Sub-manifold with boundary

Question

Let $f:M\rightarrow \mathbb{R}$ be a smooth function ($M$ is a smooth manifold). Let $a$ be a regular value of $f$. Is it true that $f^{-1}(-\infty ,a]$ is a smooth manifold with boundary $f^{-1}\{a\}$? My feel is that it is correct and the proof is very similar to the proof that $f^{-1}\{a\}$ is a smooth manifold. Am I right?

Answer 1

Yes, that's correct. You only have to check a neighbourhood of $a$ resp the counterimage of $a$ for this, since $f^{-1}((-\infty, a))$ is open and open subsets of smooth manifolds are smooth manifolds.

In a neighbourhood of $a$, if you already do know how to show that $f^{-1}(a) $ is a smooth manifold, it is more or less a restatement of (one of the possible) definitions of manifold with boundary.

Note though, that there may be values $b\in (-\infty, a)$ sucht that $f(b)$ is not regular, so in general you cannot derive results about the shape of $f^{-1}((-\infty, a])$ This is the focus of a theory called Morse theory (where you assume that possible critical points of $f$ are isolated and nondegenerate).