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I heard in class that not every fiber bundle admits a section. I am not sure why this is true, you can always pick a point on a fiber and follow it through as you glue local trivializations then you get a section, isn't this right?

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  • $\begingroup$ Other users have explained the reason why this does not work and have given counterexample. Similar to Slade's example is the covering map $E\to C$, where $C$ is the circle and $E$ is the boundary of the Möbius strip. It's not surprising that many counterexamples are covering maps: One can in fact show that every connected covering map $p:E\to B$ cannot have a section. $\endgroup$ Jun 2 '15 at 20:39
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The "point" (pun intended) is that there is no canonical identification between the fibers over various points in the base. So, it is not possible to simply "pick a point on a fiber and follow it through". In the important case of a vector bundle, there is always a section - the zero section - because the zero vector is canonical. In general, however, a fiber bundle may not admit a section - for example, the $\mathbb{Z}/2$-bundle $S^n\to \mathbb{RP}^n$ does not admit a section. (Why?)

Hope this helps!

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    $\begingroup$ The case $n=1$ is a particularly simple counterexample. It is the self-double covering map $S^1 \mapsto S^1$ given by $z \mapsto z^2$. $\endgroup$
    – Lee Mosher
    Jun 2 '15 at 20:54
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Here is a counterexample from differential topology.

Take the base space to be the two dimensional sphere $S^2$. Consider the tangent bundle $TS^2$, which of course has a section, e.g. the zero section. Now simply remove that section: let $A$ be obtained from $TS^2$ by removing the zero vector over each point of $S^2$. Then the bundle $A \mapsto S^2$ has no section: this is just a restatement of the "hairy ball" theorem from differential topology, which says that $S^2$ has no nonzero vector field.

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Let $M$ be a Möbius strip, $C\subset M$ a circle running through the center, and consider the fiber bundle $M\setminus C \to C$ given by some projection.

What happens if we try to construct a section in the way you describe? For each $c\in C$, we need to pick a point either above or below $C$.

But as you can easily verify with some paper, $M\setminus C$ is connected! So if we "follow it through", we will find that there is no consistent choice of side.

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  • $\begingroup$ I wonder if this can be generalized. Something like "If the fiber is disconnected but the total space is connected, then $p$ cannot have a section." $\endgroup$ Jun 2 '15 at 20:31
  • $\begingroup$ +1 :) Just a note for people who read both answers, Slade's answer is a great explanation of the fact that the double cover $S^1\to \mathbb{RP}^1$ (the case $n=1$ of $S^n\to\mathbb{RP}^n$ mentioned in my answer) does not admit a section. (I see that Stefan has mentioned this particular example as a comment to the question and Lee has also mentioned this example as a comment to my answer.) $\endgroup$ Jun 2 '15 at 20:57
  • $\begingroup$ @StefanHamcke any Serre fibration with connected total space which admits a section has connected fibers. This can be proved using the long exact sequence of a Serre fibration; see here. The proof shows that fibers inherit all levels of connectivity from the total space. $\endgroup$
    – Arrow
    May 31 '19 at 15:34
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For general fiber bundles it is not true as the other answers have pointed out. As you put some more structure on the bundle then it might or might not be true.

For example:

(i) Every vector bundle admits the zero section.

(ii) A principal bundle admits a (global) section if and only if is trivial.

Maybe your teacher had a more specialized class of bundles in mind when he stated that.

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  • $\begingroup$ In fact, all of the bundles mentioned in the answers (so far) are non-trivial principal bundles. :) $\endgroup$ Jun 2 '15 at 21:09

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