Does the orientation bundle not depend on the choice of cover? In Bott and Tu's book on page 84 they give a cover $(U_{\alpha},\phi_{\alpha})$ of some smooth manifold $M$ and define the "orientation bundle" to be the line bundle over $M$ given by transition functions sgn $J(g_{\alpha \beta})$ where $J(g_{\alpha \beta})$ denotes the Jacobian determinant of $g_{\alpha \beta} = \phi_{\alpha} \circ \phi_{\beta}^{-1}$.
Does this bundle not depend on the choice of open cover? I assume it must not as they go on to say "$M$ is orientable if and only if its orientation bundle is trivial". I also do not see how this claim follows either (they write that it follows "directly from the definition").
 A: I will sketch a way to prove this but you will have to fill in the details. They define in page 48

The transition functions $g_{\alpha \beta}: U_{\alpha} \cap U_{\beta} \rightarrow G$ satisfy the cocycle condition:
$$
g_{\alpha \beta} \cdot g_{\beta y}=g_{\alpha \gamma}
$$
Given a cocycle $\left\{g_{\alpha \beta}\right\}$ with values in $G$ we can construct a fiber bundle $E$ having $\left\{g_{\alpha \beta}\right\}$ as its transition functions by setting
$$E=\left(\coprod U_{\alpha} \times F\right) /(x, y) \sim\left(x, g_{\alpha \beta}(x) y\right)$$
for $(x, y)$ in $U_{\beta} \times F$ and $\left(x, g_{\alpha \beta}(x) y\right)$ in $\cdot U_{\alpha} \times F$

In page 54 they define equivelnce of cocycles $g_{\alpha\beta}$ and $g'_{\alpha \beta}$ if there exist a maps $\lambda_\alpha:U_\alpha \to GL(n,\mathbb{R})$ such that
$$
g_{\alpha \beta}=\lambda_{\alpha} g_{\alpha \beta}^{\prime} \lambda_{\beta}^{-1}\qquad \text { on } U_{\alpha} \cap U_{\beta}
$$
Then they give an exercise 6.2

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

Recall that a refinement of an open cover $\mathcal{U}=\left(U_{i}\right)_{i \in I}$ is an open cover $\mathcal{U}^{\prime}=$ $\left(U_{\alpha}^{\prime}\right)_{\alpha \in A}$ such that there exists a map $\varphi: A \rightarrow I$ with the property
$$
U_{\alpha} \subset U_{\varphi(\alpha)}, \quad \forall \alpha \in A
$$
We write this $\mathcal{U}^{\prime} \prec_{\varphi}$ U. Given a gluing cocycle $g_{i j}$ subordinated to $\mathcal{U}$ then its restriction to $\mathcal{U}^{\prime}$ is the gluing cocycle $g \mid \bullet$ defined by
$$
\left.g\right|_{\alpha \beta}=\left.g_{\varphi(\alpha) \varphi(\beta)}\right|_{U_{\alpha \beta}}
$$
You can prove using exercise 6.2 that the bundles $\left(g_{\bullet \bullet}, u, W\right)$ and $\left(g_{\bullet \bullet}^{\prime}, \mathcal{U}^{\prime}, W\right)$ are isomorphic if and only if there exist an open cover $\mathcal{V} \prec \mathcal{U}, \mathcal{U}^{\prime}$ such that the restrictions of $g$ and $g'$ to $\mathcal{V}$ are equivalent.
Let $(U_\alpha,\psi_\alpha)_{\alpha\in A}$ and $(U_{\alpha'},\psi_{\alpha'})_{\alpha'\in A'}$ be to different covers of $M$. We can take the refinement of both cover to be given by taking the intersections of the open covers and the maps of the refinement to be $\varphi(\alpha\beta')=\alpha \in A$ and $\varphi'(\alpha\beta')=\beta' \in A'$. Now we can construct the equivalence to show that both bundles are isomorphic. To construct $\lambda_{\beta \beta'}$ use the fact that the determinant is multiplicative and
$$\psi_{\alpha}\circ \psi_{\beta}^{-1}=\psi_{\alpha}\circ \psi^{-1}_{\alpha'}\circ \psi_{\alpha'}\circ\psi^{-1}_{\beta'}\circ \psi_{\beta'}\circ \psi_{\beta}^{-1}$$
so we will get $\lambda_{\beta \beta'}=sgn J(\psi_{\beta'}\circ \psi_{\beta}^{-1})$.
for the statement "M is orientable if and only if its orientation bundle is trivial". Assume you have an orientation you can choose the transition maps to be constant $1$ and this will be equivalent to the trivial line bundle by choosing the $\lambda$'s to be constant. In the other direction you can multiply the first coordinate of the maps $\psi_{\alpha}^{-1 }$ with the $\lambda_{\alpha}$ to get that the sign will be constant as
$$1=\lambda_{\alpha}\cdot \lambda_{\beta}\cdot g_{\alpha \beta}$$
by the equivalence condition.
