# If p is a covering map with B compact and fiber of b finite, E compact

## Homework Statement

Let $p: E \rightarrow B$ be a covering map.

If B is compact and$p^{-1}(b)$ is finite for each b in B, then E compact.

Note: This is a problem from Munkres pg 341, question 6b in section 54.

## The Attempt at a Solution

I begin with a cover of E denote it $\{U_\alpha\}$.

I want to reduce this to a finite subcover (thus showing that E is compact).

First I use the fact that p is a covering map and thus open to send this cover of E to a cover of B.

Denote the image of $\{U_\alpha\}$ under p by $\{W_\alpha\}$

Then since B is compact I can reduce this to a finite subcover: $\cup_{i=1}^n W_i$.

Here is where I get stuck, I'm not sure how to send this finite subcover of B back over to E. I'm not even sure if I'm going about this the right way.

Any help is greatly appreciated, thanks.

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The natural thing to do is of course looking at $p^{-1}(W_i)$. The problem is that

$$U_i\subseteq p^{-1}(W_i)$$

and equality does not hold in general because $p$ is not injective. So maybe you can do something with the fact that $p$ is a local homeomorphism and that it has finite fibers.

One thing I would do is first to replace the $U_i$ by a cover of smaller sets $V_j$ such that $p:V_j\rightarrow p(V_j)$ is a homeomorphism between open sets. Then think about how you can write $p^{-1}(p(V_j))$.

Why are we allowed to replace the $U_\alpha$ by a smaller cover? But then once we have that it's just as simple as saying we can reduce the $V_j$ to a finite subcover since after we map them to B we can take a finite subcover of them and map them back using $p^-1$ since each open set in the image is then homeomorphic to its pre-image?

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Why are we allowed to replace the $U_\alpha$ by a smaller cover?
First try to obtain a finite cover of the $V_j$. Then you can use that $V_j\subseteq U_\alpha$ for some $\alpha$ to obtain a finite cover of the $U_\alpha$.

By the way, in topology we call the $V_j$ a refinement of the $U_\alpha$. The fact that the $V_j$ has a finite subcover can then be stated as "Every open cover has a finite open refinement". This statement is equivalent to compactness.

But then once we have that it's just as simple as saying we can reduce the $V_j$ to a finite subcover since after we map them to B we can take a finite subcover of them and map them back using $p^-1$ since each open set in the image is then homeomorphic to its pre-image?
Are you using that $p^{-1}(p(V_j)) = V_j$? This is still not true. It is only true for injective maps. However, because you are dealing with a covering map, you can write $p^{-1}(p(V_j))$ in a better and more convenient form.

Sorry, I'm still confused about why we're even allowed to replace $U_\alpha$ by the $V_j$ at all.

We can write $p^{-1}(p(V_j))$ as a disjoint union of open sets in E, each of which maps homeomorphically to $p(V_j)$. But this must be finite because of the finiteness of the fibers?

Sorry, I'm still confused about why we're even allowed to replace $U_\alpha$ by the $V_j$ at all.
Yes, that is something you should think about.

We can write $p^{-1}(p(V_j))$ as a disjoint union of open sets in E, each of which maps homeomorphically to $p(V_j)$. But this must be finite because of the finiteness of the fibers?
Yes, even better: we can write $p^{-1}(p(V_j))$ as the finite union of some $V_i$. So there exist a finite set $I$ such that

$$p^{-1}(p(V_j)) = \bigcup_{i\in I} V_i$$

Using this, can you then find a open cover of the $V_j$ and hence of the $U_\alpha$?

I'm honestly so lost.

I'm honestly so lost.
You've done most of the proof! Where are you lost?