Let $M$ be a closed smooth manifold, let $N$ be a conncected smooth manifold with boundary (perhaps $\partial N = \emptyset$), and let $f:M\to N$ be a smooth map.
It is known that the number mod $2$ of points in a non-empty regular level set of $f$ does not depend on the choice of the regular value. (This number in $\mathbb{Z}/2$ is called the mapping degree mod $2$ of $f$.) Is this also true if the regular level set is empty? I mean: Can it happen that $f$ has two regular values $y,y'\in N$ such that $f^{-1}(y)$ is empty but $f^{-1}(y')$ contains an odd number of points?
I do not find a counterexample but I do not know either how to prove that this cannot occur.
Any help is much appreciated.
Whatever proof you read that $|f^{-1}(y)| \pmod 2$ is independent of the choice of regular value $y$ does not use the assumption that this set is nonempty.
One standard argument: you found a path $\gamma$ running between two regular values $y, y'$ so that $f$ is transverse to $\gamma$, and took the preimage $f^{-1}(\gamma)$. (Pick an arbitrary path and wiggle it to make it transverse.)
Then $f^{-1}(\gamma)$ is a compact 1-manifold with boundary $f^{-1}(y) \sqcup f^{-1}(y')$, and a compact 1-manifold has an even number of boundary points.
This argument does not use the assumption that $f^{-1}(y)$ is nonempty.