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.
Best Answer
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?