A maybe trivial question about fiber bundles (I'm not an expert, and I didn't find quickly an answer looking here and there). Suppose you are given fiber bundle $p\colon E\to M$, where
$E,M$ are smooth manifolds and $p$ is smoothly locally trivial. Also suppose that the fiber $F$ of the bundle is smoothly contractible. Is that true that $p$ admits a smooth section? 



smooth sections of smooth fiber bundles
The answer is: Yes (at least for finite dimensional manifolds).


In fact you only need that the fiber is contractible not smoothly contractible. Take any continuous section $s\_0 \colon B \to E$. cover $B$ by open sets $U\_i$ such that the bundle is trivializable over each $U\_i$, also make sure that the closure of each $U\_i$ is compact and that the cover is locally finite.


Furthermore, give $E$ any complete Riemannian structure. This provides each fiber $E\_x$ with an induced Riemannian structure, which is also complete. We use this to define the obvious supremum distance between any two sections of $E$ over any sub-space of $B$. We may also construct a continuous map $r \colon E \to \mathbb{R}\_+$ such that the ball of radius $r(x)$ and center $x$ in each fiber is geodesically convex. 


The construction now goes in 2 steps:


1) local construction: for each $i$ we may find a smooth section $s \colon U\_i \to E$ such that the supremum distance defined above is smaller than $r$ on all the points of $S\_0$ restricted to $U\_i$. This is easy and follows from smooth approximation of any function from $U\_i \to F$ defining a section $U\_i \to U\_i \times F$.


2) global construction: use a partion of unity to get a global construction. This is now possible because we were carefull enough to create the smooth local sections such that they lie in a geodesically convex neighborhood.


While finishing this Andrew Stacey posted a similar answer, but it seemed a waste not to poste this also. Especially since we are focussing on different details.


