A question about orientation on a Manifold Let $(U,x_1,x_2,\ldots , x_n)$ be a chart for a orientable manifold $M$, why $(U,-x_1,x_2,\ldots , x_n)$ is a chart for the Manifold $-M$, the same manifold with reversed orientation?
 A: Let's be a little careful about terminology.  "Orient-able" means that it is possible to choose an orientation, whereas "orient-ed" means that you have gone ahead and specified an orientation.
So let's say we've specified an orientation.  To answer your question, we need to know what that means.  I'll go ahead and mention three (equivalent) definitions of an "orientation" on a smooth orientable manifold $M$ and leave making definitions precise and exploring the relationships up to you.


*

*A smooth (maximal) atlas for $M$ with the property that transition maps all have positive determinant.

*A choice of component of $\Lambda^nM - \{\mbox{zero section}\}$.

*A "consistent" choice of an equivalence class of ordered bases for the tangent space at each point of $M$.


In this case, it's easiest for us to work with #1.  In the atlas for the orientable, unoriented manifold $M$, we can divide our charts up into exactly two equivalence classes, where $U\sim V$ when the transition map $\theta_{UV}$ has positive determinant.  (Why doesn't it matter that $\theta_{UV}$ varies point-by-point?)  To orient $M$, we pick one equivalence class to create a new smooth atlas, and then we forget about the other one.  (It will take a little work to show that this is indeed a smooth atlas.)
Ah-hah!  Now it's clear why if $U,(x_1,x_2,\ldots,x_n)$ is a chart for the oriented manifold $M$, $U',(-x_1,x_2,\ldots,x_n)$ is a chart for the oriented manifold $-M$.  The transition map between $U$ and $U'$ has negative determinant, so $U$ and $U'$ are not in the same equivalence class.  Since $U$ was in the atlas for $M$, $U'$ is in the atlas for $-M.$
(For a little more haranguing about orient-ed vs. orient-able, check out my answer to this question.)
