Local Preservation of Orientation Implies Global Preservation I am trying to work through some problems in differential topology, and I came across one that I can't figure out, mostly because I'm having trouble understanding some of the definitions.
Let $f \colon M \to N$ be a diffeomorphism of connected oriented smooth ($C^\infty$) manifolds. If $df_x \colon T_xM \to T_{f(x)}N$ preserves orientation at a point $x \in M$, prove that $f$ preserves orientation globally.
I have a limited knowledge of what orientation means. I understand what it means for vector spaces, and the definition that I learned for a smooth manifold is that every transition function has positive determinant. An orientation-preserving map between two smooth manifolds is one with positive determinant (in local coordinates, of course). I assume then that this question is really asking to prove that if $f$ preserves orientation at one point, then it does on all of $M$. Even so, I don't know how to prove that. I have seen this question a few different times on this website, yet never found an answer that was simple enough for me to understand at this point; everyone seemed to be using differential forms. Could somebody help me out? Thanks!
 A: First you should look for the Euclidian case, supose that $U$ and $V$ are open connected sets on $\mathbb{R}^n$ and $f:U \to V$ is diffeo. So by definition $f$ preserve orientation in $x$, if the linear transformation $df_x: \mathbb{R}^n \to \mathbb{R}^n$ preserve orientation, or in a equialent way $\mathrm{det}(df_x)>0$. Now lets proof that if exists $x_0 \in U$ s.t. $f$ preserve orientation on $x_0$ then $f$ preserve orientation in every point in $U$, but as you should know the map $d: U \ni x \mapsto \mathrm{det}(df_x) \in \mathbb{R}$ is smooth, now note that because $f$ is a diffeo. the map $d$ is never zero, so if exists $y \in U$ s.t $d(y)<0$ the facts that $d(x_0)>0$ and $U$ is connected implies that should exists $y' \in U$ s.t. $d(y')=0$, absurd so $f$ preserve orientaion on every point.
Now for the general case $f:M \to N$. Define the set:
$$A=\{x \in M : \mbox{$f$ preserve orientation on x}\},$$
we will prove that $A$ is closed, open and non-empty, so $A=M$. First $A$ is non-empty by hypothesis. Now lets proof that is open, let $x_0 \in A$, so by definition exists charts $\phi:U\subset M \to U' \subset \mathbb{R^n}$ and $\psi:V\subset N \to V' \subset \mathbb{R}^n$ s.t the map $\bar{f}: U' \to V'$, given by $ \bar{f}=\psi^{-1} \circ f \circ \phi$, satisfies $\mathrm{det}(d\bar{f}_{\phi(x_0)})>0$. So by the Euclidian case exists a connected nbhd $W$ of $\phi(x_0)$ s.t $\bar{f}$ preserve orientation on $W$, so by definition $f$ preserve orientation for every point in $\phi^{-1}(W)$. By definition of $A$ we have that $\phi^{-1}(W) \subset A$, that is, $A$ is open as claimed. The proof that $A$ is closed is pretty similar.
