The pullback orientation on $M$ induced by a local diffeomorphism $F:M\to N$ I'd like to understand the proof of the following proposition, cited from Lee's Introduction to Smooth Manifolds.

Proposition 15.15 (The Pullback Orientation). Suppose $M$ and $N$ are smooth manifolds with or without boundary. If $F:M\to N$ is a local diffeomorphism and $N$ is oriented, then $M$ has a unique orientation, called the pullback orientation induced by $F$, such that $F$ is orientation-preserving.
Proof. For each $p\in M$, there is a unique orientation on $T_p M$ that makes the isomorphism $dF_p:T_p M\to T_{F(p)}N$ orientation-preserving. This defines a pointwise orientation on $M$ ...

Why does there exist a unique orientation on $T_p M$ that makes the isomorphism $dF_p:T_p M\to T_{F(p)}N$ orientation-preserving?
Since $dF_p$ is an isomorphism, it must carry a basis for $T_p M$ to a basis for $T_{F(p)}N$. But, this is not enough. To show $dF_p$ is orientation-preserving, one must equip $T_p M$ with a choice of orientations (keep in mind that $T_p M$ is not oriented initially) and prove that $dF_p$ carries a positively-oriented basis to a positively-oriented basis, where I have been stuck.
I think the question can be taken care of properly by appealing to linear algebra, but somehow I don't know how to start it. Can anyone give me a piece of advice? If possible, please do not refer to any differential forms. Thank you.
Note. Two bases for a vector space are said to be consistently oriented if the change-of-coordinate matrix between them has positive determinant. Being consistently oriented characterizes an equivalence relation on bases, and we define an orientation on a vector space to be an equivalence class of bases.
 A: This is indeed a linear algebra problem. Let $(W, \mathcal{O})$ be an oriented vector space and $T\colon V \to W$ an isomorphism. We know that $T$ takes bases of $V$ to bases of $W$. If $\mathfrak{v} = (v_1,\ldots, v_n)$ is a basis for $V$, write $T\mathfrak{v}$ for the basis $(Tv_1,\ldots,Tv_n)$ of $W$.   Then just set $$T^*\mathcal{O} = \{ \mathfrak{v} \mid \mathfrak{v}\mbox{ is a basis for $V$ with $T\mathfrak{v}\in\mathcal{O}$}\}. $$This notation clearly justifies the name "pull-back orientation" and makes $T$ orientation-preserving by design. It is indeed an orientation, because each basis for $W$ is either in $\mathcal{O}$ or $-\mathcal{O}$, so each basis for $V$ is either in $T^*\mathcal{O}$ or in $T^*(-\mathcal{O})$; "uniqueness" then says that the expected relation $T^*(-\mathcal{O}) = -T^*\mathcal{O}$ holds.
Now, saying that $N$ is a oriented manifold is the same as saying that for each $y\in N$ there is an orientation $\mathcal{O}_y$ for $T_yN$, and that the assignment $y\mapsto \mathcal{O}_y$ is, suitably understood, smooth.
If $F\colon M\to N$ is a local diffeomorphism, then each ${\rm d}F_x\colon T_xM\to T_{F(x)}N$ is an isomorphism, so we have an orientation $({\rm d}F_x)^*(\mathcal{O}_{F(x)})$ on $T_xM$. The assignment $x\mapsto ({\rm d}F_x)^*(\mathcal{O}_{F(x)})$ is also smooth because $F$ is a local diffeomorphism (smoothness is a local notion, so requiring $F$ to be a global diffeomorphism is overkill).
A: I'm not sure if my answer is a duplicate of that by @Ivo Terek, but the notation $T^*$ is really Greek to me. Anyway, thanks for his/her kindly help, and I'd like to present a solution that is based on the idea of @Ivin Babu.
By hypothesis, an equivalence class $[\beta]$ of bases has been designated as the orientation of $T_{F(p)}N$. With the aid of $[\beta]$, $T_p M$ can now be endowed with the orientation $[(dF_p)^{-1}(\beta)]$. To make sure the orientation $[(dF_p)^{-1}(\beta)]$ is well-defined, pick another basis $\gamma$ in $[\beta]$ and show that
$$[(dF_p)^{-1}(\gamma)]=[(dF_p)^{-1}(\beta)].$$
All we have to do is show that $(dF_p)^{-1}(\gamma)$ and $(dF_p)^{-1}(\beta)$ are consistently oriented (see the note). If you grab a book on linear algebra and review the section regarding change-of-coordinate matrices, you will see
$$[(dF_p)^{-1}]_\gamma^{(dF_p)^{-1}\ (\gamma)}=\left([I_{T_p M}]_{(dF_p)^{-1}\ (\gamma)}^{(dF_p)^{-1}\ (\beta)}\right)^{-1}[(dF_p)^{-1}]_\beta^{(dF_p)^{-1}\ (\beta)}[I_{T_{F(p)}\ N}]_\gamma^\beta,$$
where $I_{T_p M}$ is the identity operator on $T_p M$, and so on. Watch closely and conclude that the matrix representations of $(dF_p)^{-1}$ are the identity matrices. Now take the determinants, and you will see
$$\det\left([I_{T_p M}]_{(dF_p)^{-1}\ (\gamma)}^{(dF_p)^{-1}\ (\beta)}\right)=\det\left([I_{T_{F(p)}\ N}]_\gamma^\beta\right)>0.$$
The last inequality holds because $\gamma\in[\beta]$.
After testifying the orientation is well-defined, we need to show that $dF_p$ is orientation-preserving, that is, prove that $\forall\alpha\in[(dF_p)^{-1}(\beta)]$, $dF_p(\alpha)\in[\beta]$. By the same argument in the previous step, you can prove that
$$[dF_p(\alpha)]=[dF_p((dF_p)^{-1}(\beta))]=[\beta],$$
which implies $dF_p(\alpha)\in[\beta]$. The differential of $F$ at $p$ indeed preserves orientation of bases.
Epilog. To address the uniqueness problem, one can observe that if $T_p M$ is instead oriented by the equivalence class of $\alpha\notin[(dF_p)^{-1}(\beta)]$, $dF_p$ will become orientation-reversing.
