$F$ is local diffeomorphism $\Leftrightarrow$ $F$ preserves or reverses orientation. Let $M$ and $N$ manifold oriented with $M$ connected. Show that $F\colon M\rightarrow N$  is local diffeomorphism if and only if $F$ preserves or reverses orientation.
My initial idea is to use orientation forms  but I do not know how.
 A: Let $F$ be a local diffeomorphism and let $\omega$ be orientation-determining non-vanishing top form on $N$. Then we have that $F^*\omega$ is a non-vanishing form on $M$ and it determines an orientation on $M$. Now because $M$ is connected the orientation determined by $F^*\omega$ and the original one on $M$ must either match at all points or at none. In the first case the both orientations are consistent, while in the latter case the postively oriented bases wrt to $F^*\omega$ are negatively oriented wrt the initial orientation. WLOG we can assume that the first case is true in our calculations.
Now let $p \in M$ and let $f:U \to F(U)$ be a diffeomorphism on some open neighbourhood around $p$. Also $p$ is the domain of some local frame, as $M$ is an orientable manifold. WLOG we can take it to be $U$. (Indeed if the domain is $V$ we can consider the intersection $U \cap V$ and the restriction of $f$ on it).
Now let $\{E_1,\dots,E_m\}$ be a positively oriented local frame on $U$. As $f$ is diffeomorphism we have that $\{df(E_1),\dots,df(E_m)\}$ is a frame on $f(U)$. Then it's positively oriented if $\phi(df(E_1),\dots,df(E_m)) > 0$. This is true as:
$$\omega(df(E_1),\dots,df(E_m)) = f^*\omega(E_1,\dots,E_m) = F^*\omega(E_1,\dots,E_m)>0$$
As this is true for every $p$ in $M$ we have that $F$ preserves the orientation.
First of all for the other direction we must have that $dF$ is non-singular in order for the orientation preserving (reversing) to be defined. Then by the Inverse Function Theorem we have that $F$ is a local diffemoprhism.
A: $(\Rightarrow)$
Let $\omega$ an oriteted positively (or negatively) form of $N$. 
We will show that $F^*\omega$ is a form of positively or negatively oriented orientation of $M$. Since $M$ is orientable, then there exists a $\eta$ form of orientation of $M$.
Since $M$ is connected, then $M$ has only two orientations, namely $\eta$ or $- \eta $. Like this,
$$F^*\omega=f\eta \text{ or } F^*\omega=f(-\eta) \text{, with }f>0$$
Therefore, $F$ preserves or reverses the orientation.
Is correct?
