Orientation under local diffeomorphism

Question

Given regular surfaces $S_1$ and $S_2$ such that $S_2$ is orientable and a local diffeomorphism $f: S_1 \rightarrow S_2$, then why is $S_1$ orientable? What I think that can be done is to choose an orientable atlas $\{ V_{\alpha} \}$ in $S_2$ and, for each $p \in U \subset S_1$ such that $f|_U$ is a diffeomorphism onto $f(U)$, choose an open cover $\{U_{\alpha}\}$ for $S_1$ where each $U_{\alpha}$ is $(f|_U)^{-1}(V_{\alpha} \cap f(U))$ and use this cover to build and orientable atlas for $S_1$. Is this possible, or there is some another construction?

Answer 1

To orient a manifold, you must know when a base of a tangent space (frame) is positive, and this choice must be continuous. In your case, first choose an orientation of $S_2$, and say that a base $B_p=(u_1,...u_n)$ of vector in $T_pS_1$ is positive if its image $Tf_p(B)$ is positive also (relative to the given orientation of $S_2$. To ckeck continutity, extend this vectors in a neighborhood of $p$ by some vector fiels $U_1,...U_n$. Note that $\det (Tf_pu_1,..,T_pfu_n) >0$ implies that the inequality $\det (Tf U_1,..,Tf U_n) >0$is still valid in the neighborhood of $p$. Here $\det$ means the choice of a determinant in a chart around $f(p)$ positive on an oriented base.