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.

The fiber is a closed subset of a compact space hence compact. It is closed because it is the inverse image of the closed set ${x}$ by a continuous function. — omar, Mar 26 '14 at 22:28
The fibres are finite, and a finite subset of any topological space is always quasicompact. (The fibres are also Hausdorff, even if the covering space isn't, so they are actually compact.) — Daniel Fischer, Mar 26 '14 at 22:56
We are proving the fibers are finite though. That is not given. — Johnny Apple, Mar 26 '14 at 22:57
The fibre is a discrete subset of a compact space, hence finite (because otherwise it would have an accumulation point). — user17786, Mar 26 '14 at 23:44
All fibers are discrete, so they are compact iff they are finite. — Stefan Hamcke, Mar 26 '14 at 23:49
Can you explain why a fiber is compact? That is the question you left unanswered. You can't say because it is finite, because that is what we are trying to prove. — Johnny Apple, Mar 26 '14 at 23:53
Some authors (Bourbaki is one, many have followed suit) define a compact space as, what you and I might call, a compact Hausdorff space. If you don't add a separation axiom, like $T_1$ or Hausdorff, I'm not sure the property is true. — Olivier Bégassat, Mar 27 '14 at 00:23
I believe it is true without any separation assumption, as (semi) seen in the link. Anyway, Stefan proved in the previous link that for CERTAIN x, the fiber is finite. He did so directly. I am confused on how to prove the OTHER x have finite (or compact) fibers. See the link, Stefan's answer, second paragraph. — Johnny Apple, Mar 27 '14 at 00:25
I actually think it is true without separation, you are right @JohnnyApple. — Olivier Bégassat, Mar 27 '14 at 00:26
How do you show that the OTHER x, the ones in more than one U_i, have compact (or finite) fibers? @Stefan Hamcke — Johnny Apple, Mar 27 '14 at 00:34
Maybe because I deleted it and then undeleted it again. You should reload the page. — Stefan Hamcke, Mar 27 '14 at 00:48

score 3 · Accepted Answer · answered Mar 27 '14 at 00:36

3

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 $\leq N$.

answered Mar 27 '14 at 00:36

Olivier Bégassat

20,248

Is it really this simple? The previous answer seemed much more intricate. – Johnny Apple Mar 27 '14 at 00:48
@JohnnyApple it looks correct to me, but I've been wrong before ^^ – Olivier Bégassat Mar 27 '14 at 00:53
Looks good to me too. I'll definitely note it. – Johnny Apple Mar 27 '14 at 00:54

Stefan Hamcke · Answer 2 · 2014-03-27T01:13:36.760

1

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.

edited Mar 27 '14 at 01:13

answered Mar 27 '14 at 00:08

Stefan Hamcke

27,733

That was my confusion. I thought we didn't need the two cases, because we were already considering a finite subcover where each member induced a homeomorphism. I just wanted the confirmation. – Johnny Apple Mar 27 '14 at 00:50
I think for my previous answer I was inspired by the proof for a more general concept, called fibre bundles, where you can show that the fibre is compact if the total space is, and there you need the point which is only in $U_i$. Covering spaces are just the fibre bundles with discrete fibers. Looks like covering spaces have there own proof for this, and you can even omit the connectedness. – Stefan Hamcke Mar 27 '14 at 00:52

score 1 · Answer 3 · answered Mar 27 '14 at 00:49

1

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.

answered Mar 27 '14 at 00:49

user17786

1,441

Why is the last line in the first paragraph true? And don't you mean C in C' gives C = C'? – Johnny Apple Mar 27 '14 at 00:52
No, I mean that $C'\in C$ implies $C'=C$. $C$ is very small, it does not contain proper non-empty closed subsets. The last line holds because if $C\setminus V\subset C$ would be closed, contradicting the maximality of $C$. – user17786 Mar 27 '14 at 00:56
Oh, maximal intersection. I misread. Sorry. – Johnny Apple Mar 27 '14 at 00:58
Anyway, this only proves that there is one point whose fibre is compact, but this was not your question. – user17786 Mar 27 '14 at 01:12

Why is the fibre of each point compact?

3 Answers3