Consider the map $f: M \to N$, transversal to the regular submanifold $S \subset N$. I.E. $df (T_x M) \oplus T_{f(x)} S = T_{f(x)} N$. We know that $f^{-1} S$ is a regular submanifold. Also we know this equality as vector spaces \begin{equation} T_{p}(f^{-1} S) \cong Df_p^{-1}(T_{f(p)} S) \end{equation}
Let now $M$, $N$ be oriented manifolds $S \subset N$ be oriented regular submanifold, all of them connected.
Show that $F^{-1}(\textit{S}) \, $ is oriented.
I don't understand the following solution sketch:
For $x \in f^{-1}(S)$ exist a subspace $H \subset T_xM$ such that it's a complement to $T_x(f^{-1}(S))$ and the sum is now direct:
$$T_x(f^{-1}(S))+H=T_x(M) \quad T_{f(x)}S+df_x(H)=T_{f(x)}N $$
- Orient $f^{-1}(S)$ point by point.
- Show that orientation is smooth, using that $f^{-1}(S)$ is a regular value.
So, I don't understand how I can orient $f^{-1}(S)$ pointwise. I tried the following: i know that the codimension of $f^{-1} S$ is the same as codimension of $S$ and so we see that $H$ should be isomorphic to $df_x H$ and we can induce orientation on $H$ having orientation on $df_x H$, which we have because $N$ and $S$ are oriented. But I am not sure that we can do it.
Also I don't understand how we use the smoothness argument to derive that the orientation itself will be smooth.
Another idea of the proof was to take the volume form $\omega_M$ on $M$ and consider the pullback, and convolve locally with the fields transversal to $f^{-1}S$, because they will naturally go to the transversal fields to $S$ after the pushforward. now if we see that $\omega_M$ convolved with the pushforwarded fields is the volume form on $S$ we somehow can derive, that this convolved pulledback form is the volume form on $f^{-1}S$.
I would like to have some clear reasoning, that may or may not follow the argument or some useful notes. Thanks!
Also somehow related question on normal bundles and induced orientation. https://math.stackexchange.com/a/3800442/477927