TITLE: Smooth atlas oriented iff transition maps have positive Jacobian determinant
QUESTION [1 upvotes]: First of all, my definition: An orientation of an $m$-dimensional smooth (connected) manifold $M$ is a choice of orientations for all tangent spaces $T_p M$, $p \in M$, such that there exists an atlas $\{(U_\alpha, h_\alpha)\}_{\alpha \in A}$ such that the maps
$$T_p h_\alpha : T_pM \to T_{h_\alpha(p)} \mathbb R^m \cong \mathbb R^m$$
are orientation preserving for all $p \in M$ and all $\alpha \in A$.  
I want to show that an atlas is oriented iff for all tranisition maps $h_{\alpha \beta} := h_\beta \circ h_\alpha^{-1}$ the Jacobian matrix $D h_{\alpha \beta} = D (h_\beta \circ h_\alpha^{-1})$ has positive determinant, i.e. $\det D h_{\alpha\beta} > 0$.  
I have struggles with both directions. If I assume that my atlas is (positive) oriented, then for every tangent space the change of basis has positive determinant, by definition. How does that imply that $\det D h_{\alpha\beta} > 0$? Similarly, if all transition maps have a positive Jacobian determinant, how do I get to $\det T_ph_\alpha$ for all $p$ and $\alpha$?  
I'd be very happy for some tips!

REPLY [1 votes]: The maps $T_p h_{\alpha}$ are precisely the differentials of $h_{\alpha}: U_{\alpha} \to h_{\alpha}(U_{\alpha})$. Now what does it mean for this map to be smooth on $U_{\alpha} \subset M$? Precisely that, $h_{\alpha} \circ h_{\beta}^{-1}:h_{\beta}(U_{\beta}) \subset \mathbb{R}^m \to \mathbb{R}^m$ is smooth where $U_{\beta} \subset U_{\alpha}$. You just needed to recall that in order to differentiate, we need coordinates for the domain. To get these coordinates we use the inverse of a local chart. 

This is an example of what it means for $f: M \to \mathbb{R}$ to be smooth. Although never said, in calculus we've always used this association of $f$ with $f \circ \phi^{-1}$, but it was hidden due to the fact that we used the identity map for our local chart.
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