TITLE: Submanifolds of oriented manifolds
QUESTION [7 upvotes]: Let $M$ be an $n$-dimensional oriented manifold. Let $f:M\to\mathbb{R}$ be a smooth function. Suppose $c$ is a regular value of $f$ with $f^{-1}(c)$ nonempty. Show that $f^{-1}(c)$ is an oriented regular submanifold of $M$.
By constant rank thoerem, I know that $f^{-1}(c)$ is an $(n-1)$-dimensional regular submafold of $M$, but how to show that it is orientable?

REPLY [3 votes]: Suppose $S$ is the manifold $f^{-1}(c)$. Prove that $\triangledown f$ is normal to $S$ in $M$. Now if $S$ is non-orientable then there exists a continuous  choice of collection of ordered basis at each point of $S$. Attach $\triangledown f$ with it. This will show that $M$ is non-orientable, a contradiction.