Total space of vector bundle deformation retracts onto 0-section of base space

I’m trying to prove the following:

Total space of vector bundle deformation retracts onto 0-section of base space.

Books seem to take this fact for granted. I checked Bott Tu and Hatcher. Online people are saying this is easy. I can’t figure it out. It’s easy to see that locally this is true. The total space has local trivializations and $\mathbb{R}^n$ deformation retracts onto $0$. I can’t think of a way to patch up these deformation retractions to get a global one on the total space. I tried using partitions of unity to no avail.

Can you help me?

Solutions Collecting From Web of "Total space of vector bundle deformation retracts onto 0-section of base space"

Scalar multiplication $v \mapsto (1 – t)v$ commutes with arbitrary linear transformations, in particular with transition functions.

Consequently, if $E$ denotes the total space of your vector bundle, $x$ denotes a local coordinate on the base, and $v$ denotes a local coordinate in the fibres, the formula $H(x, v, t) = \bigl(x, (1 – t)v\bigr)$ is independent of trivialization, and so defines a deformation retraction $H:E \times [0,1] \to E$.