# $F$ 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.

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.

• Where did you use that $M$ is connected? – Mancala Jun 8 '18 at 23:30
• @Mancala In the first paragraph. The fact that $M$ is connected means that $F^*\omega$ determines the same (or opposite) orientation on $M$ wrt to the original. – Stefan4024 Jun 8 '18 at 23:37
• @Mancala If $M$ has, let's say $2$ parts, then it's possible for $F^*\omega$ to match the original orientation on one component, while not on the other. – Stefan4024 Jun 8 '18 at 23:40
• @Mancala I've elaborated my answer a bit more. Hopefully it will help you. – Stefan4024 Jun 8 '18 at 23:48
• @ Stefan4024 thanks! – Mancala Jun 9 '18 at 20:37

$(\Rightarrow)$ 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?

• Assuming that $F^*\omega =\eta \text{ or } F^*\omega=-\eta$ is wrong, as the forms can determine the same orientation, but they are not necessarily the same. But you can say that the orientation determined by $F^*\omega$ is same as the one determined by $\eta$ or $-\eta$. – Stefan4024 Jun 10 '18 at 10:34
• @Stefan4024 Yes, I forgot to put the correction factor $f>0$! – Mancala Jun 10 '18 at 13:08