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I can show that if $E$ is compact and Hausdorff $B$ has the same properties, also I can show that if $B$ is compact and Hausdorff $E$ is Hausdorff, but I have troubles trying to prove that $E$ is also compact. Any suggestions would be appreciated.

I would like to know if there is a short way or at least a simple way to show that if E is Hausdorff so is B, I can prove it but I have to make a lot of observations and I get a really really long demostration.

This is an exercise in Hatcher (Algebraic Topology) Section 1.3, exercise 3

Frank
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  • You might also want to try doing the exercise making extensive use of ultra filters. Although they are not needed here, the resulting proof can be made much shorter. – Alexander Thumm Jun 12 '12 at 05:10

1 Answers1

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I'll try to answer the question without saying too much so that you can still work on it. I can edit my answer to give a complete solution if need be.

Let $\mathcal{U}$ be an open cover of $E$. Then for each $x\in B$ there exist $p^{-1}(x)$ is finite. Thus we can choose $U^x_1,\ldots, U^x_{n_x}\in\mathcal{U}$ such that $p^{-1}(x)$ is in the union of these sets.

Hints: Look at the image of $U^x_1,\ldots,U^x_{n_x}$ under $p$. Can you get an open set of $B$ from this containing $x$? How can you use this to get an open cover of $B$? How do you extract an open cover of $E$ from this information?

J. Loreaux
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  • Well, For any $x \in B$ I can get an evenly covered neighborhood $U_{x}$ so its inverse image is a finite union of open sets homeomorphic to $U_{x}$, $\bigcup_{k=1}^{n_{x}} A_{k}$; for each $k$, there is an unique element $ y_{k} \in A_{k}$ such that $p(y_{k})=x$. For each $y_{k}$ there is also an open set $S_{k}$ in the cover with $y_{k} \in S_{k}$, so restricting $p$ to $S_{k} \cap A_{k}$ i get an homeomorphism between $S_{k} \cap A_{k}$ and an open set contained in $U_{x}$. – Frank Jun 12 '12 at 04:12
  • By taking direct image of these open sets I can get an open cover of $B$, so I can take a finite subcover, the problem is how to extract t open cover of E when I take inverse image of this finite subcover? Do I have to modify the cover first? – Frank Jun 12 '12 at 04:12
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    For a fixed $x\in B$, consider the images $p(S_k\cap A_k)$. As you have said these are all open in $B$. What do you know about their intersection (I repeat, for the moment, $x$ is fixed)? Is it open? Is it nonempty? If we call this intersection $V_x$, what can we say about its preimage, $p^{-1}(V_x)$? – J. Loreaux Jun 12 '12 at 05:21
  • The intersection is open and obviously not empty because $x$ belongs to all of them. So with those intersections I create the open cover for $B$ right? Then when I take the inverse image of each open set in the finite subcover I get a finite union of open sets and each one is contained in at least one element of the initial cover. Finally I have obtained my finite subcover!

    Thanks a lot!

     – Frank Jun 12 '12 at 05:30
  • Good work, glad you got it. – J. Loreaux Jun 12 '12 at 12:10
  • @J.Loreaux I don't believe this will work. This cover that Frank has constructed is not a subset of the open cover which we began with; but instead seems made up of subsets of the sets in the cover. – user5826 May 21 '17 at 08:13
  • @AlJebr It will work, but the key is in the statement: "Then when I take the inverse image of each open set in the finite subcover I get a finite union of open sets and each one is contained in at least one element of the initial cover." So, instead of just taking the preimage of each element of the finite subcover of $B$, you replace it by some element of the initial cover of $E$ in which it is contained. This yields a finite subcover of the original cover of $E$. – J. Loreaux May 24 '17 at 21:15
  • I am having trouble understanding two things: a) Fix x. Given a $p(S_k\cap A_k)\ni x$, how do we know that its preimage is a finite union of open sets? All we know is that preimages of evenly covered sets are finite unions of disjoint open sets in E. But $S_k\cap A_k$ may not be evenly covered. b) How do we know that the union of these selected cover elements still covers E? It seems like you are assuming $\bigcup_{x\in B}p^{-1}(x) = E$? – Wyatt Kuehster Jan 03 '21 at 09:30
  • a) Recall that $p(S_k \cap A_k)$ is open. Also, note that $p(S_k \cap A_k) \subseteq p(A_k) \subseteq U_x$, so $p^{-1}(p(S_k \cap A_k)) \subseteq p^{-1}(U_x) = \cup_{j=1}^{n_x} A_j$. Therefore, $p^{-1}(p(S_k \cap A_k))$ is the finite union of the open sets $p^{-1}(p(S_k \cap A_k)) \cap A_j$ for $1 \le j \le n_x$. b) The union you stated is true for any function $p : E \to B$. In particular, $E = p^{-1}(B) = p^{-1}(\bigcup_{x \in B} {x}) = \bigcup_{x \in B} p^{-1}(x)$. Preimages commute with arbitrary unions or intersections. – J. Loreaux Jan 04 '21 at 20:48
  • Also, any open subset of an evenly covered open set is evenly covered (the local homeomorphisms just restrict). – J. Loreaux Jan 04 '21 at 20:55