# Let $q\colon E\to X$ be a covering map. If $E$ is compact then $X$ is compact and $q$ is a finite-sheeted covering.

Here's what I have so far: Since $$q$$ is continuous and surjective then $$X$$ is compact. For all $$x\in X$$ there exists an evenly covered neighborhood $$U_x$$, so $$\{U_x \colon x\in X\}$$ is an open cover of $$X$$. So there is a finite subcover $$\{ U_{x_1},\dots, U_{x_n}\}$$.

For a contradiction, suppose $$q$$ is not finite-sheeted. Now for for each $$x_i$$, we have that $$q^{-1}(U_{x_i}) = \bigsqcup_{\alpha \in I_i} V_{x_i,\alpha}$$ is disjoint union of open sets of $$E$$. The sets $$\{V_{x_1,\alpha}: \alpha \in I_1\},\dots,\{V_{x_n,\alpha}: \alpha \in I_n\}$$ form an open cover of $$E$$, call it $$\mathcal U$$. So by compactness have a finite subcover $$\mathcal U'\subseteq \mathcal U$$. This where I'm having trouble completing the argument. Basically I want to argue that, for say $$x_1$$, no finite collection of $$\mathcal U$$ could cover the fiber $$q^{-1}(x_1)$$ since we assumed the fiber is infinite. But I'm having trouble how to show this.

For convenience, I write $$U_k=U_{x_k}$$. Without loss of generality, choose the collection $$\{U_1,\ldots,U_n\}$$ such that $$n$$ is minimal, i.e., no collection of $$n-1$$ evenly covered open subsets of $$X$$ covers it. Let $$\mathcal U$$ be as you have described. I claim that $$\mathcal U$$ is finite. Otherwise, some $$\{V_{k,\alpha}:\alpha\in I_k\}$$ must be infinite. By compactness of $$E$$, for some $$\alpha\in I_k$$ we have $$V_{k,\alpha}\subset \cup\{V_{j,\beta}:j\neq k\}$$. Applying $$q$$, we see that $$U_k\subset\cup\{U_j:j\neq k\}$$, contradicting our assumption.

In a covering map all fibres are relatively discrete by definition. If $$X$$ has the (usually assumed) property that all singletons are closed, all fibres are also closed (and thus compact when $$E$$ is). The only discrete compact spaces are finite. QED.

• This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$ – John117 Jan 14 at 12:02