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.
About covering maps!
-
5Take 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... – 2012-11-06
-
0Possible [duplicate](http://math.stackexchange.com/questions/157208/if-pe-to-b-is-a-covering-space-and-p-1x-is-finite-for-all-x-in-b-s/157224#157224). – 2012-11-06
-
0@JacobSchlather - not quite, this is the implication E+X compact => finite sheets. The other was finite sheets implies both or neither of E and X are compact Hausdorff. – 2012-11-06
-
0@DavidRoberts I'm not sure what a 'local trivialisation' is? – 2012-11-06
-
0http://en.wikipedia.org/wiki/Fiber_bundle – 2012-11-06
-
1There 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. – 2012-11-06
-
0The definition of a covering map includes a local trivialisation, just minus the assumption that all fibres are isomorphic. – 2012-11-06
-
2Alternatively (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. – 2012-11-06
-
1@ Ronnie Brown: I don't think we can assume Hausdorff – 2012-11-06
1 Answers
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.
-
0A 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? – 2015-12-01
-
0@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. – 2015-12-01
-
0Doesn'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 $T_1$? – 2015-12-01
-
0I don't understand the question. – 2015-12-01
-
0I 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 $x_0$? – 2015-12-01
-
0We are not. Where was that said? – 2015-12-01