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Suppose I have a Morse function $f$ on a compact smooth manifold $M$, potentially with boundary, and that $h$ is an automorphism of $M$ isotopic to the identity automorphism. Then is $f\circ h$ a Morse function?

It seems clear to me that critical points of $f$ will under $h^{-1}$ be mapped to critical points of $f\circ h$ and that these points will still be locally quadratic (that is, non-degenerate.) But that these should be the only critical points is slightly mysterious to me.

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    $\begingroup$ The title says "diffeomorphism" but your question says "automorphism". I assume by "automorphism" you mean "self-diffeomorphism". $\endgroup$ Feb 29, 2020 at 21:53
  • $\begingroup$ Precisely. This is just an artifact of poor editing. $\endgroup$ Feb 29, 2020 at 22:18

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If $h$ is a diffemorphism, then $dh$ is everywhere nonsingular. Because $$ d(f \circ h) = df \circ dh $$ (depending on notation, etc. --- we're talking about the chain rule here), we have $d(f\circ h)(P)(v)$ is zero exactly when $df(Q)(w)$ is zero (where $Q = h(P)$ for some nonzero $w$, because $dh(P)(v)$ is never zero for any nonzero $v$, because $h$ is a diffeomorphism.

The one weird thing about this question is that "isotopic to the identity" is completely irrelevant; is there a part "b" or "c" that might use that assumption?

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  • $\begingroup$ Yes, this is a small part to a problem I'm working on that for one reason or another just wasn't clicking with me. Thank you for the straightforward answer. $\endgroup$ Feb 29, 2020 at 22:16
  • $\begingroup$ So why, 15 months later, did you decide to withdraw the check-mark for my answer? I don't really care about the points, but...it seems like an odd choice. $\endgroup$ Jun 23, 2021 at 17:11

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