Prove that a Covering map is proper if and only if it is finite-sheeted.

First suppose the covering map $q:E\to X$ is proper, i.e. the preimage of any compact subset of $X$ is again compact. Let $y\in X$ be any point, and let $V$ be an evenly covered nbhd of $y$. Then since $q$ is proper, and $\{y\}$ is compact, $q^{-1}( \{ y\})$ is also compact. In particular the sheets $\bigsqcup_{\alpha\in I}U_\alpha$ of V are an open cover of $q^{-1}( \{ y\})$ and must therefore contain a finite subcover $\{U_1,...,U_n\}$. Then the cardinality of the fiber $q^{-1}( \{ y\})$ is $n$, so that $q$ is finite-sheeted.

Conversely, we suppose that $q$ is finite-sheeted. Let $C\subset X$ be a compact set, and let $\{U_a\}_{a\in I}$ be an open cover of $q^{-1}(C)$...

Now how do I continue?


$\Rightarrow$ Each fiber is compact (by properness) and discrete (from definition of covering space) hence is finite.

$\Leftarrow$ You have to prove that for $K\subset X$ the inverse image $q^{-1}(K)$ is compact.
Since $\operatorname {res} q:q^{-1}(K) \to K$ is a finite covering space in its own right apply my answer to cocomi.


"$\Rightarrow$" Since a singleton $\{x\}\subset X$ is compact wrt every topology, $q^{-1}(\{x\})$ is compact. Covering gives the property for $q^{-1}(\{x\})$ to be discrete in the sense that there exists an open neighborhood $V \ni x$ such that each connected component of $q^{-1}(V)$ is homeomorphic to $V$ itself. Then clearly the connected components are disjoint open sets. This means that for each $e\in q^{-1}(\{x\})$ there exists an open neighborhood $U\ni x$ such that $U\cap q^{-1}(\{x\}) = \{e\}$ (definition of discrete set). Compactness implies sequential compactness therefore if $q^{-1}(\{x\})$ wasn't finite we could extract a sequence which does not converge to any of its points (because they are very separated by open sets). Absurdness which clearly shows that $q^{-1}(\{x\})$ is finite.

"$\Leftarrow$" Let $K\subset X$ be compact and $L=q^{-1}(K)$. If $K$ is was empty than the result would be trivial so assume $K\neq \emptyset$. Let $\{T_\alpha\}$ be an open covering for $L$. Openness of $q$ implies that $\{q(T_\alpha)\}$ is an open covering for K. For each $x\in K$ there exists $U_x \subset q(T_{\alpha})$ (for some $\alpha$) open neighborhood such that each connected component of $q^{-1}(U_x)$ is homeomorphic to $U_x$. Assume now that the components of $q^{-1}(U_x)$ are not finite. Then there are components $E_1,\ldots,E_n \subset q^{-1}(U_x)$ containing $\{e_1,\ldots,e_n\} = q^{-1}(\{x\})$. The others cannot contain them and therefore they are not bijective onto $U_x$ since they don't map anything into $x$ itself. Since these are finite we can shrink them in a way that if a component of $q^{-1}(U_x)$ has non empty intersection with $L$, than it is a subset of $T_\alpha$ for some $\alpha$. Cover $K$ by finitely many such sets $K\subset \cup_{i=1}^m U_{x_i}$. Now $L\subset q^{-1}(\cup_{i=1}^m U_{x_i}) = \cup_{i=1}^m q^{-1}(U_{x_i})$ is covered by finitely connected components and each of these is inside a $T_\alpha$, therefore $L$ is covered by finitely many $T_\alpha$, i.e. it is compact.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.