Given a differentiable map $f \colon S_1 \to S_2$ between regular surfaces such that $df_p$ is an isomorphism for each $p \in S_1$, I want to study whether the following statements are true or false:
- If $S_2$ is orientable, then so is $S_1$.
- If $S_1$ is orientable, then so is $S_2$.
I've tried to reason the first one as it follows: first of all, by the inverse function theorem we have that $f$ is locally a diffeomorphism. Given a family of charts $\{(U_i,\varphi_i)\}_{i \in I}$ that cover $S_2$, then it's easy to show that $\{(V_i,f^{-1}\circ\varphi_i)\}_{i \in I}$ is a family of charts that cover $S_1$, where $V_i \in U_i$ is an open set such that $f|_{\varphi(U_i)}$ is a diffeomorphism. Therefore, if $(V_1,f^{-1} \circ \phi_1)$ and $(V_2,f^{-1} \circ \phi_2)$ are two charts with non-empty intersection, then the cange of charts is $$(f^{-1}\circ \varphi_2)^{-1} \circ (f^{-1} \circ \varphi_1) = \varphi_2^{-1} \circ \varphi_1$$ and from here it's immediate that $S_1$ is orientable.
My problem is that I'm not convinced of the existence of the open sets $V_i$. In addition, if this proof was correct, would a similar argument prove the second statement by simply changing $f^{-1} \circ \varphi$ by $f \circ \varphi$?