Vector bundle orientable if and only if determinant bundle is (Bott and Tu Prop. 11.4)

Question

Let $E$ be a vector bundle over a smooth manifold $M$. By definition of the determinant bundle, it's clear that if $E$ is orientable then $\det E$ is as well, since $\det E$ is defined to be trivialized on the same cover as $E$ with transition functions which are the determinants of the transition functions of $E$. Bott and Tu Proposition 11.4 claims that the converse is true as well, but doesn't seem to provide any further justification. The fact that $\det E$ is orientable just means there exists a trivialization of it whose transition maps are positive numbers in $GL(1, \mathbf{R})$. For example, $\det E$ could be trivialized on all of $M$. There's no reason why $E$ needs to be trivialized on the same cover, so the existence of a trivializing cover for $\det E$ which is orientable on its own doesn't seem to imply the orientability of $E$. So is this really sufficient to deduce that $\det E$ orientable implies $E$ is? Probably I have completely misunderstood this.

Answer 1

Fix trivializations $(\varphi_i)$ of $\det E$ on an open cover $(U_i)$ of $M$ such that the transition maps are positive. Now fix trivializations $(\psi_j)$ of $E$ on an open cover $(V_j)$ such each $V_j$ is connected. Note that we can turn each $\psi_j$ into a trivialization $\det \psi_j$ of $\det E$ on $V_j$. For each $p\in V_j$, we can then consider the transition map between $\det\psi_j$ and $\varphi_i$ for any $i$ such that $p\in U_i$, and the sign of this transition map is independent of the choice of $U_i$ containing $p$. The sign of this transition map is thus a well-defined locally constant function on all of $V_j$, and thus is constant since $V_j$ is connected. Modifying $\psi_j$ if necessary (compose with a reflection), we may assume the transition map is positive.

I now claim that these trivializations $\psi_j$ (modified so that $\det\psi_j$ always has positive transition maps with each $\varphi_i$) witness the orientability of $E$. Indeed, given $p\in V_j\cap V_k$, the transition map between $\psi_j$ and $\psi_k$ at $p$ has positive determinant iff the transition map between $\det\psi_j$ and $\det\psi_k$ is positive. But we know that both $\det\psi_j$ and $\det\psi_k$ have positive transition maps to $\varphi_i$ for any $i$ such that $p\in U_i$, and so they must also have positive transition maps to each other.