Connected and simply connected neighborhoods

Question

Suppose that $E \to M$ is a (smooth) vector bundle over smooth manifold $M$. One can find the covering $\{U_i\}_i$ with the property that $E|_{U_i}$ is trivial vector bundle. The prooblem is the following: why we can assume that the sets $U_i$ have the following property:
"for every $i_1,...,i_k$ such that $U_{i_1} \cap \dots \cap U_{i_k}$ is nonempty the set $U_{i_1} \cap \dots \cap U_{i_k}$ is connected and simply connected?

Answer 1

One way to guarantee this is to pick a Riemannian metric on $M$, then pick the sets $U_i$ so that each is a very small locally convex ball in $M$. From this it follows that each set $U_{i_1} \cap \cdots \cap U_{i_k}$ is actually a convex set in a ball, and so is contractible.

Added: To address the question in the comment, it is proved in almost any differential geometry text that for any Riemannian manifold $M$ and any $x \in M$ there exists $R > 0$ such that for any $r \in (0,R)$ and for any $a,b \in B(x,r)$ there is a unique geodesic arc denoted $[a,b]$ that is contained in $B(x,r)$ and has endpoints $a,b$. That is what I mean by a "locally convex ball". If each of your sets $U_{i_1}$, …, $U_{i_k}$ is locally convex ball in this sense, it is easy to prove that the intersection $V = U_{i_1} \cap \cdots \cap U_{i_k}$ is itself convex in the same sense. From that it is easy to prove that $V$ is contractible: pick a base point $p \in V$ in the intersection, and move each point $x \in V$ along the geodesic $[x,p]$ towards $p$.