Morphism of vector bundles covering maps of the bases

Question

Let $E_{1}\to B_{1}$ and $E_{2}\to B_{2}$ be vector bundles over differentiable manifolds $B_{1}$ and $B_{2}$. What means that $F\colon E_{1}\to E_{2}$ is a morphism of bundles covering a map $f\colon B_{1}\to B_{2}$ of the bases?

Thanks.

Answer 1

Usually, a morphism of vector bundles covering a map on base spaces is a map $F:E_1\to E_2$ that fits into a commutative diagram $$ \begin{array}{ccc} E_1 & \overset{F}\to & E_2 \\ \downarrow & & \downarrow\\ B_1 & \underset{f}\to & B_2 \end{array} $$ such that for every $b\in B_1$, $F$ induces a linear map $F_b:(E_1)_{b}\to (E_2)_{f(b)}$ (both fibers are canonically vector spaces, so talking about linear maps makes sense). Some authors require the family of linear maps $(F_b)_{b\in B}$ to be of constant rank over connected components.