A problem from "Introduction to Topological Manifolds"

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 $\displaystyle q : E \rightarrow X$ such that $\displaystyle E$ is connected and locally path-connected, and every point of $\displaystyle X$ has an evenly covered

neighborhood.

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

I don't have a clue any more.

Re: A problem from "Introduction to Topological Manifolds"

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

Re: A problem from "Introduction to Topological Manifolds"

Thank you for your help.

Your mapping is not proper since the preimage of a point in $\displaystyle \mathbb{S}^1$ is an infinite discrete set in $\displaystyle \mathbb{R}$. 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.

Re: A problem from "Introduction to Topological Manifolds"

Quote:

Originally Posted by

**Aki** Thank you for your help.

Your mapping is not proper since the preimage of a point in $\displaystyle \mathbb{S}^1$ is an infinite discrete set in $\displaystyle \mathbb{R}$. 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.

Re: A problem from "Introduction to Topological Manifolds"

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$\displaystyle X$ and $\displaystyle Y$ are topological spaces, and $\displaystyle F : X \rightarrow Y$is a continuous map.

(c) If$\displaystyle F$ is a closed map with compact fibers, then $\displaystyle F$ is proper.

Re: A problem from "Introduction to Topological Manifolds"

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.

Re: A problem from "Introduction to Topological Manifolds"

Yes.

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

What troubles me is the following.

Given a compact set $\displaystyle K$ in $\displaystyle X$ and an open covering $\displaystyle \{U_\lambda\}_{\lambda\in\Lambda}$ of $\displaystyle q^{-1}(K)$,

since $\displaystyle \{q(U_\lambda\)\}_{\lambda\in\Lambda}$ is an open covering of $\displaystyle K$, one can choose from it a finite covering $\displaystyle \{q(U_\lambda_i)\}_{i=1,\cdots,N}$ of $\displaystyle K$. But, $\displaystyle \{U_\lambda_i\}_{i=1,\cdots,N}$ is not necessarily a covering of $\displaystyle q^{-1}(K)$.

Re: A problem from "Introduction to Topological Manifolds"

Quote:

Originally Posted by

**Aki** Yes.

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

What troubles me is the following.

Given a compact set $\displaystyle K$ in $\displaystyle X$ and an open covering $\displaystyle \{U_\lambda\}_{\lambda\in\Lambda}$ of $\displaystyle q^{-1}(K)$,

since $\displaystyle \{q(U_\lambda\)\}_{\lambda\in\Lambda}$ is an open covering of $\displaystyle K$, one can choose from it a finite covering $\displaystyle \{q(U_\lambda_i)\}_{i=1,\cdots,N}$ of $\displaystyle K$. But, $\displaystyle \{U_\lambda_i\}_{i=1,\cdots,N}$ is not necessarily a covering of $\displaystyle q^{-1}(K)$.

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

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

Re: A problem from "Introduction to Topological Manifolds"

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

Let $\displaystyle K \subset X$ compact and $\displaystyle \{U_\alpha\}_{\alpha\in A}$ a covering of $\displaystyle q^{-1}(K)$. One can construct a covering $\displaystyle \{V_\lambda\}_{\lambda\in \Lambda}$ of $\displaystyle K$ satisfying the condition that

(i) $\displaystyle V_\lambda$ is evenly covered by $\displaystyle q$ for each $\displaystyle \lambda \in \Lambda$,

(ii) for each $\displaystyle \lambda\in\Lambda$ there exists $\displaystyle \alpha\in A$ such that $\displaystyle V_\lambda \subset q(U_\alpha)$,

(iii)$\displaystyle \bigcup_{\alpha\in A}q(U_\alpha)=\bigcup_{\lambda\in \Lambda}V_\lambda$.

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

$\displaystyle I_\lambda^j:=\{\alpha\in A : W_\lambda^j \subset U_\alpha\}, \lambda\in\Lambda_K, j=1,\cdots,n$.

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

I'm looking forward to your comment.

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

Re: A problem from "Introduction to Topological Manifolds"

I noticed a serious error in my "proof".

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

Re: A problem from "Introduction to Topological Manifolds"

I revised my proof.

Let $\displaystyle K \subset X$ be compact and $\displaystyle \{U_\alpha\}_{\alpha\in A}$ an open cover of $\displaystyle q^{-1}(K)$. For each $\displaystyle x \in K$ there exists an open set $\displaystyle V'_x \subset X$ evenly covered by $\displaystyle q$. Let $\displaystyle y_1,\cdots,y_n$ be the preimages of $\displaystyle x$ by $\displaystyle q$. For each $\displaystyle j\in \{1, \cdots, n\}$ there exists $\displaystyle \alpha_j \in A$ such that $\displaystyle y_j \in U_{\alpha_j}$.

Define $\displaystyle V_x$ by $\displaystyle V_x := V'_x \bigcap \left(\bigcap_{j=1}^n q(U_{\alpha_j})\right)$. Then, $\displaystyle \{V_x\}_{x\in K}$ is a cover of $\displaystyle K$. Since $\displaystyle K$ is compact, there exists a finite subcover of it. Let it denoted by $\displaystyle \{V_\lambda\}_{\lambda\in \Lambda}$. Since $\displaystyle V_\lambda (\lambda\in \Lambda)$ is evenly covered by $\displaystyle q$, we have $\displaystyle q^{-1}(V_\lambda)=\bigcup_{j=1}^n W_\lambda^j$ where $\displaystyle W_\lambda^j \subset E (\lambda\in\Lambda, j=1,\cdots,n)$ are disjoint open sets such that $\displaystyle q|_{W_\lambda^j} W_\lambda^j \rightarrow V_\lambda$ is a homeomorphism. Let $\displaystyle I_\lambda^j$ be a subset of $\displaystyle A$defined by

$\displaystyle I_\lambda^j:=\{\alpha\in A : W_\lambda^j \subset U_\alpha\}, \lambda\in\Lambda, j=1,\cdots,n$. Finally, the index set $\displaystyle {\cal I}$ can be constructed by choosing one element arbitrarily from $\displaystyle I_\lambda^j$ for each $\displaystyle \lambda\in\Lambda, j=1,\cdots,n$. Then $\displaystyle \{U_\alpha\}_{\alpha \in {\cal I}}$ is a finite cover of $\displaystyle q^{-1}(K)$.