# Is always possible say that $\langle df(\frac{\partial\varphi}{\partial u})\wedge df(\frac{\partial\varphi}{\partial v}) ,N_2\rangle>0$

Let $S_2$ be an orientable regular surface and $f:S_1\to S_2$ be a differentiable map which is a local diffeomorphism at every $p\in S_1$. Prove that $S_1$ is orientable.

My attemp: Suppose that $S_2$ is oriented by $N_2$ (since all orientable surface has a vector field of unit vectors N, we can assume that $S_2$ is oriented by $N_2$), and let $\{\varphi_{\lambda}:U_\lambda\to S_1\}_{\lambda\in\Gamma}$ a family of coordinate neighborhood of $S_1$.

Since $f$ is a local diffeomorphism, then $df_p$ is injective, but in particular $T_{f(p)}(S_2)=df_p(T_p S_1)$, and if is necessary we interchange $u$ and $v$, we can assume $$\left\langle df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial u}\right)\wedge df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial v}\right) ,N_2\right\rangle_{f(p)}>0\qquad (*)$$

Now, such $\{\varphi_{\lambda}\}_{\lambda\in\Gamma}$ is an orientation of $S_1$.

Suppose that $\varphi_{\alpha},\varphi_{\beta}\in\{\varphi_{\lambda}\}_{\lambda\in\Gamma}$ and such that $\varphi_{\alpha}(r,s)\cap\varphi_{\alpha}(u,v)\neq0$. Now, for the change of cordinates we have $$(\varphi^{-1}_{\alpha}\circ\varphi_{\beta})(r,s)=(u(r,s),v(r,s))\implies \varphi_{\beta}(r,s)=\varphi_{\alpha}((u(r,s),v(r,s)))$$ Then, $$\dfrac{\partial\varphi_{\beta}}{\partial r}=\dfrac{\partial\varphi_{\alpha}}{\partial u}\dfrac{\partial u}{\partial r}+\dfrac{\partial\varphi_{\alpha}}{\partial v}\dfrac{\partial v}{\partial r}$$

$$\dfrac{\partial\varphi_{\beta}}{\partial s}=\dfrac{\partial\varphi_{\alpha}}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial\varphi_{\alpha}}{\partial v}\dfrac{\partial v}{\partial s}$$

So, $df\left(\dfrac{\partial\varphi_{\beta}}{\partial r}\right)\wedge df\left(\dfrac{\partial\varphi_{\beta}}{\partial s}\right)=\dfrac{\partial(u,v)}{\partial(r,s)}df\left(\dfrac{\partial\varphi_{\alpha}}{\partial u}\right)\wedge df\left(\dfrac{\partial\varphi_{\alpha}}{\partial v}\right)$. Therefore,

$$0<\left\langle df\left(\dfrac{\partial\varphi_{\beta}}{\partial r}\right)\wedge df\left(\dfrac{\partial\varphi_{\beta}}{\partial s}\right) ,N_2\right\rangle=\dfrac{\partial(u,v)}{\partial(r,s)}\left\langle df\left(\dfrac{\partial\varphi_{\alpha}}{\partial u}\right)\wedge df\left(\dfrac{\partial\varphi_{\alpha}}{\partial v}\right),N_2\right\rangle$$ Which implies that $\dfrac{\partial(u,v)}{\partial(r,s)}>0$ and a regular surface is say "orientable" if is possible cover it with a family of coordinates neighborhood such that if a point belongs two differents coordinates family then, the change of coordinates has positive jacobian in such point.

The solution of this problem looks fine to me, but I have the little problem that is, if is always possible say that $\left\langle df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial u}\right)\wedge df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial v}\right) ,N_2\right\rangle_{f(p)}>0$.

My Attemp: If I have a basis for $T_{p}(S_1)$, $\{\varphi_u,\varphi_v\}$, and $df_p:T_p(S_1)\to T_{f(p)}(S_2)$ is a injective linear map, then $\{df_p(\varphi_u),df_p(\varphi_v)\}$ is linearly independent in $T_{f(p)}(S_2)$, i.e., a basis for $T_{f(p)}(S_2)$. Now, we can define a positive basis if $$\left\langle df_p(\varphi_u)\wedge df_p(\varphi_v), N_2 \right\rangle>0$$ Because a orientation $N_2$ on $S_2$ induces an orientation on each tangent space $T_{f(p)}(S_2)$. This is right?. Thanks!

• I don't know what you mean by "oriented by N," but the fact that $f$ is a local diffeomorphism tells you that $df_p$ maps $T_pS_1$ isomorphically to $T_{f(p)}S_2$. – Ted Shifrin Oct 21 '16 at 2:13
• @TedShifrin Hi! thanks for your answer, this is exactly what I wanna prove. If $f$ is a local diffeomorphism then $df_p$ mas $T_pS_1$ isomorphically to $T_{f(p)}S_2$, thanks – MathUser Oct 21 '16 at 3:43
• A local diffeomorphism $f$ has a local inverse $g$, so write down the chain rule (for both $g\circ f$ and $f\circ g$) and see that $df_p$ is an invertible linear map onto its image. (Clearly, $df_p(T_pS_1\subset T_{f(p)}S_2$, but you'll show they have the same dimension. Indeed, you only need $df_p$ maps onto, so you'll need only one of the equations you get from the chain rule.) – Ted Shifrin Oct 21 '16 at 3:47
• @TedShifrin I wrote that mean "Oriented by $N_2$", You believe that the answer is right? – MathUser Oct 24 '16 at 3:01
• OK. Then it seems fine to me :) – Ted Shifrin Oct 24 '16 at 14:43

Orientability can be defined in various ways. I'll stick to the following (alluded to in your question): An abstract two-dimensional surface $S$ is orientable if we can select an atlas for $S$ such that all transition maps have positive Jacobian wherever defined.
We are given an orientable surface $S_2$ (maybe embedded in ${\mathbb R}^3$). Let $\bigl(V_\beta,\psi_\beta\bigr)_{\beta\in J}$ be an atlas for $S_2$ with patches $V_\beta\subset S_2$ and coordinate functions $\psi_\beta:\>V_\beta\to{\mathbb R}^2$, such that all transition maps $\psi_{\beta'}\circ\psi_\beta^{-1}$ have positive Jacobian wherever defined.
In addition we are given a surface $S_1$ and a differentiable map $f:\>S_1\to S_2$ which is locally a diffeomorphism. We then may choose an atlas $\bigl(U_\alpha,\phi_\alpha\bigr)_{\alpha\in I}$ for $S_1$, such that for each $\alpha\in I$ the restriction $f\restriction U_\alpha$ maps $U_\alpha$ diffeomorphically into $V_\beta$ for a suitably chosen $\beta=\beta_\alpha$. The map $$f_{\alpha\beta}:=\psi_\beta\circ( f\restriction U_\alpha)\circ \phi_\alpha^{-1}: \qquad(x,y)\mapsto(u,v)$$ has nonzero Jacobian throughout, and after renumbering the output variables $(x,y)$ of $\phi_\alpha$ (if necessary) we may assume that this Jacobian is positive. Note that our assumption on the $\psi_\beta$ guarantees that the resulting numbering is independent of the $\beta$ we have chosen for this particular $\alpha$.
When the necessary renumberings have been performed all transition maps $\phi_{\alpha'}\circ\phi_\alpha^{-1}$, being compositions of ${\mathbb R}^2\to{\mathbb R}^2$ maps with positive Jacobian, have positive Jacobian as well.
• Thanks for your answer, but I think, if I prove that $(*)$ is always true, then my answer will be correct, can give me some hint to show this, thanks! – MathUser Oct 25 '16 at 19:30