# A fiber bundle over Euclidean space is trivial.

What’s the easiest way to see this? The only thing I could think to do was 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 proved in the same way the Kunneth theorem is proved:  Induct on the size of a finite good cover for the base space, applying the Mayer Vietoris sequence and the Poincare lemma for the induction step. Its assumed that there exists a finite good cover for the base space but it’s not assumed 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 may be there was a short, elementary proof that the authors had taken for granted.

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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.