6
$\begingroup$

Let $f:X\rightarrow Y$ be a diffeomorphism between connected oriented manifolds. $f$ is orientation-preserving at $p\in X$ if the induced map $df_{p}:T_{p}X\rightarrow T_{f(p)}Y$ is orientation-preserving; similarly $f$ is orientation-reversing at $p$ if the derivative is orientation-reversing. Why must $f$ be either orientation-preserving everywhere or orientation-reversing everywhere?

I think it is true that the sets of points where $p$ is orientation-preserving and orientation-reversing are both open (which implies the result), but I can't prove this.

  • 1
    Do you need to proof "A connected orientable manifold has exactly two orientation"?2012-05-01
  • 3
    The intuition is pretty clear: to say that $df_p$ is orientation-preserving is to say that its "determinant" is positive and so it must be positive in a neighborhood of $p$ by continuity. But of course this needs to be made precise and I am not sure what your definition of orientation is so...2012-05-01
  • 0
    An orientation of an $n$-dimensional vector space $V$ is a partition of the 1-dimensional space $\Lambda^{n}(V^{\times})$ in to of 'positive' and 'negative' vectors, and $f$ is orientation preserving at $p$ if under the map $(df_{p})^{*}$ positive vectors are mapped to positive vectors.2012-05-01

1 Answers 1