Let $M$ be a smooth manifold, and $E_1\xrightarrow{\pi_1}M,E_2\xrightarrow{\pi_2}M$ are two vector bundles of rank $m$ and $n$. By definition, there exist two open cover $\{U_i\}_{i\in I}$ and $\{V_j\}_{j\in J}$ such that $\pi_1^{-1}(U_i)\cong U_i\times \mathbb{R}^{m}$ and $\pi_2^{-1}(V_j)\cong V_j\times \mathbb{R}^{n}$.
My question is that, can we choose an open cover $\{W_k\}$ such that $\pi_1^{-1}(W_k)\cong W_k\times \mathbb{R}^{m}$ and $\pi_2^{-1}(W_k)\cong W_k\times \mathbb{R}^{n}$ for any $k$? I guess we can take all $U_i\cap V_j$, but in this case, I don't know if we can prove $\pi_1^{-1}(U_i\cap V_j)\cong (U_i\cap V_j)\times \mathbb{R}^{m}$.
Could someone explain this? Thanks!
The homeomorphism $\phi_{1,i}:\pi_1^{-1}(U_i)\cong U_i\times \mathbb R^m$ is just $\pi_1$ under the projection : $$\mathrm{pr}_{U_i}(\phi_{1,i}(x))=\pi_1(x)$$
This is one of the axioms on a vector bundle. So, $$\mathrm{pr}_{U_i}\phi_{1,i}(\pi_1^{-1}(U_i\cap U_j))=\pi_1(\pi_1^{-1}(U_i\cap U_j))=U_i\cap U_i$$ because $\pi_1$ is onto. So, restricted to $\pi_1^{-1}(U_i\cap U_j)$, $\phi_{1,i}$ is a homeomorphism to $U_i\cap U_j\times \mathbb R^m$.