Why is the fibre of each point compact?

Question

For a compact covering space, the fibres of the covering map are finite.

I am working on the same question as the one posed in this link, but there was an unanswered question at the end, namely, why is the fibre of EACH point compact. This just so happens to be the question I cannot answer, so I am looking for an answer to this. I didn't want to repost however, so maybe a mod will appropriately be able to deal with this post.

Also, where did the top answerer use the fact that preimages of fibers are compact to complete the problem? I never saw the fact used in his proof.

Answer 1

Let $p:E\to B$ a covering map. Assume $E$ is compact. Let us prove that the fibers are finite. By definition, every point $b\in B$ is contained in an open neighborhood $V_b$ in $B$ that is evenly covered. Let us write $$p^{-1}(V_b)=\bigsqcup_{x\in p^{-1}(b)}V_{b,x}$$ We get an open cover of $E$ $$E=\bigcup_{b\in B}\bigcup_{x\in p^{-1}(b)}V_{b,x}$$ By compactness of $E$, there is a finite subcover $(V_j=V_{b_j,x_j})_{1\leq j\leq N}$. Let $b_0\in B$. The since $p$ induces an homeomorphism between every $V_j$ and $p(V_j)$, every $V_j$ meets the fiber $p^{-1}(b_0)$ at most once. Also, $p^{-1}(b_0)\subset \cup_{j=1}^N V_j=E$ so the fiber is finite of cardinality $\leq N$.

Answer 2

I see now that the minimality of the cover of $X$ isn't necessary. For a point $x\in U_i$, its fiber $f^{-1}(x)$ is contained in the union of the finitely many $V_k^l$ covering $Y$, and each $V_k^l$ maps injectively to $X$, so it can contain at most one point from the fiber of $x$. Hence $p^{-1}(x)$ must be finite.

You can still complete it the way I started it in my previous post. The idea is that the fibers are bijective to each other among the points in an evenly covered $U$. See my edited answer there.

Answer 3

Let $Y\to X$ be the covering map, and let $C$ be a maximal intersection of non-empty closed subsets (that is, if $C'\subset C$ is closed, then $C=C'$). This implies that for all $x\in C$, and for all open neighbourhoods $V$ of $x$, it holds that $C\subset V$.

Then $f^{-1}(C)$ is closed, and therefore compact (because $Y$ is compact).

Let $\{x_i\}_{i=0}^n$ be an infinite sequence of points with $f(x_i) = x$. It is contained in $f^{-1}(C)$, so it has an accumulation point $y$ with $f(y)=z\in C$. Let $g\colon U\to Y$ be a local inverse with $g(z) = y$ and $g(x)=q$. Then any open neighbourhood of $q$ contains $y$ (because they always contain a $g(V)$ with $V$ a neighbourhood of $x$ and $V$ contains $C$), so $\{x_i\}$ converges to $q$ too.

Therefore $f^{-1}(x)$ has an accumulation point $q$, but we assumed that it was discrete, so we arrive to a contradiction.