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What's the easiest way to see this? The only thing I could think of was to try to patch together trivializations. I couldn't find a way to make that work. Thank you!

edit: For the record, here's why I asked about this special case of the more general result about fiber bundles over contractible spaces.

In the much beloved book by Bott and Tu, it's claimed that the Leray Hirsch theorem can be proven in the same way the Kunneth theorem is proven:  induct on the size of a finite good cover for the base space, then apply the Mayer Vietoris sequence and the Poincare lemma for the induction step. It's assumed that there exists a finite good cover for the base space, but it's not assumed that this cover is a refinement of the cover of the base space, which gives the local trivializations of the fiber bundle. Therefore, to apply the Poincare lemma in the induction step, it seems that you need to know that the result I asked about is true. Since fiber bundles had just been introduced in the text, I thought there may be a short, elementary proof that the authors had taken for granted.

cngzz1
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Leray Hirsch
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    let me know the reason for the downvote so I don't do it again. Is it because I didn't tell you what I've tried? – Leray Hirsch Aug 24 '12 at 03:06
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    I did not downvote, but allow me to hazard a guess. (If I'm wrong, perhaps the downvoters will explain their thinking). My guess is that basically every resource on fiber bundles contains a proof of the fact that "A fiber bundle over a contractible space is trivial", and you are trying to prove a special case of this. They, perhaps, think you should have done more of your own research before asking here. – Jason DeVito - on hiatus Aug 24 '12 at 04:02
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    Thanks, Jason. I guess that's fair? I did see on Wikipedia that the bundle is trivial if the base space is a contractible CW complex. I guess I asked because I thought perhaps the proof of the special case might be easier. – Leray Hirsch Aug 24 '12 at 04:09
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    @JasonDeVito There are usually several different correct answers to a mathematical question. Discouraging to post a question in this site just because it is easy to find an answer on the internet may not be a good idea. There might be a very good answer which is not well known. – Makoto Kato Aug 24 '12 at 04:14
  • Well, again, I was only speculating - I could be way off the mark. The special case may well have an easier proof, but I'm not familiar with one. Have you tried the extra special case of $\mathbb{R}$? My (completely unfounded) intuition says the $\mathbb{R}^2$ case should be similar to the general case. – Jason DeVito - on hiatus Aug 24 '12 at 04:15
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    Put a connection on the fiber-bundle, then parallel transport from the zero of your vector space along a straight line in the Euclidean space gives you your trivialization. – Ryan Budney Aug 24 '12 at 06:28

2 Answers2

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Since you quote Bott--Tu, I'll use that:

They define on p.56 the pullback of a vector bundle. More generally you can define in the same manner the pullback of a fiber bundle $E$ on a space $B$ with fiber $F$ and structure group $G$: given a map $f : B' \to B$, there exists a new fiber bundle $f^{-1}E$ on $B'$ with again fiber $F$ and structure group $G$.

  • Theorem 6.8 (p.57) extends to this setting: if $f \sim g$ ($f$ is homotope to $g$), then $f^{-1}E \cong g^{-1}E$ (the proof is essentially the same).

  • It is also true that $(g \circ f)^{-1}E \cong f^{-1}(g^{-1}E)$, just like for vector bundles.

Now if $B$ is a Euclidean space, it's contractible, meaning there are map $f : B \to \{*\}$ and $g : \{*\} \to B$ such that $f \circ g \sim id_{\{*\}}$ and $g \circ f \sim id_B$. Now if you apply that last equality to any fiber bundle $E$ on $B$, you get $f^{-1}(g^{-1}E) \cong id^{-1}E = E$. But a fiber bundle on a singleton is necessarily trivial (directly from the definitions), thus $g^{-1}E$ is trivial. And the pullback of a trivial bundle is trivial again, so $E = f^{-1}(g^{-1}E)$ is trivial again.

Najib Idrissi
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Here is an easy proof of the fact that a smooth fiber bundle over a contractible base is trivial. Bott and Tu's book, mentioned in the original question, work in this framework of smooth manifolds and maps. The proof uses Moser's trick and relies on the following general basic result.

Theorem: Every smooth fiber bundle admits a complete Ehresmann connection.

This was proposed as an exercise without proof in [Greub, Halperin, Vanstone: Connections, Curvature, and Cohomology I, Academic Press (1972)], it was presented as a theorem in [Michor: Topics in Differential Geometry, (GSM 93), American Mathematical Society (2008)] and other references with a gap in the proof, and it was finally established in del Hoyo: Complete connections on fiber bundles. Indag. Math. 27, 985–990 (2016).

Let $p:E\to B$ be a smooth fiber bundle, and let $h:f_0\cong f_1:B'\times I\to B$ be a smooth homotopy. Let us pick a complete Ehresmann connection for the pullback bundle $h^*E\to B'\times I$. Let $X\in\mathfrak{X}(h^*E)$ be the horizontal lift of the vector field $(0,\frac{d}{dt})$. Then the time 1 flow $\phi^X_1$ yields an isomorphism between $f_0^*E=h^*E|_{B'\times 0}$ and $f_1^*E=h^*E|_{B'\times 0}$. In particular, if $f_0=\rm{id}_{B}$ and $f_1$ is constant then $f_1^*E$ is trivial and so does $E=f_0^*E$.

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