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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

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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.2012-06-12

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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?

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    @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 $E$.2017-05-24