Prove that a surface is orientable.

Let S2S_2 be an orientable regular surface and f:S1→S2f:S_1\to S_2 be a differentiable map which is a local diffeomorphism at every p∈S1p\in S_1. Prove that S1S_1 is orientable

My attemp: Suppose that S2S_2 is oriented by N2N_2, and let {φλ:Uλ→S1}λ∈Γ\{\varphi_{\lambda}:U_\lambda\to S_1\}_{\lambda\in\Gamma} a family of coordinate neighborhood of S1S_1.

Since ff is a local diffeomorphism, then dfpdf_p is injective, but in particular Tf(p)(S2)=dfp(TpS1)T_{f(p)}(S_2)=df_p(T_p S_1), and if is necessary we interchange uu and vv, we can assume ⟨dfp(∂φλ∂u)∧dfp(∂φλ∂v),N2⟩f(p)>0\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

Now, such {φλ}λ∈Γ\{\varphi_{\lambda}\}_{\lambda\in\Gamma} is an orientation of S1S_1.

Suppose that φα,φβ∈{φλ}λ∈Γ\varphi_{\alpha},\varphi_{\beta}\in\{\varphi_{\lambda}\}_{\lambda\in\Gamma} and such that φα(r,s)∩φα(u,v)≠0\varphi_{\alpha}(r,s)\cap\varphi_{\alpha}(u,v)\neq0. Now, for the change of cordinates we have (φ−1α∘φβ)(r,s)=(u(r,s),v(r,s))⟹φβ(r,s)=φα((u(r,s),v(r,s)))(\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, ∂φβ∂r=∂φα∂u∂u∂r+∂φα∂v∂v∂r\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}

∂φβ∂s=∂φα∂u∂u∂s+∂φα∂v∂v∂s\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(∂φβ∂r)∧df(∂φβ∂s)=∂(u,v)∂(r,s)df(∂φα∂u)∧df(∂φα∂v)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\rangle0<\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\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.

This proof is correct??; Because, my teacher says the answer is wrong, since the answer is incomplete and more things are missing, can give me some help pls, thanks!

Since all orientable surface has a vector field of unit vectors N, we can assume that S_2S_2 is oriented by N_2N_2.

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