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Can someone post a proof of the statement that if $X$ is compact then the covering map $q:E\rightarrow X$ is finitely sheeted given that $E$ is compact as well.

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Take a local trivialisation over a finite cover of $X$, hence an open cover of $E$ which has a finite subcover. Then since I presume you are using classical logic, a subset of a finite set is finite... – David Roberts Nov 6 '12 at 0:29
Possible duplicate. – JSchlather Nov 6 '12 at 0:52
There should be some other way without mention of local trivialisation...I guess I'm looking for an answer that relies only on the elementary definitions and properties of covering maps. – frost Nov 6 '12 at 4:38
Alternatively (and assuming Hausdorff): $ E_x = q^{-1}(x) $ is closed and so compact in $E$, and is also discrete, since $q$ is a covering. Hence $E_x$ is finite. – Ronnie Brown Nov 6 '12 at 16:51
@ Ronnie Brown: I don't think we can assume Hausdorff – frost Nov 6 '12 at 17:03

Suppose $p$ is an infinite sheeted cover. Let $x\in X$ and let $U_x$ be a neighbourhood of $x$ such that $p^{-1}(U_x)$ is homeomorphic to a disjoint union of infinitely many copies of $U_x$. Index these subsets by some infinite indexing set $I$ so $$p^{-1}(U_x)\cong\bigsqcup_{i\in I}V^{(i)}_x$$ where $V^{(i)}_x$ is homeomosphic to $U_x$ for all $i\in I$. Further, suppose that the restriction of $p$, $p|V_{x}^{(i)}\colon V_{x}^{(i)}\rightarrow U_x$ is a homeomorphism. Such a set $U_x$ is guaranteed by the definition of a covering space.

Note that the collection of sets $\{V^{(i)}_x \mid \forall x\in X,\forall i\in I\}$ is a cover for $E$ and so has a finite subcover. Suppose such a finite subcover is given by the set $A=\{V^{(i_0)}_{x_0},\ldots, V^{(i_n)}_{x_n}\}$. Now, the open set $V^{(i_0)}_{x_0}$ only covers a single point in the fiber of the point $x_0$ and, because $A$ is finite, there exists a $k$ such that $V_{x_k}^{(i_k)}$ covers an infinite number of points in the fiber of $x_0$.

This is clearly a contradiction however, as the definition of a covering map says that $p$ restricted to any one of the homeomorphic copies of $U_x$ in the preimage of $U_x$ is itself a homeomorphism. But $p|{V_{x_k}^{(i_k)}}$ isn't a homeomorphism because it is not injective (an infinite number of points in $V_{x_k}^{(i_k)}$ get mapped to $x_0$). We conclude that $p$ is not infinite-sheeted.

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A bit late, but I have 2 questions: 1) why does $V_{x_0}^{(i_0)}$ contain a single point in the fiber of $x_0$? 2) Why is the fiber of $x_0$ infinite? – Kevin Sheng Dec 1 '15 at 0:01
@KevinSheng The sets $V_{x_0}^{(i)}$ were chosen so that they were disjoint and each, by construction, contains a point in the fiber of $x_0$. As $V_{x_0}^{(i_0)}$ contains a point in the fiber of $x_0$, it cannot contain any other point $z$ in the fiber of $x_0$ because some other $V_{x_0}^{(j)}$ contains $z$ and so $z\in V_{x_0}^{(i_0)}\cap V_{x_0}^{(j)}$ contradicting disjointedness. The fiber of $x_0$ is infinite because we assumed (in order to reach a contradiction) that $p$ is an infinite sheeted cover. – Dan Rust Dec 1 '15 at 0:37
Doesn't an infinite sheet cover give infinite copies of a neighborhood of $x_0$? How can $x_0$ be it's own neighborhood unless $X $ is $T_1$? – Kevin Sheng Dec 1 '15 at 0:41
I don't understand the question. – Dan Rust Dec 1 '15 at 0:41
I mean that from the definition of a covering map, every $x \in X $ has a neighborhood that has homeomorphic copies in $E $, why are we assuming $\{x_0\} $ is a neighborhood of $x_0$? – Kevin Sheng Dec 1 '15 at 0:49

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