Verify proof for continuity of pointwise orientation This has been asked here, but there is no answer.

Prove that $[(X_1,...,X_n)]$ on a manifold $M$ is continuous if and only if every point $p$ in $M$ has a coordinate neighborhood $(U,\phi) = (U,x^1,...,x^n)$ such that for all $ q \in U$, the differential $ \phi_{ \ast,q}: T_qM \to T_{f(q)}\mathbb{R} \simeq \mathbb{R}^n $ carries the orientation of $T_qM$ to the stander orientation of $\mathbb{R}^n $ in the following sense: $(\phi_{ \ast}X_{1,q},..., \phi_{ \ast}X_{n,q}) \sim (\frac{\partial}{\partial r^1},..., \frac{\partial}{\partial r^n})$



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*This is an exercise from An Introduction to Manifolds by Loring W. Tu, but are there some mistakes? I think:


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*$\color{red}{\phi(q)}$ and not $f(q)$ in "$T_qM \to T_{f(q)}\mathbb{R}$"

*$(\phi_{ \ast}X_{1,q},..., \phi_{ \ast}X_{n,q}) \sim (\frac{\partial}{\partial r^1}\color{red}{\mid_{\phi(q)}},..., \frac{\partial}{\partial r^n}\color{red}{\mid_{\phi(q)}})$
Here is my proof: What is the correct proof if I am wrong?
$\rightarrow$
I'm not sure continuity is needed here. A pointwise orientation is a frame and so $X_{1,q},..., X_{n,q}$ is an ordered basis for $T_qU \cong T_qM$, by Remark 8.2. The $(U,\phi)$ in question is the same as the one in Lemma 21.4. Since $\phi$ is a diffeomorphism by Proposition 6.10, $\phi_{ \ast}$ is an isomorphism of vector spaces by Corollary 8.6 and so carries basis element to basis element.


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*Perhaps this is wrong because I am supposed 


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*to deduce that locally $X_{1,q} = \frac{\partial}{\partial x^n}\mid_{q}$, perhaps with Proposition 8.9, something like what was done in the proofs of Theorem 21.5 or Theorem 21.10

*to somehow use Proposition 8.8.

*to somehow use to the $dx^1 \wedge ... \wedge dx^n (X_1, ..., X_n) > 0$ on $U$ from Lemma 21.4
$\leftarrow$
Again, without continuity assumed (all the more relevant because continuity is what we're trying to prove this time), a pointwise orientation is a frame and so $X_{1,q},..., X_{n,q}$ is an ordered basis for $T_qU \cong T_qM$, by Remark 8.2.
Again, $\phi$ is a diffeomorphism by Proposition 6.10, $\phi_{ \ast}$ is an isomorphism of vector spaces by Corollary 8.6 and so carries basis element to basis element.
The $\frac{\partial}{\partial r^1}\color{red}{\mid_{\phi(q)}},..., \frac{\partial}{\partial r^n}\color{red}{\mid_{\phi(q)}}$ are smooth and so by assumption of equivalence, the $\phi_{ \ast}X_{1,q},..., \phi_{ \ast}X_{n,q}$ are smooth. Since $\phi_{ \ast}$ is an isomorphism, I guess the $(X_{1,q},...,X_{n,q})$ are smooth and thus continuous.


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*The last part might be wrong because I assume smoothness is preserved under a vector space isomorphism.

 A: A pointwise orientation is not a frame but rather a equivalence class of of (possible discontinous) frames, where two frames are equivalent iff at each point the corresponding ordered bases are equivalent. 
Furthermore if a pointwise orientation is continous on $U$ then this just means one can  find a frame which is continous on $U$ and represents the pointwise orientation. So for example even if 
$(\frac{\partial}{\partial r^1},\dots, \frac{\partial}{\partial r^n})$ is continous and $(\frac{\partial}{\partial r^1},\dots, \frac{\partial}{\partial r^n})\sim (\phi_{ \ast}X_{1},\dots, \phi_{ \ast}X_n)$ on $\phi(U)$ this does not necessarily imply that $(\phi_{ \ast}X_{1},\dots, \phi_{ \ast}X_n)$ is continous.
But i think the approach of using Lemma 21.4 is good:
Set $\beta=dr^1\wedge\dots\wedge dr^n$. Then

  
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*$\beta(\frac{\partial}{\partial r^1}|_{\phi(q)},\dots, \frac{\partial}{\partial r^n}|_{\phi(q)})=1$ 
  
*$\beta(\phi_{ \ast}X_{1,q},\dots, \phi_{ \ast}X_{n,q})=dx^1\wedge\dots\wedge dx^n(X_{1,q},\dots,X_{n,q})$

Now by 1., 2. and the corollary of Lemma 21.1  $(\phi_{ \ast}X_{1,q},\dots, \phi_{ \ast}X_{n,q}) \sim (\frac{\partial}{\partial r^1}|_{\phi(q)},\dots, \frac{\partial}{\partial r^n}|_{\phi(q)})$ is equivalent  to $dx^1\wedge\dots\wedge dx^n(X_{1,q},\dots,X_{n,q})>0$ and hence we can apply Lemma 21.4 to prove the desired equivalence.
