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Given functions $f, g,$ and $h$ on some smooth manifold M such that $f \circ g$ and $g \circ h$ are $C^\infty$, must $f\circ g\circ h$ be $C^\infty$?

I can't imagine this is true, but I'm having trouble coming up with a counterexample. I've also had no luck in coming up with a proof, but I haven't been focusing on it for too long.

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  • $\begingroup$ Does "smooth" mean $C^\infty$? $\endgroup$
    – Angina Seng
    Commented Dec 18, 2019 at 3:44
  • $\begingroup$ @LordSharktheUnknown Yes, $C^\infty$ and smooth are synonymous here. $\endgroup$
    – Logan Clark
    Commented Dec 18, 2019 at 3:45
  • $\begingroup$ Aren't composites of $C^\infty$ functions always $C^\infty$? $\endgroup$
    – Angina Seng
    Commented Dec 18, 2019 at 3:46
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    $\begingroup$ How about taking $f=g=h$ to be a non-smooth involution? $\endgroup$
    – Angina Seng
    Commented Dec 18, 2019 at 3:50
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    $\begingroup$ @CharlesHudgins Yes, take $f:\mathbb{R}\to\mathbb{R}$ such that on the irrationals $f(x)=x$ and on the rationals $f(x)=-x$. Then $f$ isn't even continuous, but $f(f(x))=x$ $\endgroup$
    – Logan Clark
    Commented Dec 18, 2019 at 3:56

1 Answer 1

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As @Lord Shark the Unknown suggested, pick $$f=g=h= \left\{ \begin{array}{lc} x & \mbox{ if } x \in \mathbb Q \\ -x & \mbox{ if } x \notin \mathbb Q \\ \end{array} \right. $$

More generally, pick any non-smooth bijection $f$ and set $g=f^{-1}$ and $h=f$.

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