How a local diffeomorphism preserves orientation

[catellanos]

Hi, I'm currently reading Manfredo do Carmo's "Differential Geometry of curves and surfaces" and I'm stocked with problem 2 of section 2-6 which is "Let $S_2$ be an orientable regular surface and $\varphi:S_1 \rightarrow S_2$ be a differentiable map which is a local diffeomorphism at every $p \in S_1$.Prove that $S_1$ is orientable".

 

[lippyka]

The proof of this problem is based on the fact that a local diffeomorphism preserves orientability. Since \varphi:S_1 \rightarrow S_2 is a local diffeomorphism, it will also preserve orientability. That is, if S_2 is orientable, then S_1 must also be orientable.To prove this, let's assume that S_2 is orientable and let \theta be an orientation on S_2. Then, since \varphi is a local diffeomorphism, it will have a derivative at each point p \in S_1. This means that \varphi'(p) is an invertible linear map. Now, since \varphi'(p) is invertible, it has a non-zero determinant. This implies that \varphi'(p) preserves orientation. That is, if \theta is an orientation on S_2, then \varphi'(p)\theta is an orientation on S_1. Since \varphi'(p) preserves orientation, it follows that S_1 must be orientable. Thus, we have shown that if S_2 is orientable, then S_1 must also be orientable.