Suppose there is a vector bundle (smooth, with constant rank finite-dimensional fibres) over a (smooth, second-countable, Hausdorff, not necessarily connected) manifold $B$ of dimension $n$.


(a) Is it true, that the manifold B can be covered by a finite number of sets $U\_1,\dots,U\_N$ s.t. the vector bundle, restricted to $U\_i$, is isomorphic to a trivial one for every $i=1,\dots,N$?


(b) If yes, can $N$ be taken to be $n+1$?


P.S. Some observations:


1. It's proven in the book by Milnor and Stasheff, that every bundle allows a countable locally finite trivialization cover.
2. Part (a) is obviously trivial for compact manifolds.
3. It seems, that (b) is true if $B$ is an $n$-dimensional CW-complex. **Proof:** Denote with $B\_k$ union of cells of dimension $0,\dots,k$. Prove by induction in $k$, that there are subsets $U\_0,\dots,U\_k$ of $B$, which cover $B\_k$, s.t. the restriction of the bundle to each of them is trivializable. Start with case $k=0$: construct contractible neighbourhoods of each 0-cell, which do not intersect with each other. Take there union. Now to prove the claim for the next value of $k$ it is enough to construct a contractible non-intersecting neighbourhoods of each $X\_\alpha=e\_\alpha\setminus (U\_0\cup \dots\cup U\_{k-1})$. Call the desired neighbourhood with $V\_\alpha$. First, note, that $X\_\alpha$ is closed in $e\_\alpha^k$ and doesn't intersect with its boundary $\partial e\_\alpha^k$, so we can find its neighbourhood in $e\_\alpha^k$, which doesn't intersect with $\partial e\_\alpha^k$. This set is our candidate for $V\_\alpha\cap B\_k$. Extending it to an open set in $B\_{k+1}$ can be done cell by cell: interpreting $e^{k+1}\_\beta$ as a unit ball with the center in the origin, we can write every its point as $r\theta$, where $\theta\in S^k$ and $r\in [0,1]$. We include $r\theta$ in $V\_\alpha\cap B\_{k+1}$ iff $\theta$ is already there and $r>0.99$. Repeating this procedure we extend it to $B$.


**Edit:** Open sets $U\_1,\dots, U\_N$ are assumed to be open. I don't ask them to be connected.



Does every vector bundle allow a finite trivialization cover?
The answer (to both questions (a) and (b)) is **YES** (assuming $B$ is a **smooth manifold**). A proof can be found on Walschap's book "[Metric Structures in Differential geometry](http://rads.stackoverflow.com/amzn/click/038720430X)", p. 77, Lemma 7.1.


For the OP's convenience, here's a sketch of the proof. Choose an open cover of $B$ such that your vector bundle is trivial over each element. From general results in topology, this (and in fact any) cover of an $n$-dim manifold $B$ admits a refinement $\{ V\_\alpha\}\_{\alpha\in A}$ such that any point in $B$ belong to at most $n+1$ $V\_\alpha$'s. Let $\{\phi\_\alpha\}$ be a partition of unity subordinate to this cover and denote by $A\_i$ the collection of subsets of $A$ with $i+1$ elements. Given $a=\{\alpha\_0,\dots,\alpha\_i\}\in A\_i$, denote by $W\_a$ the set consisting of those $b\in B$ such that $\phi\_\alpha(b)\lt\phi\_{\alpha\_0}(b),\dots,\phi\_{\alpha\_i}(b)$ for all $\alpha\neq\alpha\_0,\dots,\alpha\_i$. Then the collection of $n+1$ open subsets $U\_i:=\cup\_{a\in A\_i} W\_a$ covers $B$ and is such that your bundle restricted to each $U\_i$ is trivial.


