Show there is an oriented atlas. 
Suppose $M$ is a manifold with the following property:
Every point $p \in M$ has a neighborhood $U$ and a continuous ordered
  frame on $TM\vert_U = TU$ which evaluated at every $q \in U$ forms a
  positive basis of $T_q M$.
Show that an orientation on each tangent space satisfying the above
  property determines an oriented atlas on $M$.

Attempt:
Well somehow, I will have to produce an atlas of charts such that for any two such charts $\phi, \psi$, we have that $\det (D(\psi \circ \phi^{-1})(\phi(p))  > 0$.
I'm not sure how I will have to do this. 
Given $p \in M$, I know there is a chart $(U, \phi)$ near $p$. I guess I then could look at $T_q U = T_qM$ and see if has the same orientation as the ordered frame. If yes, then we keep the chart. If not, then we change one coordinate with a minus sign. But I'm not sure if this is the right way to proceed.
 A: We are given a family $\omega$ of orientations $\omega_p, p \in M$, of the tangent spaces $T_p M$ such that every point $p \in M$ has a neighborhood $U$ and a continuous ordered frame on $TM\vert_U = TU$ which evaluated at every $q \in U$ forms a positive basis of $T_q M$. Call such $U$ admissible and the continuous ordered frame $\sigma$ on $TU$ positively oriented rel. $\omega$.
Positive basis of $T_q M$ means that the ordered frame represents the orientation $\omega_p$ (recall that an orientation is an equivalence class of ordered frames = ordered bases).
Now let $\phi : U \to V \subset \mathbb R^n$ be a chart on $M$ such that $U$ is admissible and connected. Note that each $p \in M$ has an open neighborhood $U$ carrying such a chart. We know that $\phi$ induces a bundle isomorphism $T \phi : TU \to TV$. The latter bundle admits a canonical bundle isomorphism $\iota_V : TV \to V \times \mathbb R^n$.
The bundle isomorphism $\tilde T \phi = \iota_V \circ T \phi$ establishes a bijective correspondence between continuous ordered frames on $TU$ and continuous ordered frames on $V \times \mathbb R^n$. The frames on $\mathbb R^n$ can be identfied with matrices in $GL(n,\mathbb R)$ (each vector in a frame is identified with a column in the associated matrix). Thus each continuous ordered frame $\gamma$ on $V \times \mathbb R^n$ can be identified with a continuous map $\gamma^* : V \to GL(n,\mathbb R)$. Since $V$ is connected, $\gamma^*(V)$ is contained either in $GL_+(n,\mathbb R)$ = set of matrices with positive determinant or in $GL_-(n,\mathbb R)$ = set of matrices with negative determinant. Let us say that $\gamma$ is positively oriented if $\gamma^*(V) \subset GL_+(n,\mathbb R)$. This means that $\gamma(x)$ represents the standard orientation $o_x$ of $\{x\} \times \mathbb R^n$ for all $x \in V$.
Now let $\mathcal O$ denote the set of all charts $\phi : U \to V$ on $M$ having the following properties:


*

*$U$ is connected.

*For each $q \in U$, the orientation $\omega_q$ of $T_qM$ is mapped by $\tilde T_q \phi$ to the standard orientation $o_{\phi(q)}$ on $\{\phi(q)\} \times \mathbb R^n$.
For each $p \in M$ there exists a chart in $\mathcal O$ such that $p \in U$. To see this, choose any chart $\phi' : U' \to V'$ around $p$. Let $U''$ be an admissible open neigborhood of $p$ and $U \subset U' \cap U''$ be a connected open neigborhood of $p$. Then $\phi'$ restricts to a chart $\phi : U \to V$. Of course $U$ is again admissible. Let $\sigma$ be a positively oriented continuous ordered frame on $TU$ and let $\gamma$ be the continuous ordered frame on $V \times \mathbb R^n$ which corresponds to $\sigma$ under $\tilde T \phi$. If it is positively oriented, we are done. If not, then $\gamma^*(V) \subset GL_-(n,\mathbb R)$. But then then the chart $\psi = R \circ \phi : U  \to R(V)$, where $R$ is a reflection on $\mathbb R^n$, satisfies 1. and 2.
Thus $\mathcal O$ is an atlas on $M$. 
We claim that $\mathcal O$ an oriented atlas. So let $\phi_1 : U_1 \to V_1$ and $\phi_2 : U_2 \to V_2$ be charts in $\mathcal O$. Let $\psi = \phi_2 \circ \phi_1^{-1}  : \phi_1(U_1 \cap U_2) \to \phi_2(U_1 \cap U_2)$ be the transition function. On the tangent spaces it satisfies $T_x \psi = T_{\phi_1^{-1}(x)} \phi_2 \circ T_x \phi_1^{-1}$. Via the $\iota_{V_i}$ it corresponds to the "Euclidean" derivative $d \psi(x) : \mathbb R^n \to \mathbb R^n$. But by construction it is orientation preserving which means that the determinant of $d \psi(x)$ is positive.
