$q$ is a regular value of $f^\circ=f\mid_{M^\circ}$ or $\partial M$, but $S=f^{-1}(q)$ is not a manifold;
I just learned the following Sard's theorem:
Sard's theorem: For any smooth map $f$ of a manifold $X$ with boundary into a boundaryless manifold $Y$, almost every point of $Y$ is a regular value of both $f: X \to Y $and $ \partial f: \partial X \to Y$.
I want to find some examples showing that "almost every" condition cannot be strengthened. To be more specific,
Suppose $M$ is a manifold with boundary, $\dim M\geq 2$ and $f: M\to N$ is smooth. $q\in N$, construct two examples:
- $q$ is a regular value of $f^\circ=f\mid_{M^\circ}$, but $S=f^{-1}(q)$ is not a manifold;
- $q$ is a regular value of $\partial f=f\mid_{\partial M}$, but $S=f^{-1}(q)$ is not a manifold;
Appreciate any help!
This is the version of Sard's theorem specifically when dealing with manifolds with boundary (as Guillemin and Pollack themselves state). The usual version appears earlier on p. 39.
When you are looking for examples, you need to be talking now about manifolds with boundary, not manifolds. The point of the two regular value conditions is that then you will get a submanifold of $X$ with boundary whose boundary is the intersection of the preimage with $\partial X$. Here's an example to contemplate: Consider $f\colon\Bbb H^2\to\Bbb R$, $f(x,y) = x^2+(y-1)^2-1$. Look at $f^{-1}(0)$. Is $0$ a regular value of both maps?
For your second question, it's far easier. Take $f(x,y) = x(y-1)$, still mapping $\Bbb H^2\to\Bbb R$. Now $0$ is a regular value of $\partial f$, but is $f^{-1} (0)$ a submanifold with boundary of $\Bbb H^2$?