Showing linear independence of sections of a vector bundle

Question

Let $\pi:E\to B$ be a smooth vector bundle and $B$ be a paracomapact manifold. Let $U\subset B$ be an open set and $(U,\phi)$ be a local trivialization. Then we know that the sections defined as $\sigma_i(b)=\phi^{-1}(b,e_i), i=1,2,\cdots, r$ forms a local $r$-frame defined from $U$ to $E$ (here $r$ is the rank of E). Now, suppose $(U_\alpha, \phi_\alpha)$ is a collection of local trivialization such that $(U_\alpha)$ covers $B$. Let $(\lambda_\alpha)$ be a partition of unity subordinate to $(U_\alpha)$. Define the local sections $\sigma_{\alpha,i}$ on $U_\alpha$ as before. Now let, $$\tilde{\sigma}_{\alpha,i}=\lambda_\alpha\sigma_{\alpha,i} \ \ \ \mbox{on $U_\alpha$} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 \ \ \ \ \mbox{on $B\setminus Supp(\lambda_\alpha)$}. $$ Define $\sigma_i=\sum_\alpha\tilde{\sigma}_{\alpha,i}$. Does the restriction $(\sigma_1,\sigma_2,\cdots, \sigma_r)\mid_{U_\alpha}$ span $\Gamma(E,U_\alpha)$?

Answer 1

No, it is not always true, since this will imply that every manifold is parallelizable, and the sphere $S^2$ is not parallelizable.

https://en.wikipedia.org/wiki/Parallelizable_manifold