How to prove that all smooth vector bundles on a given vector bundle are the pull back of a vector bundle on the base

Question

Recently, during a conversation, I heard about the result (previously mentioned also here on MO), whose statement is reported below. Not having the specific background necessary to reconstruct a proof of such a (probably standard) result by myself, I would like to have some hints, or, at least, know references containing its proof.

Notations

All the considered objects and maps are smooth. Let $E\longrightarrow B$ and $F\longrightarrow E$ be vector bundles.
Embedd $B$ into $E$ through the zero section of $E\longrightarrow B$, and let $F_B\longrightarrow B$ be the vector bundle obtained by restriction of $F\longrightarrow E$ to $B$.

Statement

There exists $\rho:F\longrightarrow E\times_B F_B$, a non-canonical vector bundle isomorphism over $\operatorname{id}_E$, such that $\rho|_{F_B}=\operatorname{id}_{F_B}$.

Answer 1

As a follow up to the comment placed by the OP after the answer of Mark Grant, I would like to add some details. I hope the OP will find them an helpful complement to the response by Mark Grant.

Let $\pi:E\longrightarrow M$ be a vector bundle, and $H$ an homotopy from $N$ to $M$ $$\begin{align}H:N\times[0,1]&\longrightarrow M\\(x,t)&\longmapsto H_t(x)=H^x(t).\end{align}$$ Then a vector bundle isomorphism $\rho:H_0^\ast E\longrightarrow H_1^\ast E$ can be constructed as follows.

Fix a linear connection $\nabla$ on the vector bundle $E\longrightarrow M$. For any smooth curve $\gamma:[0,1]\to M$, let the linear isomorphism $\operatorname{Pt}[\gamma]:E_{\gamma(0)}\longrightarrow E_{\gamma(1)}$ be the associated parallel trasport on $(E,\nabla)$.

Finally, for any $(x,u)\in H_0^\ast(E):=N\times_{(H_0,\pi)}E$, define $$\rho(x,u)=(x,\operatorname{Pt}[H^x](u)\in H_1^\ast(E):=N\times_{(H_1,\pi)}E.$$

Furthermore, for any $A\subset N$, if $H$ is an homotopy rel $A$, i.e. $H_t(x)=H_0(x)$, for all $x\in A$, then $\rho$ acts like the identity map on $(H_0|_A)^\ast E$.

Note that the arbitrariness of this choice of $\nabla$ will effect the construction of $\rho$, so that this procedure does not establish at all a canonical isomorphism among the pull-backs.

Edit I found this presentation by Scott Morrison much more complete than my short answer.

Answer 2

This follows from the fact that pulling back by homotopic maps produces isomorphic bundles.

Let $ p:E\to B $ and $ s:B\to E $ denote the projection and zero section, respectively. These are homotopy inverses. In particular $ sp $ is homotopic to the identity, and so $(sp)^*F=p^*s^*F =E\times_B F_B $ is isomorphic to $ F$.