How do I show that every covering space $f:X \rightarrow Y$ is a fiber bundle with a discrete fiber?

Was wondering is this really trivial

So a fiber space is a short exact sequence $ F \rightarrow E \rightarrow B$

So do you just let X=E=F and B=Y. See having trouble seeing what a fiber bundle is. Just seems really trivial.

Fiber space of the Möbius strip example hasn't really cleared it up on what it is.

  • 2
    $\begingroup$ The challenge for you seems to be understanding the definition of Fibre Bundle. I suggest Steenrod's "Topology of Fibre Bundles" for a thorough treatment, but I'm pretty sure it's done more intuitively in Hatcher. Once you know what a fibre bundle is, then yes it's trivial. $\endgroup$
    – William
    Apr 18 '12 at 16:45
  • $\begingroup$ So what is the structure group? $\endgroup$
    – Aolong Li
    Nov 4 '18 at 9:43

Maybe I can help give you some intuition and then you can try the problem again.

Think algebraically for a second.

A short exact sequence $0\to A\to B\to C\to 0$ of abelian groups is, secretly, just a description that $B$ is "put together" in some sort of way from $A$ and $C$. In a perfect world this sequence is split which, among other things, tells us that $B\cong A\oplus C$ in the "most natural way". Thus, what we can see is that, in general, the existence of a short exact sequence like the one mentioned above is that $B$ is "almost" a product of $A$ and $C$ where, in general, this almost is due to some kind of obstruction.

Topologically, a fiber bundle is an expression of the form $F\to E\to B$. Analogizing before this tells us that $E$ is somehow put together from $B$ And $F$. And, once again, in a perfect world $E$ would be put together in the simplest possible way--it would just be the product $B\times F$. And, once again, we don't live in a perfect world and so there exists fiber bundles that aren't trivial. That said, what we do get is that the middle space in a fiber bundle is "almost" a product of $F$ and $B$. Now though we have a simpler, nicer explanation of what "almost" means--namely, locally. Intuitively, putting $F\to E\to B$ means that locally $E$ looks like the product of $B\times F$ even if the two globally differ.

The question then is how do we phrase this rigorously. Well, of course, just as in the case of algebra, we want our middle guy to surject onto our right guy. In particular we are going to want some surjective map $\pi:E\to B$ (this is why $B$ is a B--it's the base space) called a projection. We see then that what "looks locally like a product" should mean is that for each point $x\in E$ there exist a neighborhood $U$ of $\pi(x)$ such that $\pi^{-1}(U)$ is homeomorphic to $U\times F$ by some homeomorphic $\varphi:\pi^{-1}(U)\xrightarrow{\approx}U\times F$ in a nice way. Namely, we want our homeomorphism to respect $\pi$ so that $\pi$ is the same thing as $\pi_1\circ \varphi$ (where $\pi_1$ is the standard projection $U\times F\to U$). This tells us that $\pi^{-1}(U)$ looks like $U\times F$ the "same everywhere".

So, now if you can figure out why $X$ looks locally like a product of $X$ and $Y$.

EDIT: I hope I didn't misunderstand your question. I thought you were asking for intuition about fiber bundles, and it may have been just that you were trying to figure out why the problem was so easy. If so, I apologize for this wishy-washy response.


Yes, I think you're right about the triviality of the question. By definition, a covering space is a (continuous, surjective) map $p: X \to Y$ such that for every $y \in Y$, there exists an open neighborhood $U$ containing $y$ such that $p^{-1}(U)$ is homeomorphic to a disjoint union of open sets in $X$, each being mapped homeomorphically onto $U$ by $p$. This is the same as saying that $p^{-1}(U) \cong F \times U$, where $F$ is some discrete space, so the local triviality condition is satisfied. You immediately get that $p$ is a fibre bundle with discrete fiber.  

  • $\begingroup$ Thought I would add some detail after thinking about this myself. Since F is a discrete space, sets $ \{f\}\times U$ are open (and for different $f$ disjoint). They are of course also homeomorphic to $U$. Thus $F\times U = \cup_{f\in F} \{f\}\times U$ satisfies the part of the definition pertaining to the preimages under $p$. $\endgroup$ Nov 15 '21 at 10:28

This basically follows almost straight from the definitions. You've probably proven at some point that if $Y$ is connected, then there is a discrete space $F$ such that for all $y \in Y$, $f^{-1}(y) \cong F$, and furthermore there is a neighborhood $U$ of $y$ such that $f^{-1}(U) \cong U \times F$. So the covering $f: X \longrightarrow Y$ is the same as the fiber bundle $$F \hookrightarrow X \xrightarrow{~f~} Y.$$ You should verify that the fact that $f: X \longrightarrow Y$ is a covering indeed makes $F \hookrightarrow X \xrightarrow{~f~} Y$ into a fiber bundle; it is just a matter of using the definitions.

  • $\begingroup$ Why is $f^{-1}(y)\cong F$ for all $y\in Y$ $\endgroup$
    – Babai
    Mar 13 '15 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.