# Showing linear independence of sections of a vector bundle

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 $$\Gamma(E,U_\alpha)$$?

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