vector bundles and cocycles I need a detailed solution to a self-study book's exercise (Bott & Tu's Differential forms in algebraic topology, exercise 6.2).

Show that two vector bundles on $M$ are isomorphic iff their cocycles
  relative to some open cover are equivalent.

I can show it in one direction: isomorphism implies cocycles equivalent, but not the other. 
Please give detail answer; I am totally burnt out by and do not need "hints" :)
 A: The proof below is a rewriting of the proof of Theorem 2.7 in Fibre Bundles by Dale Husemoller.
Let $M$ be a manifold, $\{U_\alpha\}$ an open cover of $M$, and $\pi:E\to M$ and $\pi':E'\to M$ two vector bundles of rank $n$ over $M$, with trivializations $\{\phi_\alpha\}$ and $\{\phi_\alpha'\}$, whose transition functions are denoted by $g_{\alpha\beta}$ and $g_{\alpha\beta}'$, respectively. By assumption, $g_{\alpha\beta} (x) = \lambda_\alpha g_{\alpha\beta}' \lambda_\beta^{-1} (x)$ for all $x\in U_\alpha \cap U_\beta$; here $\lambda_\alpha:U_\alpha\to GL(n,\mathbb{R})$ for all $\alpha$. We wish to construct a bundle isomorphism $f:E\to E'$.
For each $\alpha$, we define $f_\alpha:U_\alpha \times \mathbb{R}^n \to U_\alpha \times \mathbb{R}^n$ by $f_\alpha(x,y) = (x,\lambda_\alpha(x)^{-1}y)$, and we define $f:E\to E'$ by requiring that $f=\phi_\alpha' f_\alpha \phi_\alpha^{-1}$, or $\phi_\alpha' f_\alpha = f \phi_\alpha$ on $U_\alpha \times \mathbb{R}^n$. By construction, $f$ is linear on corresponding fibers.
So far $f$ is only defined locally; to prove that $f$ is a globally defined bundle map, let $(x,y)\in (U_\alpha \cap U_\beta) \times \mathbb{R}^n$, and we will check two definitions of $f$ (using the indices $\alpha$ and $\beta$, respectively) agree. That is, we want to show $\phi_\beta' f_\beta(x,y) = f \phi_\beta(x,y)$ implies (hence equivalent to) $\phi_\alpha' f_\alpha(x,y) = f \phi_\alpha(x,y)$. For this we make the following calculation:
$$
  \phi_\beta' f_\beta(x,y) = \phi_\beta'(x,\lambda_\beta(x)^{-1} y)
  =\phi_\alpha'(x,g_{\alpha\beta}'(x) \lambda_\beta(x)^{-1} y)
  =\phi_\alpha'(x,\lambda_\alpha(x)^{-1} g_{\alpha\beta}(x) y)
  =\phi_\alpha' f_\alpha(x, g_{\alpha\beta}(x) y) .
$$
Using $f\phi_\beta(x,y) = f\phi_\alpha(x,g_{\alpha\beta}(x) y)$, we have $f\phi_\alpha(x,g_{\alpha\beta}(x) y) = \phi_\alpha' f_\alpha(x,g_{\alpha\beta}(x) y)$, or $f \phi_\alpha = \phi_\alpha' f_\alpha$. Therefore, $f$ is a well-defined global bundle map.
It is clear that $f$ is locally invertible (all $\phi_\alpha$, $\phi_\alpha'$ and $\lambda_\alpha$ are), and its local inverses paste together to give a bundle map by the same consideration. Hence $f$ is a bundle isomorphism.
