# Surface orientation

Let $S_{1}$ and $S_{2}$ be two oriented surfaces ($N_{1}$ and $N_{2}$ their normal fields, respectively). We say that a local diffeomorphism $f$ : $S_{1}$$\rightarrow$$S_{2}$ preserves orientation if its differential takes, for each $p$ $\in$ $S_{1}$, positively-oriented basis on the tangent plane $T_{p}S_{1}$ onto positive-oriented basis on $T_{p}S_{2}$ (meaning that $det(a,b,N(p))$ $>$ $0$ implies $det(df_{p}(a),df_{p}(b),N(p))$ $>$ $0$ for {$a$,$b$} basis of $T_{p}S_{1}$). Also, define the Jacobian at a point $p$ $\in$ $S_{1}$ to be:

($Jac$ $f$)($p$) = $det$$(df_{p}(e_{1}), df_{p}(e_{2}), N_2(f(p)))$, for a positively-orientated orthonormal basis {$e_{1}$,$e_{2}$} on $T_{p}S_{1}$.

Problem If $S_{1}$ and $S_{2}$ are connected, show that $f$ preserves orientation if and only if $Jac$ $f$ is positive everywhere.

Almost complete proof

Suppose $f$ preserves orientation. This way, for any positively-oriented basis {$e_{1}$,$e_{2}$} on $T_{p}S_{1}$ we have $Jac$ $f$ $>$ $0$. Conversely, suppose $Jac$ $f$ $>$ $0$ everywhere. As the Jacobian is independent of the choice of a basis, we have that, for every positively-oriented basis on {$u_{1}$,$u_{2}$} on $T_{p}S_{1}$, $Jac$ $f$ $>$ $0$. What am I missing? I mean, there is still the hypotesis of connectedness to be used.

Are you requiring your bases to be orthonormal? If not then your $Jac$ is not independent of the positive-oriented basis chosen; but its sign is, so that's not really an issue. It does seem that as written this problem is trivial and the connectedness is not necessary. –  Anthony Carapetis Sep 27 '13 at 21:57