Equivalence between orientation of the tangent bundle and orientation of manifolds

Question

If $M^{n}$ is a manifold then the following statement are equivalent.

1. The tangent bundle $(TM,\pi,M)$ is an orientable $n$-dimensional vector bundle.
2. $M$ has an $\lbrace (U,h)\rbrace$ atlas on $M$ such that $\det(D(g\circ h^{-1})(\bar{x}))>0$ for any charts $(U,h)$ and $(V,g)$ with $\bar x\in h(U\cap V)$

2) $\Rightarrow$ 1) is simple because $\lbrace (U,Id_{M}\times Dh)\rbrace$ is an atlas of the tangent bundle vector such that transition function have positive determinant , but how to find an atlas for $M$ if I have an atlas of the vector bundle $TM$ with this property?

Answer 1

Take any vector bundle $E\to M$. An orientation of $M$ is given by a cover of $M$ with open sets $V_i$ and on each $V_i$ a frame $(e_{i,1}, \ldots, e_{i, n})$ so that the on $V_i \cap V_j$ the linear change of coordinates between $(e_{i,1}, \ldots, e_{i, n})$ and $(e_{j,1}, \ldots, e_{j, n})$ has positive determinant.

Consider an atlas of $M$ as a manifold with charts $(U_{\alpha}, \phi_{\alpha})$. Let $p$ a point in $U_{\alpha}$. Consider the sign of the determinant of the linear change of coordinates from $(\frac{\partial}{\partial x_{\alpha,1}},\ldots \frac{\partial}{\partial x_{\alpha,n}}) $ to $(e_{i,1}, \ldots, e_{i, n})$ where $i$ is so that $V_i$ contains $p$. This sign does not depend on $i$ (since the changes from $i$ to $j$ have positive $\det$). Moreover, it is continuous with $p$. Therefore, it is constant on each component of $U_{\alpha}$. We can change the atlas to another one with same $U_{\alpha}$ but $\phi_{\alpha}$ is flipped accordingly to make the sign positive ( flip two coordinates to change sign from $-1$ to $+1$, do that on each component of $U_{\alpha}$ that requires it).

It is easy to see that we have now an atlas so that the signs of the $\det$ of the linear changes from $(\frac{\partial}{\partial x_{\alpha,1}},\ldots \frac{\partial}{\partial x_{\alpha,n}}) $ to $(\frac{\partial}{\partial x_{\beta,1}},\ldots \frac{\partial}{\partial x_{\beta,n}}) $ is $+1$. But the linear change of coordinates is the jacobian matrix. Done.