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How to prove the following theorem (or where the detailed proof can be seen, since it isn't proved in lectures I've attended):

Two basis of tangent space of manifold have same orientation if matrix of transition map from one base to another has positive determinant. This defines relation of equivalence among basis of tangent space, with two classes of orientation. Choice of class of orientation is continuous in point $p$ if exists neighborhood of that point $U_p$ and locally defined movable frame $X_1,\dots X_n$ in neighborhood $U_p$, such that is for every $q\in U_p$ one represent of chosen equivalence class $X_{1q}\dots X_{nq}$.

Show that this definition is correct and prove that manifold is orientable iff exists choice of class of orientation on manifold which depends continuously of points on manifold.

Thanks in advance.

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    $\begingroup$ Do you define orientability in terms of the existence of a nowhere-zero top form, or in some other way? If you do define it this way, you can show that if $w_p(X_1,X_2,..,X_n) \neq 0$ , then $w_q(T(X_1),..,T(X_n)) \neq 0$ , forand $T$ is a linear map with non-zero determinant any $p,q$. $\endgroup$ – user99680 Nov 24 '13 at 21:15
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If you define orientability in terms of the existence of a nowhere-zero top $n$ -form $w$ , then you can show (as a result of multilinear algebra), that if $w_p(X_p1,X_p2,..,X_pn) \neq 0$ (where $X_p1, X_p2,...,X_pn$) , and $T$ is a linear map between $T_p , T_q $ with $Det (T) \neq 0$ , then $w_q(T(X_q1),..,T(X_qn)) \neq 0$ , for any $p,q$ in the manifold. Then the existence of this linear map $T$ between tangent spaces at any two different points guarantees the existence of a nowhere-zero top form.

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