# Prove that if a smooth manifold $M$ is contractible then every vector bundle over $M$ is trivial

I've seen that this can be proved by using that if two functions are homotopic then the pullbacks of such functions are isomorphic, but the only "easy" proof of this I found is in Hatcher's vector bundles book and I don't find this proof very clear. This was left to me as a homework exercise and all we've seen of vector bundles are the basic definitions, constrictions by cocycles and that every vector bundle has a riemmanian metric so I don't know how to proceed with only this, any help would be appreciated

You might also try Hussemoller's book "Fiber Bundles", Chapter 2, Corollary 4.8.

Usually one proves, in the same breath, that if $$f,g : X \to Y$$ are two homotopic maps and if $$B$$ is a vector bundle over $$Y$$ then $$f^*(B)$$, $$g^*(B)$$ are isomorphic vector bundles over $$X$$. In Hussemoller's book this is is Chapter 2, Theorem 4.7.

• The proof provided in that book is only for topological vector bundles because lemma 4.1 uses the gluing lemma for continuous functions (over closed sets) but such statement of the lemma is not true for smooth functions
– H4z3
Mar 26 at 20:26
• @H4z3 A smooth bundle is smoothly trivial if and only if it is topologically trivial. Indeed, the latter implies you can find $n$ (=rank) linearly independent continuous vector fields on $M$, but you can approximate these arbitrarily close by smooth vector fields and if you're close enough, these will again be linearly independent (since $\mathrm{Gl}(n)$ is open), hence witness smooth triviality. Mar 27 at 13:52
• Why can you approximate the vector fields by smooth vector fields?
– H4z3
Mar 27 at 19:04
• @H4z3: See Steenrod's book "Topology of fiber bundles", p. 25. Mar 30 at 23:28
• I will leave this as a coment to anyone who might need it. There's a proof of this in Hirsch book that works for smooth manifolds
– H4z3
Apr 1 at 14:35