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Let $M$ be a compact smooth manifold and $f\in\mathscr{C}^\infty(M)$ be a Morse function with distinct critical values. Such functions are exactly stable smooth functions on $M$, that is, those smooth functions $f\in\mathscr{C}^\infty(M)$ such that there exists a neighborhood $U_f\in\mathscr{C}_W^\infty(M)$ (endowed with the Whitney topology) such that for every $g\in U_f$ there are diffeomorphisms $\phi:M\to M$ and $\psi:\mathbb{R}\to\mathbb{R}$ such that $$\psi\circ f=g\circ\phi.$$

I was wondering whether we can or not suppose the two diffeomorphisms $\phi$ and $\psi$ in the definition of stable function are orientation preserving if $M$ is oriented. A sufficient condition would be to prove that $\phi$ can be chosen close to $id_M$ in $\mathscr{C}_W^\infty(M,M)$ and $\psi$ close to $id_\mathbb{R}$ in $\mathscr{C}_W^\infty(\mathbb{R})$, since in this case they are forced to be orientation preserving. Thanks in advance for any answer.

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This follows from a general theorem of Mather. See the book of M. Golubitsky and V.Guillemin,

Stable Mappings and Their Singularities, Springer Verlag, 1973,

Mather's thorem is Theorem 1.5, Chapter III. The rather involved proof is carried in Chapter V.

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  • $\begingroup$ Thank you very much for your answer and the reference. If I understood well, you mean that the fact that the diffeomorphisms $\phi$ and $\psi$ as above are close to the identities follows from infinitesimal stability, is it? I noticed that the proof of Mather's theorem is quite involved since it deals with many different (but finally equivalent) stability conditions, that's the reason why I feel I am missing the idea on the heuristics behind the equivalences. $\endgroup$ Commented May 29 at 15:08
  • $\begingroup$ The orientation preserving part iis tricky. If you replace a function f by -f there are equivalent but they may not by equivalent via orientation preserving maps $\endgroup$ Commented May 30 at 7:53
  • $\begingroup$ Absolutely, I agree with you but they are not close in the Whitney topology, indeed my question refers to a sufficiently close neighborhood of $f$. Do you know whether the result I am looking for is known to be true or are you awere of a counterexample? $\endgroup$ Commented May 31 at 8:26
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    $\begingroup$ @EnricoSavi Prop. 2.2, Chap. II of the book Golubitsky and Gullemin is the result you want. It implies that if $f$ is a stable Morse function and any nearby function is conjugarte to it by diffeomorphisms in the connected components of the identity. Stability means that the orbit of the action by diffeomorphisms of $M$ and $\mathbb{R}$ is open in $C^\infty(M)$. The openess is preserved is preseerved if only the identity components act on $C^\infty(M)$. $\endgroup$ Commented Jun 2 at 11:37

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