Let $M$ be a smooth, connected manifold of dimension $n>1$, and $\mathscr O_p$ an orientation of $T_pM$. define $\widehat M=\{(p,\mathscr O_p)\}_{p\in M}$. Then, $\widehat M$ can be made into an oriented manifold. To do this, define the projection $\hat \pi:(p,\mathscr O_p)\mapsto p,$ let $\mathscr O$ be an orientation on $U\in \tau_M$ and declare the basis elements in $\widehat M$ to be $\widehat U_{\mathscr O}=\{(p,\mathscr O_p):p\in U\}.$ This assignment induces a topology on $\widehat M$.
No problems so far.
We have
$1).\ \widehat M$ is basically two copies of $M$, because there are two orentations for $T_pM$, which are in turn equivalence classes of bases for $T_pM:(X_i)\sim (Y_i)\Leftrightarrow $ the change of basis matrix beteween them has positive determinant.
$2).\ $ The orientation $\mathscr O$ on $U$ is induced by a chart $(\phi,U)$ on $M$.
$3).\hat \pi$ is a generalized covering map ($\widehat M$ need not be connected).
So far so good.
Now, paraphrasing Lee,
$4).\ $ Using $\hat \pi$ we can define a pointwise orientation on $T_{(p,\mathscr O_p)}\widehat M$: we have that $(\pi_*)_p:T_{(p,\mathscr O_p)}\widehat M\to T_pM$. We can define a pointwise orientation as the "unique orientation such that $(\pi_*)_p$ os orientation-preserving."
The orientation on $T_pM$ is either $\mathscr O_p$ or $-\mathscr O_p$. Is Lee saying that we pick one, compute the det of $\hat \pi_*$ and if it comes out positive, then we assign that as the orientation of $T_{(p,\mathscr O_p)}\widehat M$, and if not then we take the opposite orientation? I am confused.
$5).\ $ The orientation defined in $3).$ agrees with the pullback induced by $\hat \pi$ from $(U,\mathscr O)$ so the pointwise orientation is continuous.
I computed the pullback but have not been able to show that the orientation it induces agrees with the previous one.
Here is the paragraph from Lee's book that I am having trouble with:
I have seen this construction using top forms (Jeffrey Lee), and homology (Hatcher) both of which I understand (I think!), but I want to understand clearly what is going on in Lee's construction, because it is very basic which means that I am missing some fundamental ideas, which I want to get straight.
Best Answer
I think it is only a notational issue.
In my opinion it is misleading to write $(p,\mathcal O_p)$ for the points of $\hat M$ because that suggests that we have a function assigning to $p \in M$ an orientation $\mathcal O_p$ of $T_pM$. I would neutrally write $(p, \mathcal O)$, where $\mathcal O$ is one of the two orientations of $T_pM$.
Via the map $\hat \pi : \hat M \to M, \hat \pi( p, \mathcal O) = p$, we obtain a topology on $\hat M$ such that $\hat \pi$ becomes a covering projection with two sheets. This is due to the fact that $M$ is locally orientable. That is, if we take an orientation $\Omega_U = (\omega_p)_{p \in U}$ on some open $U \subset M$ (such $\Omega_U$ always exists if $U$ carries a chart), then the two sheets over $U$ are given by $U_\pm = U_\pm(\Omega_U) = \{ (p,\pm \omega_p) \mid p \in U \}$.
Moreover, we get a smooth atlas on $\hat M$ such that $\hat \pi$ is smooth.
Now give $T_{(p,\mathcal O)} \hat M$ the unique orientation $\omega_{(p,\mathcal O)}$ such that such $T_{(p,\mathcal O)} \hat \pi : (T_{(p,\mathcal O)} \hat M, \omega_{(p,\mathcal O)}) \to (T_p M, \mathcal O)$ is orientation preserving. Intuitively, the orientation on $T_{(p,\mathcal O)} \hat M$ is just the given orientation $\mathcal O$ at the point $(p,\mathcal O)$. This procedure has nothing to do with any local (or global) orientation on $M$.
Then one can show that $\hat \Omega = (\omega_{(p,\mathcal O)})_{(p,\mathcal O)}$ is an orientation on $\hat M$. It is defined canonically without making any choices.
Edited:
Let $\mathfrak A$ be the smooth subatlas on $M$ consisting of all charts $\phi : U \to V$ with connected $U$. Connectedness assures that $U$ has exactly two orientations. Let us call an open $U \subset M$ nice if it is occurs as the domain of a chart in $\mathfrak A$. For nice $U$, the set $(\hat \pi)^{-1}(U)$ can be decomposed uniquely as the disjoint union of two subsets $U_\pm = U_\pm(\Omega_U)$ as above. This decomposition does not depend on the choice of $\Omega_U$. In fact, we have $U_\pm(-\Omega_U) = U_\mp(\Omega_U)$ which yields the same two sets. Only the indexing by $\pm$ depends on $\Omega_U$. By construction of $U_\pm$, $T_{(p,\mathcal O)} \hat \pi : (T_{(p,\mathcal O)} \hat M, \omega_{(p,\mathcal O)}) \to (T_p M, \omega_p)$ is orientation preserving for $(p,\mathcal O) \in U_+(\Omega_U)$ and orientation reversing for $(p,\mathcal O) \in U_-(\Omega_U)$.
The set of all $U_+$ and all $U_-$ forms the basis of a topology on $\hat M$ and it is easy to see that $\hat \pi$ becomes a covering projection with two sheets. An atlas $\hat{\mathfrak A}$ on $\hat M$ is given by the maps $\phi_\pm = \phi \circ \hat \pi \mid_{U_\pm} : U_\pm \to V$. This atlas is clearly smooth. That is, $\hat M$ has been given the structure of a smooth manifold such that $\hat \pi$ becomes smooth.
The family $\hat \Omega = (\omega_{(p,\mathcal O)})_{(p,\mathcal O)}$ is by construction locally an orientation. In fact, it is an orientation on each of the sets $U_\pm(\Omega_U)$. Thus it is an orientation on $\hat M$. This is true for any smooth manifold $N$ and any family of orientations $(\omega_q)_{q \in N}$ of the tangent spaces $T_qN$ which is locally an orientation. Observe that the existence of such a (global) family does not follow from the obvious fact that each smooth manifold $N$ is locally orientable. Indeed, there is no reason to assume that we can paste a suitable set of local orientations to a global family of orientations $(\omega_q)_{q \in N}$ which restricts to all given local orientations.
Note also that $\hat{\mathfrak A}$ is not an oriented atlas. It only contains oriented atlases.