covering map with finite fibres and preimage of a compact set Let $f:X\to Y$ be a covering map (covering maps are surjective) , Y be compact set. And suppose that $f^{-1}(y) $ is finite for each $y\in Y$. Prove that $X$ is also compact.
I think that this problem does not need the hyphotesis of covering map , I think that only is necessary that $f$ is an open map (covering maps are open maps)
Well I dunno how to start this problem :/
If I consider an open covering $ X = \bigcup U_a $ , then $f(X)=Y = f(\bigcup U_a) = \bigcup f(U_a) $ 
I don't know if it's convenient to consider a finite subcovering of $Y$ at this moment, or I need some tricky argument ... Please help me )=
 A: You do need more than the fact that the map is open and has finite fibres and compact range.
For $n\in\Bbb Z^+$ let $$X_n=\left\{\left\langle\frac1n,k\right\rangle\in\Bbb R\times\Bbb Z:0\le k\le n\right\}\;,$$ and let $$X=\{\langle0,0\rangle\}\cup\bigcup_{n\in\Bbb Z^+}X_n\;,$$ topologized as a subset of $\Bbb R^2$ with the usual topology. Let $f:X\to\Bbb R:\langle x,y\rangle\mapsto x$ be the projection to the first coordinate, and let
$$Y=f[X]=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}\;;$$
clearly $Y$ is compact, and $f$ has finite fibres. I claim that $f$ is open.
To see this, let $U\subseteq X$ be open. If $\langle0,0\rangle\notin U$, then $f[U]\subseteq Y\setminus\{0\}$, so $f[U]$ is open in $Y$. And if $\langle0,0\rangle\in U$, there is an $m\in\Bbb Z^+$ such that $\left\langle\frac1n,0\right\rangle\in U$ for each $n\ge m$, in which case $f[U]$ contains a nbhd of $0$ in $Y$ and must again be open in $Y$.
Finally, $X$ is clearly not compact, as $\left\{\left\langle\frac1n,n\right\rangle:n\in\Bbb Z^+\right\}$ is a closed set in $X$ with no limit point.
Notice that this $f$ is not a covering map: the point $0$ in $Y$ has no evenly covered open nbhd. You really will need to use the fact that you’re dealing with a covering map.
Added: A map $f:X\to Y$ is perfect if it is continuous and closed and has compact fibres. If you prove that your covering map is closed, you’ll have shown that it is perfect. Now use the following lemma.

Lemma. If $f:X\to Y$ is perfect, and $Y$ is compact, then $X$ is compact.
Start of Proof. Let $\mathscr{U}$ be an open cover of $X$. For each $y\in Y$ let $X_y=f^{-1}\big[\{y\}\big]$; by hypothesis each $X_y$ is compact, so there is a finite subset $\mathscr{U}_y$ of $\mathscr{U}$ that covers $X_y$. Let $V_y=\bigcup\mathscr{U}_y$; clearly $X_y\subseteq V_y$. Use the fact that $f$ is closed to show that there is an open $W_y\subseteq Y$ such that $X_y\subseteq f^{-1}[W_y]\subseteq V_y$. $\mathscr{W}=\{W_y:y\in Y\}$ is then an open cover of $Y$, so it has a finite subcover $\mathscr{W}_0$. Use $\{f^{-1}[W]:W\in\mathscr{W}_0\}$ and the appropriate collections $\mathscr{U}_y$ to find a finite subcover of $\mathscr{U}$.

