The problem statement:
Let $M$ be a connected nonorientable smooth manifold with or without boundary, and let $\widehat{\pi}: \widehat{M} \to M$ be its orientation covering. Prove that if $X$ is any oriented smooth manifold with or without boundary, and $F: X \to M$ is any local diffeomorphism, then there exists a unique orientation-preserving local diffeomorphism $\widehat{F}: X \to \widehat{M}$ such that $\widehat{\pi} \circ \widehat{F} = F$.
My attempt:
I defined the map as $\widehat{F}(p) = \left(F(p), \left[dF_p(E_1), …, dF_p(E_n)\right]\right)$ where $(E_1, …, E_n)$ is any oriented basis for $T_pX$. Since $F$ is a local diffeomorphism, $\left(dF_p(E_1), …, dF_p(E_n)\right)$ is a basis for $T_{F(p)}M$, and if $(E_1', …, E_n')$ is another oriented basis for $T_pX$ then the transition matrix relating $\left(dF_p(E_1'), …, dF_p(E_n')\right)$ to $\left(dF_p(E_1), …, dF_p(E_n)\right)$ is the same as the transition matrix relating $(E_1', …, E_n')$ to $(E_1, …, E_n)$ so it has positive determinant and the orientations determined by these bases are the same. So $\widehat{F}$ is well-defined.
It's clear from the construction that $\widehat{\pi} \circ \widehat{F} = F$.
I want to show that $\widehat{F}$ is a local diffeomorphism.
First pick a smooth chart $(U, \varphi)$ for $X$ containing $p$. Since $F$ is a local diffeomorphism, we may assume that $F\vert_U: U \to F(U)$ is a diffeomorphism onto an open subset $F(U)$ of $M$ (after possibly shrinking $U$). Shrinking $U$ again if necessary, we may also assume that $F(U)$ is evenly covered by $\widehat{\pi}$. Then $\left(F(U), \varphi \circ (F\vert_U)^{-1}\right)$ is a smooth chart for $M$ containing $F(p)$. Also if $V$ is the component of $\widehat{\pi}^{-1}(F(U))$ containing $\widehat{F}(p)$, then $\left(V, \varphi \circ (F\vert_U)^{-1} \circ \widehat{\pi}\vert_V\right)$ is a smooth chart for $\widehat{M}$ containing $\widehat{F}(p)$.
Since $F(U)$ is evenly covered, it must be connected so $U$ is also connected (consequence of the definition of evenly covered used in the book, or else we could just choose $U$ to be connected already). Then if we already know that $\widehat{F}$ is continuous, $\widehat{F}(U)$ is a connected subset of $\widehat{\pi}^{-1}\left(F(U)\right)$ containing $\widehat{F}(p)$ so $\widehat{F}(U) \subseteq V$.
The coordinate representation of $\widehat{F}$ with respect to the smooth charts defined above is $$\left(\varphi \circ (F\vert_U)^{-1} \circ \widehat{\pi}\vert_V\right)\circ\widehat{F}\circ\varphi^{-1} = \varphi \circ (F\vert_U)^{-1} \circ (\widehat{\pi}\circ\widehat{F})\vert_U\circ\varphi^{-1}$$ $$ = \varphi \circ (F\vert_U)^{-1} \circ F\vert_U\circ\varphi^{-1} = \text{id}_{\varphi(U)}$$
And this is enough to show that $\widehat{F}$ is a local diffeomorphism.
But I can't really prove that $\widehat{F}$ is continuous or to prove in any other way that $\widehat{F}(U) \subseteq V$. I'm probably missing something trivial or looking at it the wrong way.
Best Answer
We do not need to show that $\hat F$ is continuous to see that $\hat F(U) \subset V$.
Indeed, the orientation of $X$ induces an orientation on the open subset $F(U) \subset M$ such that the diffeomorphism $F \mid_U : U \to F(U)$ is orientation preserving. If $[E_1^p, \ldots, E_n^p]$ is the orientation of $T_p X$ with $p \in U$, then $[\tilde E_1^{F(p)}, ..., \tilde E_n^{F(p)}]$ with $\tilde E_i^{F(p)} = dF_p(E_i^p)$ is the induced orientation on $T_{F(p)} F(U) = T_{F(p)} M$.
It follows from the construction of the orientation covering that the set $W = \bigcup_{q \in F(U)} \left\{ (q, [\tilde E_1^{q}, ..., \tilde E_n^{q}]) \right\}$ is one of the two sheets over $F(U)$. But by definition of $\hat F$ we have
$\hat F (U) \subset W$
$W = V$ since $\hat F (p) = (q, [\tilde E_1^{F(p)}, ..., \tilde E_n^{F(p)}]) \in W$.