets. We may pull back these sets to obtain $kn$ sets in the refinement that cover the covering space.


Let's show that $C=K_1\cup\cdots \cup K_n$ is a finite union of compact subsets $K_i\subset C $, which suffices to prove that $C$ is compact.

For that let's choose a finite trivializing cover $U_1,\cdots,U_n\subset X$ of the covering $p:C\to X$.
This is possible by compactness of $X$.
Choose then a shrinking by open subsets $V_i\subset U_i$ of this cover, meaning that $(V_i)$ is still an open cover of $X$ and that $\bar V_i\subset U_i$ : this is possible because $X$ is normal, like all compact spaces, and thus we may apply this shrinking process: Dugundji page 152.

The rest is then clear: since every $\bar V_i$ is compact (since it is closed in $X$), so is $K_i=p^{-1}(\bar V_i)$, which is the union of a finite number of homeomorphic copies of the compact space $\bar V_i$, and we have kept our promise to write $C=K_1\cup\cdots \cup K_n$ as a finite union of compact subsets $K_i$.

  • $\begingroup$ More generally a finite covering space is proper : see here $\endgroup$ – Georges Elencwajg Dec 19 '14 at 14:53

$\mathcal{O}$ be an open cover of $C$. For $x \in X$, choose an evenly covered open neighborhood $U_x$ around $x$ such that each open cover in $p^{-1}(U_x)$ is contained in an open set of $E$ which is union of open sets of $\mathcal{O}$. $(*)$ Doing this for each $x \in X$ constructs an open cover $\mathcal{O'}$ of $X$ which has a finite subcover $\mathcal{U}$ of $X$ by compactness. Lifting $\mathcal{U}$ to $C$ gives a finite cover (since $p$ is a finite-sheeted cover) of $C$ which has a refines to $\mathcal{O}$ by construction. Thus $C$ is also compact.

$(*)$ Let $G_x$ be an evenly covered neighborhood around $x$. So $p^{-1}(G_x) = \bigsqcup V_{i}$ where the disjoint union is finite, as $p$ is finite-sheeted. $\{O_1, \cdots, O_n\}$ be a finite subcover of $\mathcal{O}$ covering $p^{-1}(x)$ and construct $O = \bigcup O_i$. Then consider $U_i = p(O \cap V_i)$. $U_x = \bigcap U_i$ is the desired neighborhood around $x$, as $p^{-1}(U_x) = \bigsqcup A_i$ where $A_i \subseteq O \cap V_i \subset O$, implying $p^{-1}(U_x) \subset O$.

  • $\begingroup$ HOw can I choose a such neighborhood $U_x$? $\endgroup$ – cococomi Dec 17 '14 at 14:41
  • $\begingroup$ @cococomi I've left it for you to work out. $\endgroup$ – Balarka Sen Dec 17 '14 at 14:42
  • $\begingroup$ WHy each sheet must be contained in a member of an open cover??? $\endgroup$ – cococomi Dec 17 '14 at 14:46
  • $\begingroup$ @cococomi I have no idea what do you mean. You are picking up a nbhd around $x$ in $X$ which is evenly covered and every slice is contained in the open cover. That such an nbhd is possible is left as an exercise for you. Prove it. $\endgroup$ – Balarka Sen Dec 17 '14 at 14:49
  • $\begingroup$ Well.. Please don't get offended but really I feel like I'm asking for help since I have no idea, but you keep saying me to prove it.. Fix $x$ and say $\{c_1,...,c_k\}$ are the inverses of $x$. WHy all these $c_i's$ contained in a single member of an open cover? $\endgroup$ – cococomi Dec 17 '14 at 14:56

Let $\{V_\alpha\}_{\alpha\in I}$ be an open cover of $C$.

Let $U_x$ be an open evenly covered neighborhood of $x$ for each $x\in X$.

Since $C$ is a finite sheeted covering, $p^{-1}(U_x)$ can be decomposed into finite open sets $A^{(x)}_1,...,A^{(x)}_{n_x}$each of which is homeomorphic to $U_x$.

THen, $\bigcup_{\alpha\in I}\bigcup_{i}p(A^{(x)}_i\cap V_\alpha)$ is an open neighborhood of $x$.

Collect all $p(A^{(x)}_i\cap V_\alpha)$ for every $x$ and make a finite subcover.

Then there are finite $x_1,...,x_n$ such that $p(A^{(x_j)}_i\cap V_{\alpha_{x_j}})$'s form a finite subcover of $X$.

Since $p$ is an embedding on each $A^{(x_j)}_i$, $p^{-1}(p(A^{(x_j)}_i\cap V_{\alpha_{x_j}}) $is a finite union of $\bigcup_i A^{(x_j)}_i\cap V_{\alpha_{x_j}}$.

Each of this union with respect to $x_j$ is a subset of $V_{\alpha_{x_j}}$, hence all these collections are finite.

  • $\begingroup$ More generally a finite covering space is proper : see here $\endgroup$ – Georges Elencwajg Dec 19 '14 at 14:53

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