I need to prove the following result: Let $f: M \rightarrow N$ be a smooth function, if $M, N$ are orientable manifolds and $c$ is a regular value of $f$ then $f^{-1} (c)$ is a orientable manifold.
Using the implicit function theorem I was able to show that $f^{-1} (c)$ is a manifold.
Indeed, let $p$ be a point in $f^{-1} (c) $. Consider the coordinates $\phi: U_1 \rightarrow M \cap V_1 $ and $\psi: U_2 \rightarrow N \cap V_2$, such that $p$ $\in$ $V_1$ and $c$ $\in$ $V_2$.
Since by hipoteses $\psi^{-1} (c)$ is a regular value of $\psi^{-1} \circ f \circ \phi$ we can reorder coordinates and find, by implicit function theorem, a function $\varepsilon: V_{0} \rightarrow U_0$ such that $\psi^{-1} \circ f \circ \phi ( x , \varepsilon (x)) = \psi^{-1}(c)$.
So we can define the local coordinate $(\Phi , V_0)$ as $\Phi(x) = \phi(x,\varepsilon(x))$. Showing that $f^{-1}(c)$ is a manifold. My problem is prove that this manifold is orientable.
Can anyone help me?