A problem from "Introduction to Topological Manifolds"

**Aki** · August 5th 2011, 10:55 PM

Could anyone help me solve the following problem ?

Show that a covering map is proper if and only if it is finite-sheeted.

This is the problem 11-10 from "Introduction to Topological Manifolds" by John M. Lee. The definition of a covering map in this book is the following.

A covering map is a continuous surjective map $q : E \rightarrow X$ such that $E$ is connected and locally path-connected, and every point of $X$ has an evenly covered
neighborhood.

My first attempt was to show that the image of any sequence in $E$ that diverges to infinity diverges to infinity in $X$ . But, I noticed that for this to be a sufficient condition for properness of $q$ , $E$ should be a second countable Hausdorff space, and that no additional condition is assumed for $E$ other than connectedness and local path-connectedness.

I don't have a clue any more.

**Jose27** · August 6th 2011, 11:28 AM

Try to prove that $e^{2\pi i(\cdot)} : \mathbb{R} \to \mathbb{S}^1$ is not proper using as little special properties of $\mathbb{R}$ as possible, the general idea should follow from there.

**Aki** · August 6th 2011, 06:21 PM

Thank you for your help.

Your mapping is not proper since the preimage of a point in $\mathbb{S}^1$ is an infinite discrete set in $\mathbb{R}$ . I understand that to prove a continuous map being proper, it is necessary to show that it is a closed map in addition to showing that it has compact fibers. The former is harder for me.

**Jose27** · August 6th 2011, 09:44 PM

Originally Posted by Aki

Thank you for your help.

Your mapping is not proper since the preimage of a point in $\mathbb{S}^1$ is an infinite discrete set in $\mathbb{R}$ . I understand that to prove a continuous map being proper, it is necessary to show that it is a closed map in addition to showing that it has compact fibers. The former is harder for me.

According to the first edition of the book a proper map is one such that the preimage of compact sets are compact.

**Aki** · August 7th 2011, 10:29 PM

The definition of proper maps in the second edition is the same.
I referred to Proposition 4.93 (c).

Proposition 4.93 (Sufficient Conditions for Properness)
Suppose $X$ and $Y$ are topological spaces, and $F : X \rightarrow Y$ is a continuous map.
(c) If $F$ is a closed map with compact fibers, then $F$ is proper.

**Jose27** · August 7th 2011, 11:12 PM

You don't need that proposition: The idea is that you associate with a covering of the preimage of a compact set in X a finite refinement (here's where you use the finite-sheeted-ness) using that the set in X is compact and the special open sets one can define associated with points in X.

Edit: I'm assuming you have trouble with finite-sheeted implies proper.

**Aki** · August 8th 2011, 12:46 AM

Yes.
But, I'm not sure about the converse too.

What troubles me is the following.
Given a compact set $K$ in $X$ and an open covering $\{U_\lambda\}_{\lambda\in\Lambda}$ of $q^{-1}(K)$ ,
since $\{q(U_\lambda\)\}_{\lambda\in\Lambda}$ is an open covering of $K$ , one can choose from it a finite covering $\{q(U_\lambda_i)\}_{i=1,\cdots,N}$ of $K$ . But, $\{U_\lambda_i\}_{i=1,\cdots,N}$ is not necessarily a covering of $q^{-1}(K)$ .

**Jose27** · August 8th 2011, 08:12 AM

Originally Posted by Aki

Yes.
But, I'm not sure about the converse too.

What troubles me is the following.
Given a compact set $K$ in $X$ and an open covering $\{U_\lambda\}_{\lambda\in\Lambda}$ of $q^{-1}(K)$ ,
since $\{q(U_\lambda\)\}_{\lambda\in\Lambda}$ is an open covering of $K$ , one can choose from it a finite covering $\{q(U_\lambda_i)\}_{i=1,\cdots,N}$ of $K$ . But, $\{U_\lambda_i\}_{i=1,\cdots,N}$ is not necessarily a covering of $q^{-1}(K)$ .

Where are you using the fact that $q$ is a covering?

Okay, here's a bigger hint: Let $K \subset X$ compact and a covering $(V_a)_{a\in A}$ of the inverse image. For every $x\in K$ there is an $U_x$ such that $W_x=q^{-1}(U_x) =\cup_{j=1}^n U_j$ where the $U_j$ are open and disjoint and $q:U_j\to U$ is an homeomorphism (this is just the definition of a covering map). Try to shrink the $W_j$ so that they are inside one $V_a$ but in a way that they are still preimages of a set and have analogous properties to $W_j$ . Now use the compacteness of $K$ to finish the argument.

**Aki** · August 8th 2011, 11:56 PM

Thanks to your guidance, I managed to come up with something resembling a proof of the first half of the problem.

Let $K \subset X$ compact and $\{U_\alpha\}_{\alpha\in A}$ a covering of $q^{-1}(K)$ . One can construct a covering $\{V_\lambda\}_{\lambda\in \Lambda}$ of $K$ satisfying the condition that
(i) $V_\lambda$ is evenly covered by $q$ for each $\lambda \in \Lambda$ ,
(ii) for each $\lambda\in\Lambda$ there exists $\alpha\in A$ such that $V_\lambda \subset q(U_\alpha)$ ,
(iii) $\bigcup_{\alpha\in A}q(U_\alpha)=\bigcup_{\lambda\in \Lambda}V_\lambda$ .
For the condition (i), one can use the fact that every connected open
subset of an evenly covered open subset is itself evenly covered.

It follows from the compactness of $K$ that there exists a finite set $\Lambda_K \subset \Lambda$ such that $K\subset\bigcup_{\lambda\in \Lambda_K}V_\lambda$ . Since $V_\lambda (\lambda\in\Lambda_K)$ are evenly covered by $q$ , we have $q^{-1}(V_\lambda)=\bigcup_{j=1}^n W_\lambda^j$ where $W_\lambda^j \subset E (\lambda\in\Lambda_K, j=1,\cdots,n)$ are disjoint open sets such that $q|_{W_\lambda^j} W_\lambda^j \rightarrow V_\lambda$ is a homeomorphism. Let $I_\lambda^j$ be a subset of $A$ defined by
$I_\lambda^j:=\{\alpha\in A : W_\lambda^j \subset U_\alpha\}, \lambda\in\Lambda_K, j=1,\cdots,n$ .

Finally, the index set ${\cal I}$ can be constructed by choosing one element arbitrarily from $I_\lambda^j$ that is not empty for $\lambda\in\Lambda_K, j=1,\cdots,n$ . Then $\{U_\alpha\}_{\alpha \in {\cal I}}$ is a finite covering of $q^{-1}(K)$ .

I'm looking forward to your comment.

[I corrected a mistake in the definition of the index set ${\cal I}$ .]

**Aki** · August 9th 2011, 09:45 PM

I noticed a serious error in my "proof".
By the definition of $I_\lambda^j$ , only one necessary open set in $\{U_\alpha\}_{\alpha\in A}$ is guaranteed to be chosen for a finite cover of $q^{-1}(K)$ . Other open sets necessary in the finite cover on the other sheets can be lost, if they intersect with $W_\lambda^j$ but don't contain it.

**Aki** · August 10th 2011, 03:25 AM

I revised my proof.

Let $K \subset X$ be compact and $\{U_\alpha\}_{\alpha\in A}$ an open cover of $q^{-1}(K)$ . For each $x \in K$ there exists an open set $V'_x \subset X$ evenly covered by $q$ . Let $y_1,\cdots,y_n$ be the preimages of $x$ by $q$ . For each $j\in \{1, \cdots, n\}$ there exists $\alpha_j \in A$ such that $y_j \in U_{\alpha_j}$ .
Define $V_x$ by $V_x := V'_x \bigcap \left(\bigcap_{j=1}^n q(U_{\alpha_j})\right)$ . Then, $\{V_x\}_{x\in K}$ is a cover of $K$ . Since $K$ is compact, there exists a finite subcover of it. Let it denoted by $\{V_\lambda\}_{\lambda\in \Lambda}$ . Since $V_\lambda (\lambda\in \Lambda)$ is evenly covered by $q$ , we have $q^{-1}(V_\lambda)=\bigcup_{j=1}^n W_\lambda^j$ where $W_\lambda^j \subset E (\lambda\in\Lambda, j=1,\cdots,n)$ are disjoint open sets such that $q|_{W_\lambda^j} W_\lambda^j \rightarrow V_\lambda$ is a homeomorphism. Let $I_\lambda^j$ be a subset of $A$ defined by
$I_\lambda^j:=\{\alpha\in A : W_\lambda^j \subset U_\alpha\}, \lambda\in\Lambda, j=1,\cdots,n$ . Finally, the index set ${\cal I}$ can be constructed by choosing one element arbitrarily from $I_\lambda^j$ for each $\lambda\in\Lambda, j=1,\cdots,n$ . Then $\{U_\alpha\}_{\alpha \in {\cal I}}$ is a finite cover of $q^{-1}(K)$ .

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