# Composition of Morse function with diffeomorphism isotopic to identity is a morse function

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.

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

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.