Let M and N be oriented manifold and let $f:M\to N$ be a smooth map between them. Suppose $y \in N$ is a regular value for $f$, how can we show that $f^{-1}(y)$ is orientable?
I've seen a solution for the case $N = \mathbb{R}$, but I fail in generalizing that proof.
Since somebody upvoted this old question of mine maybe it's worthy to put a quick answer.
Let $Z= f^{-1}(y)$. At a point $z\in Z$ the tangent bundle of $M$ splits $T_zM = T_z Z \oplus \nu_z Z$ where $\nu_z Z$ is the normal bundle of $Z$ (we can use a Riemannian metric for simplicity and consider the orthogonal of $T_z Z$).
Consider the differential $d_z f: T_z Z \oplus \nu_z Z \to T_y N $. Since $y$ is regular $d_z f|_{\nu_z Z}^{T_y N}$ is an isomorphism ($\ker d_z f = T_z Z$). Consequently the orientation of $T_y N$ induces an orientation of $\nu_z Z$.
We define can now define the orientation of $Z$ as the unique orientation of $T_z Z$ such that $$T_z M \simeq T_z Z \oplus \nu_z Z$$ as oriented vector spaces.