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Let $M$ be a manifold (not necessarily compact) , for the sake of clearness embedded in $\mathbb{R^n}$ and $f\colon M\rightarrow \mathbb{R}$ a smooth function.

The theorem of Sard gives us that $$f+\langle\ \cdot\ ,a\rangle \colon M\rightarrow \mathbb{R}, \ x\mapsto f(x)+a_1x_1+...+a_nx_n$$ is a Morse function for almost all $a\in\mathbb{R}^n$.

Now suppose I have a finite set of regular values $c_1,...,c_n$ of $f$, so $f^{-1}(c_1),...,f^{-1}(c_n)$ do not contain critical points. Can I deform $f$ slightly to $\tilde f$, such that it becomes a Morse function, but the level sets of $c_1,..,c_n$ remain unchanged, i.e. $f^{-1}(c_i)=\tilde f^{-1}(c_i)$?

This is somehow a relative version of the density of Morse functions in the space of smooth functions.


1 Answer 1


Take any smooth nonnegative function $h\colon M\to \mathbb R$ such that all points of $h^{-1}(0)$ are regular for $f$ and $h^{-1}(0)$ includes an open neighborhood of all your level sets $f^{-1}(c_i)$.

Assume $\iota\colon M\to \mathbb R^n$ is an embedding. Note that $M\to \mathbb R^{n+1}$ defined as $$\hat\iota=x\mapsto(h(x),h(x)\cdot \iota(x))$$ is an embedding of $M\backslash h^{-1}(0)$.

Fix $a\in\mathbb R^{n+1}$ and pass to $$\tilde f=f+\langle \hat\iota,a\rangle.$$ For a generic choice of vector $a$, you get a Morse functions and $\tilde f=f$ in a neighborhood of your level sets.

(That was the first thing came to my mind, I am sure there are better solutions.)

  • $\begingroup$ I expect you mean $\tilde{f}=f+<h\cdot id_M,a>$? $\endgroup$
    – Francis
    May 22, 2015 at 9:46
  • $\begingroup$ @Francis: Now it is corrected. $\endgroup$ May 22, 2015 at 16:23

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