Suppose $f:X\to Y$ is a diffeomorphism of connected oriented manifolds with boundary. Show that if $df_x: T_x(X)\to T_{f(x)}(Y)$ preserves orientation at one point, then $f$ preserves orientation globally.

What I can see is that since $f$ is a diffeomorphism, each $df_x$ is an isomorphism. But the concept of orientation preserving linear map remains not so clear to me (see this question). Is the fact that $df_x$ preserves orientation amounts to saying that $\det df_x > 0$? After that, am I supposed to use some kind of continuity of $x\mapsto df_x$?

  • $\begingroup$ «$\det df_x$» makes no sense, because $df_x$ is not an endomorphism of a vector space but just a linear map between two different vector spaces: such a thing does not have a determinant. $\endgroup$ Apr 3 '18 at 4:22
  • $\begingroup$ Notice that that is exactly the same problem that was signalled to you in the question you linked to! $\endgroup$ Apr 3 '18 at 4:23
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    $\begingroup$ Answering your deleted comment: because by choosing different bases you can get the determinant to be anything you want (except zero) $\endgroup$ Apr 3 '18 at 4:33
  • $\begingroup$ @MarianoSuárez-Álvarez Thanks for clarifications. (The reason I deleted my previous comment is that I realized that it didn't make sense.) So I guess I need to show that any positively oriented basis of $T_xX$ maps to a positively oriented basis of $T_{f(x)} Y$ for every $x\in X$ (where orientations are specified since $X,Y$ are oriented), right? If so, could you give at least some hints for this? I believe this should be obvious, but not at my level of understanding of orientations. $\endgroup$
    – user500094
    Apr 3 '18 at 4:41
  • $\begingroup$ Please do no use the word "obvious": everything is obvious once it is obvious and never before that, so claiming that something «should be obvious» is never helpful to anyone… In any case, consider the set of points $x$ of $X$ such that $df_x$ preserves the orientation, and show that it is open and closed. $\endgroup$ Apr 3 '18 at 4:45

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