Even if a manifold $M$ is not orientable, does every point have an open orientable neighborhood?
This question comes up when I read a proof of Theorem 15.40, which shows the existence of orientation covering of a manifold in the J. M. Lee's book Introduction to Smooth Manifolds. In the proof, the author assumes that we can have an orientable open neighborhood for any point of the manifold.
By the definition of manifold, given atlas $\{(U,\phi)\}$ since $U$ is diffeomorphic to $\mathbb{R}^{k}$, maybe it gives some orientation, but I cannot show it rigorously. Could anyone help me?
I think we can simply use Lee's more basic definition. We only need to know that $p\in U$ for some chart $(U,x^i:1\le i\le n).$ Because in this case, we can give $U$ the pointwise orientation $\mathscr O$ determined by the vectors $\{\frac{\partial}{\partial x^i}\Big|_p:1\le i\le n,\ p\in U\}.$ Then by construction $\left(\frac{\partial}{\partial x^i}\Big|_p:1\le i\le n\right)$ is a positively oriented basis for $T_pU$ at every point $p\in U$ and by definition, $\mathscr O$ is continuous, and so again, by definition, $\mathscr O$ is an orientation on $U.$
I think this approach has the benefit of making it clear that problems with orientability can only arise from incompatibility of the charts, where the Jacobian of at least one transition function fails to be positive.